SHEET 20 TRIGONOMETRIY (COLLECTION # 1)
Single Correct Type
cos A cos C + cos(A + B) cos(B + C)
Que. 1. If A + B + C = 180o then cos A sin C − sin(A + B)cos(B + C) simplifies to
(a) − cot C
(b) 0
(c) tan C
(d) cot C
(code-V1T2PAQ2)
Que. 2 Let 3a = 4, 4b = 5,5c = 6, 6d = 7, 7e = 8 and 8f = 9. The value of product (abcdef), is
(a) 1
(b) 2
(c)
6
(d) 3
(code-V1T2PAQ3)
(d) cos ec15o
(code-V1T2PAQ4)
Que. 3. Which of the following numbers is the largest ?
(a) cos15ο
(c) sec15o
(b) tan 60 o
sin α − sin γ
Que. 4. If α + γ = 2β then the expression
simplifies to
cos γ − cos α
(a) tan β
(b) − tan β
Que. 5. If A = 110o then
(a) tan A
(code-V1T2PAQ6)
(c) cot β
(d) − cot β
(c) cot A
(d) – cot A
(code-V1T4PAQ2)
(d) 776
(code-V1T4PAQ4)
1 + 1 + tan 2 2A
equals
tan 2A
(b) – tan B
Que. 6. Minimum value of y = 256sin 2 x + 324 cos ec2 x ∀ x ∈ R is
(a) 432
(b) 504
Que. 7. If A = 320o then
(a) tan
A
2
−1 + 1 + tan 2 A
is equal to
tan A
(b) − tan


(c) 580
x 

A
2
(code-V1T5PAQ1)
(c) cot
x 

x


A
2
(d) − cot
A
2
x 

Que. 8. The product  cos  . cos  .  cos  .......  cos
 is equal to
2
4
8
256
sin x
(a)
128sin
x
256
(b)
sin x
256sin
x
256

sin x
(c)
128sin
x
128
(d)
(code-V1T5PAQ3)
sin x
512sin
x
512
Que. 9. In a triangle ABC∠A = 60o , ∠B = 40o and ∠C = 80o. If P is the centre of the circumcircle of triangle ABC with radius unity, then the radius of the circumcircle of triangle BPC is (code-V1T5PAQ5)
(a) 1
(b) 3
(c) 2
(d) 3 2
Que. 10. In a triangle ABC if angle C is 90o and area of triangle is 30, then the minimum possible value of
the hypotenuse c is equal to
(code-V1T5PAQ9)
(a) 30 2
(b) 60 2
(c) 120 2
Que. 11. Which one of the following relations does not hold good ?
(a) sin
π
13π
1
sin
=−
10
10
4
(c) 4 ( sin 24o + cos 6o ) = 15 + 3
(b) sin
(d) 2 30
(code-V1T5PAQ13)
π
13π
1
+ sin
=−
10
10
2
(d) sin 2 72o − sin 2 60o =
5 −1
4
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Question. & Solution. Trigo. Page: A - 1 of A - 34
Que.12. With usual notations, in a triangle ABC, a cos ( B − C ) + b cos ( C − A ) + c cos ( A − B ) is equal to
(a)
abc
R2
abc
4R 2
(b)
(c)
4abc
R2
(d)
abc
2R 2
(code-V1T5PAQ15)
Que. 13. General solution of the equation, 2sin 2 x + sin 2 2x = 2, is
(a) ( 2n + 1)
π
4
(b) nπ ±
π
4
(c) nπ ±
π
2
(code-V1T5PAQ16)
π
4
(d) nπ ± ∪ nπ ±
π
2
Que. 14. In a triangle ABC, ∠A = 60o and b : c = 3 + 1: 2 then ( ∠B − ∠C ) has the value equal to
(a) 15o
(b) 30o
(c) 22.5o
(d) 45O
(code-V1T5PAQ17)
Que. 15. If the two roots of the equation, x 3 − px 2 + qx − r = 0 are equal in magnitude but opposite sign then
(a) pr = q
(b) qr = p
2
Que. 16. The equation, sin θ −
(a) no root
(c) pq = r
(d) pq + r = 0
4
4
= 1− 3
has
sin θ − 1
sin θ − 1
(code-V1T5PAQ20)
2
(b) one root
(c) two roots
(code-V1T5PAQ18)
(d) infinite roots
a 2 + b2 + c2
lies in the interval
Que. 17. If a,b,c are the sides of a triangle then the expression
ab + bc + ca
(a) (1,2)
(b) [1,2]
(c) [1,2)
(d) (1,2]
(code-V1T5PAQ23)
Que. 18. If α and β are the roots of the equation a cos 2θ + bsin 2θ = c then cos 2 α + cos 2 β is equal to
(a)
a 2 + ac + b 2
a 2 + b2
(b)
a 2 − ac + b 2
a 2 + b2
(c)
2b 2
a 2 + c2
(d)
2a 2
b2 + c2
b
Que. 19. If tan θ = , then the value of a cos 2θ + b sin 2θ is equal to
a
(a) a
(b) b
Que. 20. Let f k ( x ) =
(a)
(b)
2
Que. 21. In a triangle ABC, cos
r
2R
(code-V1T7PAQ5)
(d) a 2 b
sin k x + cos k x
then f 4 (x) − f 6 (x) is equal to
k
1 1
− cos 2 2x
4 6
(a) 2 +
(c) ab 2
1 1 2
+ sin 2x
12 4
(c)
1 1
− cos 2 x
3 4
(code-V1T7PAQ6)
(d)
1
12
A
B
C
+ cos 2 + cos 2 equal
2
2
2
(b) 4 −
7r
2R
(c) 2 +
(code-V1T5PAQ24)
r
4R
(code-V1T7PAQ7)
(d)
3  4r 
 + 1
4 R

where r and R have their usual meaning.
Que. 22. A cricle is inscribed in a triangle ABC touches the side AB at D such that AD = 5 and BD = 5 and BD
= 3. If ∠A = 60o then the length of BC equals [Advise - Que. of properties of triangle but tru by 2D.]
(a) 9
(b)
120
13
(c) 12
(d) 13
(code-V1T7PAQ8)
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Question. & Solution. Trigo. Page: A - 2 of A - 34
Que. 23. If in a trianlge ABC, sin A, sin B, sin C are in A.P., then
(code-V1T7PAQ10)
(a) The altitudes are A.P.
(b) The altitudes are in H.P
(c) The medians are in G.P.
(d) The medians are in A.P.
Que. 24. Which of the following inequalitie(s) hold(s) true in any triangle ABC ?
(code-V1T7PAQ13)
(a) sin
A
B
C 1
sin sin ≤
2
2
2 8
2
(c) sin
(b) cos
A 2B 2C 3
sin
sin
<
2
2
2 4
Que. 25. If α =
A
B
C 3 3
cos cos ≤
2
2
2
8
2
(d) cos
A
B
C 9
+ cos 2 + cos 2 ≤
2
2
2 4
π
which of the following hold(s) good ?
7
(code-V1T7PAQ14)
(a) tan α.tan 2α.tan 3α = tan 3α − tan 2α − tan α
(b) cos ec α = cos ec 2α + cos ec 4α
(c) cos α − cos 2α + cos3α has the value equal to 1/2
(d) 8cos α.cos 2α.cos 4α has the value equal to 1.
Que. 26. Identify which of the following are correct ?
(a) ( tan x )
n ( sin x )
> ( cot x )
n ( sin x )
(code-V1T7PAQ16)
 π
, ∀ x ∈  0, 
 4
π
(b) 4n cosec x < 5n cosec x, ∀ x ∈  0, 
2

1
(c)  
2
n ( cos x )
1
< 
3
n ( cos x )

 π
, ∀ x ∈  0, 
 2
(d) 2n ( tan x ) < 2n (sin x ) , ∀ x ∈  0, π 
2


Que. 27. Which of the following is always equal to cos 2 A − sin 2 A ?
(code-V1T10PAQ4)
(a) sin 2A
(b) cos ( A + B ) cos ( A − B) − sin ( A + B ) sin ( A − B)
(c) sin ( A + B ) cos ( A − B) − cos ( A + B ) sin ( A − B)
(d) cos ( A + B ) sin ( A − B) − sin ( A + B ) cos ( A − B)
Que. 28. Number of values of x ∈ [ 0, π] satisfying cos 2 5x − cos 2 x + sin 4x.sin 6x = 0, is(code-V1T10PAQ5)
(a) 2
(b) 3
(c) 5
(d) infinitely many
sin α + sin β + sin ( α + β )
Que. 29. sin α + sin β − sin α + β (wherever difined) simplifies to
(
)
(a) cot
α
β
cot
2
2
(b) cot
α
β
tan
2
2
(c) tan
α
β
cot
2
2
(code-V1T10PAQ6)
(d) tan
α
β
tan
2
2
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Question. & Solution. Trigo. Page: A - 3 of A - 34
Que. 30. Let A, B, C be three angles such that A + B + C = π. If tan A. tan B = csc
cos A cos B
π
then the vlaue of
cos C
6
is equal to
1
2
(a)
(code-V1T12PAQ5)
1
3
(b)
(c) 1
(d)
1
2
Que. 31. Which value of θ listed below leads to 2sin θ > 1 and 3cos θ < 1 ?
(a) 70o
(b) 140o
(c) 210o
(code-V1T12PAQ7)
(d) 280o
Que.32. In a trinalge ABC, a = 3, b = 4 and c = 5. The value of sin A + sin 2B + sin 3C equals
(a)
24
25
14
25
(b)
(c)
64
25
(d) None.
(code-V1T13PAQ1)
Que.33. Let y = ( sin x + cos ecx ) + ( cos x + sec x ) , then the minimum value of y,, ∀ x ∈ R is
2
(a) 7
2
(b) 8
(c) 9
(code-V1T13PAQ4)
(d) 10
Que.34. If the equation cot 4 x − 2 cosec 2 x + a 2 = 0 has atleast one solution then, sum of all possible integral
values of ‘a’ is equal to
(code-V1T13PAQ5)
(a) 4
(b) 3
(c) 2
(d) 0
Que.35. If θ be an acute angle satistying the equation 8cos 2θ + 8sec 2θ = 65, then value of cos θ is equal to
(a)
1
8
(b)
2
3
(c)
2 3
3
(d)
3
4
Que.36. If 2sin x + 7 cos px = 9 has atleast one solution then p must be
(code-V1T13PAQ8)
(a) an odd integer
(b) an even integer
(c) a rational number
(d) an irrational number
Que.37. If θ ∈ ( π / 4, π / 2 ) and
∞
n =1
(a ) 3
1
∑ tan
n
θ
= sin θ + cos θ then the value of tan θ is
(b) 2 + 1
(c) 2 + 3
(b) a 2 + b 2 − 1
(c)
a 2 − b2 − 1
(code-V1T13PAQ11)
(d) a 2 − b2 + 1
Que.39. The least value of x for 0 < x < π / 2, such that cos(2x) = 3 sin(2x), is
(a)
π
12
(b)
2π
12
(c)
3π
12
(code-V1T13PAQ10)
(d) 2
Que.38. If sin x + a cos x = b then the value of a sin x − cos x is equal to
(a) a 2 + b 2 + 1
(code-V1T13PAQ7)
(d)
(code-V1T13PAQ14)
4π
12
Que.40. Let θ ∈ [ 0, 4π] satisfying the equation ( sin θ + 2 )( sin θ + 3)( sin θ + 4 ) = 6. if the sum of all value of θ
is of the form kπ then the value of ‘k’, is
(code-V1T13PAQ18)
(a) 6
(b) 5
(c) 4
(d) 2
Que.41. Let f (x) = a sin x + c, where a and c are real numbers and a > 0. Then f (x) < 0 ∀ x ∈ R if
(a) c < −a
(b) c > −a
(c) − a < c < a
(d) c < a
(code-V1T13PAQ19)
Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
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Question. & Solution. Trigo. Page: A - 4 of A - 34
Que.42. In which one of the following intervals the inequality, sin x < cos x < tan x < cot x can hold good ?

π
 3π
 5π 3π 

(b)  , π 
 4 
(a)  0, 
 4
 7π

(d)  , 2π 
 4

(c)  , 
 4 2 
(code-V1T13PAQ21)
Que.43. If sin x + sin y + sin z = 0 = cos x + cos y + cos z then the expression, cos ( θ − x ) + cos ( θ − y )
+ cos ( θ − z ) , for θ =∈ R is
(code-V1T15PAQ1)
(a) independent of θ but dependent on x, y, z
(b) dependent on θ but independent of x, y, z
(c) dependent on x, y, z and θ
(d) independent of x, y, z and θ
Que.44. In ∆ABC, AB = 1, BC = 1 and AC = 1 2. In ∆MNP, MN = 1, and ∠MNP − 2∠ABC. The side MP
equals
(b) 7 4
(a) 3 2
9
Que.45. The sum
∑ sin
n =1
2
(c) 2 2
(d) 7 2
nπ
equals
18
(a) 5
(code-V1T15PAQ6)
(b) 4
Que.46. Let area of a triangle ABC is
(code-V1T15PAQ5)
(c)
(
)
5 +1
3 −1
, b = 2 and c =
2
(
(d) 3 5
)
3 − 1 and ∠A is acute. The measure of the
angle C is
(code-V1T15PAQ10)
(a) 15o
(b) 30o
(c) 60o
(d) 75o
Que.47. If the inequality sin 2 x + a cos x + a 2 > 1 + cos x holds for any x ∈ R then the largest negative intergral
value of ‘a’ is
(code-V1T18PAQ2)
(a) – 4
(b) – 3
(c) – 2
a 1
(d) – 1

n
Que. 48. The value of the expression sin   .  + ∑ cos ( ka )  , is
 2   2 k =1


1
1 
sin   n −  a 
2 
2 
(a) 1 sin   n + 1  a 
(b)
(c) sin ( ( n + 1) a )
(b) cos n ( a )
2

2 
(code-V1T18PAQ3)
Que. 49. Suppose ABC is a triangle with 3 acute angle A,B and C. The point whose coordinates are
( cos B − sin A,sin B − cos A ) can be in the
(code-V1T18PAQ4)
(a) first and 2nd quadrant
(b) second the 3rd quadrant
(c) third and 4th quadrant
(d) second quadrant only
Que. 50. Number of solution of the equation, sin 4 x − cos 2 x sin x + 2sin 2 x + sin x = 0 is 0 ≤ x ≤ 3π, is
(a) 3
(b) 4
(c) 5
(d) 6
(code-V1T19PAQ3)
Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
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Question. & Solution. Trigo. Page: A - 5 of A - 34
Que. 51. Let α, β and γ be the angles of a triangle with 0 < α <
π
π
and 0 < β < , satisfying
2
2
sin 2 α + sin 2 β = sin γ, then
(a) γ ∈ ( 0,30ο )
(code-V1T20PAQ3)
(b) γ ∈ ( 30ο , 60o )
(c) γ ∈ ( 60ο ,90o )
(d) γ = 90o
Que. 52. Let L and M be the respective intersections of the internal and external angle bisectors of the
triangle ABC at C and the side AB produced. If CL = CM, then the value of ( a 2 + b 2 ) is (where a & b
have their usual meanings)
(a) 2R2
(code-V2T1PAQ4)
(b) 2 2 R 2
(c) 4R 2
(d) 4 2 R 2
Que. 53. In a triangle ABC, if A − B = 120o and R = 8r where R and r have their usual meaning then cos C
equals
(a) 3/4
(b) 2/3
(c) 5/6
(d) 7/8
(code-V2T3PAQ9)
Que. 54. The system of equations
(code-V2T5PAQ13)
x − y cos θ + z cos 2θ = 0
− x cos θ + y − z cos θ = 0
.
x cos 2θ − y cos θ + z = 0
has non trivial solution for θ equals
(b) nπ +
(a) nπ only, n ∈ I.
(c) ( 2n − 1)

π
only, n ∈ I.
2
1
π
only, n ∈ I.
4
(d) all value of θ
1
1
1
a
−1
+ tan −1 + tan −1 + tan −1  is expressed as a rational
Que. 55. If tan  tan
in lowest form ( a + b ) has
b
2
3
4
5

the value equal to
(code-V2T8PAQ1)
(a) 19
(b) 27
(c) 38
(d) 45
Que. 56. A sector OABO of central angle θ is constructed in a circle with centre O and of radius 6. The
radius of the circle that is circumscribed about the triangle OAB, is
(code-V2T8PAQ7)
(a) 6 cos
θ
2
(b) 6 sec
θ
2


θ


(c) 3  cos + 2 
2
(d) 3sec
θ
2
Que. 57. Let s, r, R respectively specify the semiperimeter, inradius and circumradius of a triangle ABC.
Then ( ab + bc + ca ) in terms of s, r and R given by
(a) sr + rR + Rs
(b) R 2 + r 2 + rs
(c) s 2 + R 2 + 2rs
(code-V2T8PAQ8)
(d) r 2 + s 2 + 4Rr
Que. 58. The value of the expression (1 + tan A )(1 + tan B ) when A = 20o and B = 25o reduces to
(a) prime number
(b) composite number
(c) irrational number
(d) rational which is not an integer.
(code-V2T13PAQ5)
Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
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Question. & Solution. Trigo. Page: A - 6 of A - 34
Que. 59. The least value of the expression
(a) 1
(b) 2
cot 2x − tan 2x
 π
in  0,  equals
8
 5π

1 + sin  − 8x  
 2

(c) 4
(code-V2T13PAQ11)
(d) none.
Que. 60. The value of x satisfying the equation sin ( tan −1 x ) = cos ( cos −1 (x + 1) ) is
(a)
1
2
(b) −
1
2
(c)
(code-V2T13PAQ22)
(d) No finite value
2 −1
Que. 61. Which one of the following quantities is negative ?
(code-V2T13PAQ23)
(a) cos ( tan −1 (tan 4) ) (b) sin ( cot −1 (cot 4) ) (c) tan ( cos −1 (cos 5) ) (d) cot ( sin −1 (sin 4) )
Que. 62. The product of all real values of x satisfying the equation

 2x 2 + 10 x + 4 
 2 − 81 x
sin −1 cos  2
= cot  cot −1 

 x +5 x +3 
 9x





(a) 9
(b) – 9
(code-V2T13PAQ29)
 π
  + is
 2
(c) – 3
(d) – 1


2π

−1 x − 1
Que. 63. Product of all the solution of the equation tan −1  2x
, is
=
 + cot 
2
 x −1 
 2x  3
(a) 1
(b) –1
(c) 3
(a) 1
a − b + c − d where a, b, c, d ∈ N and
ad
is equal to
bc
(b) 2
(code-V2T14PAQ4)
(d) − 3
Que. 64. If the value of tan ( 37.5o ) can be expressed as
( a > b > c > d ) then the value of
2
(code-V2T14PAQ13)
(c) 3
(d) 4
Que. 65. If the minimum value of expression y = ( 27 )
cos x
+ ( 81)
sin x
can be expressed in the form
where a, b ∈ N and are in their lowest term then the value of ( a + b ) equals
a/b
(code-V2T17PAQ10)
Que. 66. Let f (x) = sin x + cos x + tan x + arc sin x + arc cos x + arc tan x. If M and m are maximum and minimum values of f(x) then their arithmetic mean is equal to
(code-V2T18PAQ3)
(a)
π
+ cos1
2
(b)
π
+ sin1
2
(c)
π
+ tan1 + cos1
4
(d)
π
+ tan1 + sin1
4
Que. 67. The sum ∑ cos  rπ  equals
2009
r =1
(a) 0
3π 

Que. 68.  tan 
8 

(code-V2T19PAQ1)
 6
(b) 1
2009
3π 

+  − cot 
8 

(c)
3 +1
2 2
(d)
2+ 3
2
2009
is
(a) even integer
(c) rational which is not an integer
(code-V2T19PAQ3)
(b) odd integer
(d) irraional
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Question. & Solution. Trigo. Page: A - 7 of A - 34
Comprehesion Type
# 1 Paragraph for Q. 1 to Q. 3
Let r and R denote the radii of the incircle and the circumcircle of the triangle ABC with sides a, b, c
and a + b + c = 2s. Also ∆ denotes the area of the trianlge.
(code-V1T16PAQ7,8,9)
1.
The value of the product
A
∏ cot 2
equals
cyclic
ABC
∆
(b) 2
s
r2
(a)
∆
2.
The vlaue of the sum
R (a + b + c)
(c)
A
B
∏ cot 2 cot 2
2
abc
(d)
r
s
(d)
4R
r
equals
cyclic
A,B,C
(a)
3.
R + 4r
r
(b)
4R + r
r
4R + r
R
(c)
Let f (x) = 0 denotes a cubic whose roots are cot
A
B
C
, cot , cot . If the triangle ABC is such that one
2
2
2
of its anlge is 90o then which one of the following holds good ?
(a) r + 2R = s
(b) 3r + 2R = s + 2
(c) 1 + r + 4R = 2s
(d) 4r + R = s
# 2 Paragraph for Q. 4 to Q. 6
Let ABC be an acute triangle with orthocenter H.D,E,F are the feet of the perpendiculars from A,B,and
C on the opposite sides. Also R is the circumradius of the triangle ABC.
(code-V1T18PAQ6,7,8)
Given ( AH )( BH )( CH ) = 3 and ( AH ) + ( BH ) + ( CH ) = 7
2
4.
The ratio
(a)
5.
2
3
14R
2
has the value equal to
(b)
3
7R
(c)
7
3R
(d)
14
3R
The product ( HD )( HE )( HF ) has the value equal to
(a)
6.
∏ cos A
∑ cos A
2
9
64R 3
(b)
9
8R 3
(c)
8
9R 3
(d)
9
32R 3
(c)
3
2
(d)
5
2
The value of R is
(a) 1
(b) 31/ 3
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Question. & Solution. Trigo. Page: A - 8 of A - 34
Assertion & Reason Type
In this section each que. contains STATEMENT-1 (Assertion) & STATEMENT-2(Reason).Each
question has 4 choices (A), (B), (C) and (D), out of which only one is correct.
Bubble (A) STATEMENT-1 is true, STATEMENT-2 is True; STATEMENT-2 is a correct
explanation for STATEMENT-1.
Bubble (B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct
explanation for STATEMENT-1.
Bubble (C) STATEMENT-1 is True, STATEMENT-2 is False.
Bubble (D) STATEMENT-1 is False, STATEMENT-2 is True.
Que. 1. Statement - 1 :
(code-V1T4PAQ8)
In a triangle ABC if A is obtuse then tan B tan C > 1
because
Statement - 2 :
In any triangle ABC, tan A =
tan B + tan C
tan B tan C − 1
Que. 2. Statement - 1 :
(code-V1T4PAQ9)
cos (10 ) and cos ( −10 ) both are negative and have the same value.
c
c
because
Statement - 2 :
cos θ = cos ( −θ ) and the real numbers (10 ) and ( −10 ) both lie in the third quadrant.
c
c
Que. 3. Let α, β and γ satisfy 0 < α < β < γ < 2π and cos ( x + α ) + cos ( x +β) + cos ( x + γ ) = 0 ∀x ∈ R
γ −α =
Statement - 1 :
2π
.
3
(code-V1T6PAQ1)
because
cos α + cos β + cos γ = 0 and sin α + sin β + sin γ = 0.
Statement - 2 :
Que. 4. If A + B + C = π then
Statement - 1 :
(code-V1T6PAQ3)
cos 2 A + cos 2 B + cos 2 C has its minimum value
3
.
4
because
Statement - 2 :
1
8
Maximum value of cos A cos Bcos C = .
Que. 5. Let f (x) = 3sin 2 x + 4sin x cos x + 4 cos 2 x, x ∈ R.
Statement - 1 :
Greatest and least values of f (x) ∀ x ∈ R are
(code-V1T8PAQ11)
7 + 17
7 − 17
and
respectively..
2
2
because
Statemtnt 2 :
a + b − a 2 + b 2 + c2 − 2ab
a + b + a 2 + b 2 + c 2 − 2ab
≤ a sin 2 x + b sin x cos x + c cos 2 x ≤
2
2
where a, b, c ∈ R
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Question. & Solution. Trigo. Page: A - 9 of A - 34
Que. 6. Statement - 1 :
tan
6π
5π
π
6π
5π
π
− tan
− tan = tan .tan . tan
7
7
7
7
7
7
(code-V1T10PAQ7)
because
Statement - 2 :
If θ = α + β, then tan θ− tan α − tan β = tan θ.tan α.tan β.
Que. 7. Consider the following statements
Statement - 1 :
(code-V1T12PAQ8)
In any right angled triangle ABC, sin 2 A + sin 2 B + sin 2 C = 2
because
Statement - 2 :
Que. 8. Statement-1:
= n cot
In any triangle ABC, sin 2 A + sin 2 B + sin 2 C = 2 − 2 cos A cos B cos C ]
In
any
triangle
A
B
C

n  cot
+ co t + c o t 
2
2
2 

ABC
A
B
C
+ n cot + n cot
2
2
2
because
Statement - 2 :
Que. 9. Statement - 1 :
(code-V1T14PAQ5)
(
(
n 1 + 3 + 2 + 3
)) = n1 + n
General solution of
(
3 + n 2 + 3
)
tan 4x − tan 2x
nπ π
= 1 is x =
+ ,n ∈I
1 + tan 4x tan 2x
2 8
because
Statement - 2 :
Que. 10. Statement - 1 :
(code-V1T18PAQ11)
π
4
General solution of tan α = 1 is α = nπ + , n ∈ I.
The equation sin ( cos x ) = cos ( sin x ) has no real solution (code-V1T19PAQ10)
because
Statement - 2 :
sin x ± cos x is bounded in  − 2, 2 
Que. 11. Statement 1:
In any triangle ABC, cot A + cot B + cot C > 0
(code-V2T7PAQ10)
because
Statement 2: Minimum value of cot A + cot B + cot C in any triangle ABC is 1.
Que. 12. Let P be the point lying inside the acute triangle ABC such that angles sutended by each side at P
is 120o. Equilateral triangles AFB, BDC,CEA are constructed outwardly on sides AB, BC, CA of
∆ABC
Statement 1: Lines AD, BE, CF are concurrent at P.
(code-V2T18PAQ8)
because
Statement 2: P is the radical centre of circumcircles of triangles ABF, BDE, CEA
Que. 13. Let ABC be an acute triangle whose orthocentre is at H. Altitude from A is produced to meet the
circumcircle of the triangle ABC at D.
(code-V2T19PAQ7)
Statement 1: The distange HD = 4R cos Bcos C where R is the circumradius of the triangle ABC.
because
Statement 2: Image of orthocentre H in any side of an acute triangle lies on its circumcircle.
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Question. & Solution. Trigo. Page: A - 10 of A - 34
More than One May Correct Type
Que. 1. If 2cos θ + 2 2 = 3sec θ where θ ∈ ( 0, 2π ) then which of the following can be correct ?
1
2
(a) cos θ =
(b) tan θ = 1
(c) sin θ = −
1
2
(d) cot θ = −1
(code-V1T2PAQ9)
Que. 2 Which of the following real numbers when simplified are neither terminating nor repeating decimal ?
(a) sin 75o.cos 75o
(b) log 2 28
(d) 8−(log
(c) log 3 5.log 5 6
)
27 3
(code-V1T2PAQ10)
Que. 3. Suppose ABCD (in order) is a quadrilateral inscribed in a circle. Which of the following is/are
always True ?
(code-V1T2PAQ11)
(a) sec B = sec D
(b) cot A + cosC = 0
(c) cosecA = cosecC
(d) tan B + tan D = 0
Que. 4. which of the following quantities are rational ?
(code-V1T4PAQ15)
 9π   4 π 
 sec  
 10   5 
11π   5π 
 sin  
  12 
(b) cos ec 
π
 
(d) 1 + cos 1 + cos 1 + cos 
9 
9 
9 

(a) sin 
 12
π
 

4
4
(c) sin   + cos  
8
8
2π 
4π 
8π 
Que. 5. In a triangle ABC, a semicircle is inscribed, whose diameter lies on the side c. If x is the length of
the angle bisector through angle C then the radius of the semicircle is
(code-V1T6PAQ6)
abc
(a) 4R 2 (sin A + sin B)
(c) x sin
C
2
(b)
∆
x
(d)
2 s(s − a)(s − b)(s − c)
s
Que. 6. The possible value(s) of x satisfying the equation sin x − 3sin 2x + sin 3x = cos x − 3cos 2x + cos 3x, is/
are.
(code-V1T6PAQ9)
(a) −
7π
8
(b) −
π
8
(c)
π
8
(d)
π
4
Que. 7. In which of the following sets the inequality sin 6 x + cos 6 x > 5 holds good ?
(code-V1T6PAQ12)
8
 π π


(a)  − , 
8 8
 3π 5 π 
,
(b) 

 8 8 
 π 3π 
(c)  , 
4 4 
 7 π 9π 
(d)  , 
 8 8 
Que. 8. The value of x satisfying the equation cos ( n x ) = 0, is
(a) e π / 2
(b) e−π / 2
(c) eπ
(code-V1T10PAQ9)
(d) e3π / 2
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Question. & Solution. Trigo. Page: A - 11 of A - 34
Que. 9. Which of the following do/does not reduce to unity ?
(code-V1T10PAQ11)
sin (180o + A ) cot ( 90o + A ) cos ( 360o − A ) cos ecA
(a) tan 180o + A . tan 90o + A .
sin ( − A )
(
) (
)
sin ( −A )
(b) tan 180o + A
(
)
(c)
−
tan ( 90o + A )
cot A
+
cos A
sin ( 90o + A )
sin 24o cos 6o − sin 6o sin 66o
sin 21o cos 39o − cos 51o sin 69o
cos ( 90o + A ) sec ( − A ) tan (180o − A )
(d) sec 360o + A sin 180o + A cot 90o − A
(
) (
) (
)
Que. 10. Which of the following identities wherever difined hold(s) good ?
(code-V1T10PAQ12)
(a) cot α − tan α = 2cot 2α
(b) tan ( 45o + α ) − tan ( 45o − α ) = 2 cos ec2α
(c) tan ( 45o + α ) + tan ( 45o − α ) = 2sec2α
(d) tan α + cot α = 2 tan 2α.
Que. 11. Let α, β and γ are some angles in the1st quadrant satisfying tan ( α + β ) =
15
17
and cos ecγ =
then
8
8
which of the following holds good ?
(code-V1T14PAQ7)
(a) α + β + γ = π
(b) cot α.cot β.cot γ = cot α + cot β + cot γ
(c) tan α + tan β + tan γ = tan α.tan β. tan γ
(d) tan α. tan β + tan β. tan γ + tan γ. tan α = 1
Que. 12. Which of the following statements are always correct ? (where Q denotes the set of rationals)
⇒
(a) cos 2θ ∈ Q and sin 2θ ∈ Q
⇒
(b) tan θ ∈ Q
⇒
(code-V1T14PAQ8)
sin 2θ, cos 2θ and tan 2θ ∈ Q (if defined)
⇒
(c) If sin θ ∈ Q and cos θ ∈ Q
(d) if sin θ ∈ Q
tan θ ∈ Q (in difinde)
tan 3θ ∈ Q (if defined)
cos 3θ ∈ Q
Que. 13. Given that sin 3θ = sin 3α, then which of the following angles will be equal to cos θ ?
π
3


(a) cos  + α 
π
3


(b) cos  − α 
 2π

+ α
 3

(c) cos 
(
)
 2π

− α  (code-V1T14PAQ9)
 3

(d) cos 
Que. 14. If the quadratic equation ( cos ec2 θ − 4 ) x 2 + cot θ + 3 x + cos 2
3π
= 0 holds true for all real x then
2
the most general values of θ can be given by
(a) 2nπ +
11π
6
(b) 2nπ +
5π
6
(c) 2nπ ±
(code-V1T15PAQ12)
7π
6
(d) nπ ±
11π
6
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Question. & Solution. Trigo. Page: A - 12 of A - 34
Que. 15. If α and β are two different solution of a cos θ + b sin θ = c, then which of the following hold(s)
good ?
(code-V1T15PAQ13)
(a) sin α + sin β =
(c) cos α cos β =
2bc
2
a + b2
(b) sin α sin β =
2ac
2
a + b2
c2 − a 2
a 2 + b2
(d) cos α cos β =
c2 + a 2
a 2 + b2
Que. 16. If in a triangle ABC, cos 3A + cos 3B + cos 3C = 1, then
(code-V1T15PAQ14)
(a) ∆ABC is a right angled triangle.
(b) ∆ABC is an obtuse angle triangle
(c) ∆ABC is an acute angled triangle
(d) Inradius ‘r’ of the triangle ABC is either 3 ( s − a ) or
Que. 17. The value of x in (0, π / 2) satisfying the equation,
(a)
π
12
(b)
5π
12
(c)
3 ( s − b ) or
3 (s − c ).
3 −1
3 +1
+
= 4 2 is (code-V2T2PAQ11)
sin x
cos x
7π
24
(d)
11π
36
Que. 18. In a triangle ABC, 3sin A + 4cos B = 6 and 3cos A + 4sin B = 1 then ∠C can be (code-V2T8PAQ12)
(a) 30o
(b) 60o
(c) 90o
(d) 150o
Que. 19. If cos 3θ = cos 3α then the value of sin θ can be given by
(a) ± sin α
π

(b) sin  ± α 
3

 2π

(c) sin  + α 
 3

(code-V2T15PAQ9)
 2π

(d) sin  − α 
 3

Que. 20. Let ABC be a triangle with AA1, BB1, CC1 as their medians and G be the centriod. If the points A, C1,
G, B1 be concyclic then which one of the following relations do/does not hold good ? (code-V2T19PAQ10)
(a) 2a 2 = b 2 + c2
(b) 3b 2 + c 2 + a 2
(c) 4c 2 + a 2 + b 2
(d) 3a 2 + b 2 + c 2
Match Matrix Type
Que. 1. Match the general solution of the trigonometric equation given in Column - I with their corre(code-V1T14PBQ1)
sponding entries given Column - II.
Column - I
Column - II
A.
cos 2 2x + cos 2 x = 1
P.
π
π
x = nπ+ ∪ nπ+ ,n ∈ I.
4
6
B.
cos x = 3 (1 − sin x )
Q.
x=
C.
1 + 3 tan 2 x = 1 + 3 tan x
R.
π
x = ( 2n − 1) , n ∈ I.
6
D.
tan 3x − tan 2x − tan x = 0
S.
π
π
x = 2nπ+ ∪ 2nπ + ,n ∈ I.
2
6
(
)
nπ
,n ∈I
3
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Question. & Solution. Trigo. Page: A - 13 of A - 34
Que. 1. The expression 2cos
π
9π
7π
9π
.cos + cos
+ cos
simplifies to an integer P. Find the value of P..
17
17
17
17
(code-V1T1PAQ1)
sin ( α + β ) − 2sin α + sin ( α − β )
Que. 2. Show that the expression cos α + β − 2cos α + cos α − β is independent of β. (code-V1T1PAQ2)
(
)
(
)
Que. 3. If the expression
k.
sin θ.sin 2θ + sin 3θ sin 6θ + sin 4θ sin13θ
= tan kθ, where k ∈ N. Find the value of
sin θ cos 2θ + sin 3θ cos 6θ + sin 4θ cos13θ
(code-V1T1PAQ4)
8
Que. 4. If y = cos
x
x
π
π
− sin 8 . Find the value of y when x = and also when x = . (code-V1T3PAQ4)
2
2
4
6
Que. 5. In a trianlge ABC, given sin A : sin B : sin C = 4 : 5 : 6 and cos A : cos B : cos C = x : y : z. If the ordered
pair (x, y) satisfies this, then compute the value of ( x 2 + y 2 + z 2 ) where x, y, z ∈ N and are in their
lowest form. (code-V1T7PBQ2)
Que. 6. Let A = cos 360o.sin 2 270o − 2cos180o.tan 225o , B = 3sin 540o.sec 720o + 2 cos ec450o − cos 3600o
C = 2 sec 2 2π.cos 0c + 3sin 3
3π
5π
3π
3π
− cos ec
and D = tan π.cos + sec 2π − cos ec . Find the value of
2
2
2
2
A + B − C ÷ D.
(code-V1T9PAQ1)
Que. 7. If the value of the expression E = cos 4 x − k 2 cos 2 2x + sin 4 x, is independent of x then find the set of
values of k.
(code-V1T9PAQ3)
Que. 8. If cot
π
= p + q + r + s where p, q, r, s ∈ N, find the value of ( p + q + r + s ) . (code-V1T9PAQ5)
24
Que. 9. Let L denotes the value of the expression,
sin 2θ − sin 6θ + cos 2θ − cos 6θ
, when θ = 27 o
sin 4θ − cos 4θ
tan x tan 2x
when x = 9o.
tan 2x − tan x
and
M denotes the value of
and
N denotes the numerical value of the expression (wherever difined)
(code-V1T9PAQ7)
1 − cos 4α
1 + cos 4α
+
2
sec 2α − 1 cos ec2 2α − 1
when it is simplified.
Find the value of the product (LMN).
4
5
Que. 10. If cos ( α + β ) = ;sin ( α − β ) =
5
π
and α, β lie between 0 and , then find the value of tan 2β.
13
4
(code-V1T11PAQ1)
Que. 11. Find the range of values of k for which the equation 2 cos 4 x − sin 4 x + k = 0, has atleast one solution. (code-V1T11PAQ3)


Que. 12. Prove that, sin 3 θ + sin 3  θ +
2π 
4π 
3
3
 + sin  θ +
 − = − sin 3θ.
3 
3 
4

 tan 2r −1 
Que. 13. Compute the value of the sum ∑ 
r 
r =1  cos 2 
(code-V1T11PAQ5)
n
(code-V1T11PAQ6)
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Question. & Solution. Trigo. Page: A - 14 of A - 34
Que. 14. If 15sin 4 α + 10 cos 4 α = 6 then find the value of 8cosec6 α + 27 sec6 α.
(code-V1T15PDQ1)
Que. 15. Given that x + sin y = 2008 and x + 2008cos y = 2007 where 0 ≤ y ≤ π / 2. Find the value [ x + y] .
(Here [x] denotes greatest interger function)
(code-V2T1PDQ2)
φ 2 (θ n )
Que. 16. The sum ∑ 2 sin x.sin1[1 + sec(x − 1).sec(x + 1) ] can be written in the form as ∑ (−1)
where
ψ (θ n )
n =1
x=2
4
44
n
φ and ψ are trigonometric functions and θ1 , θ2 , θ3 , θ4 are in degrees ∈ [0, 45]. Find ( θ1 + θ2 + θ3 + θ4 ) .
(code-V2T17PDQ1)
Que. 17. If the total between the curves f (x) = cos −1 (sin x) and g(x) = sin −1 (cos x) on the interval [ −7 π, 7 π] is
A, find the value of 49A. (Take π = 22 / 7 )
(code-V2T17PDQ3)
[SOLUTION]
Single Correct Type
Que. 1. (D)
cos A cos C + ( − cos C )( − cos A )
cos A sin C − ( sin C )( − cos A )
Que. 2 (B) 3a ; = 4;a = log 3 4;
|||1y
⇒
2 cos A cos C
= + cos C
2 cos A sin C
b = log 4 etc.
Hence abcdef = log 3 4; log 4 5; log 5 6.log 6 7.log 7 8.log 8 9 = log 3 9 = 2
⇒
2.
Que. 3. (D)
cos15o = 2 + 3 ≅ 3.732;
cos ec15o =
tan 60o = 3 ≅ 1.732; sec15o =
4
= 6 − 2 = 1.035;
6+ 2
4
= 6 + 2 = 3.86 which is largest
6− 2
Que. 4. (C)
α−γ
α+γ
2sin 
cos 


sin α − sin γ
 2 
 2  = cot  α + γ  = cot β
=


cos γ − cos α
α−γ α+γ
 2 
2sin 
sin
 

 2   2 
2
Que. 5. (B) 1 + 1 + tan 2A = 1 + sec 2A
tan 2A
tan 2A
( 2A = 220 )
o
=
1 − sec 2A
 1 − cos 2A 
= −
 = − tan A.
tan 2A
 sin 2A 
Que. 6. (C) y = 256 ( sin 2 x + cos ec 2 x ) + 68cos ec 2 x, 256 ( (sin x − cos ecx)2 + 2 ) + 68cos ec2 x
Minimum when x =
Que. 7. (a)
E=
π
π
or − and minimum vlue = 512 + 68 = 580
2
2
−1 + sec A 1 − cos A
A
=
= tan
tan A
sin A
2
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Question. & Solution. Trigo. Page: A - 15 of A - 34
Que. 8. (B).
Que. 9. (A) Let R ' be the radius of the circumcircle of triangle ABC using sine law in trianlge BPC
a
= 2R1
sin120o
a
= 2R
sin 60o
................. (1)
also
(R = 1
given ) a = 3;
form (1)
Que. 10. (D) ∆ = ab ⇒ ab = 60
c = a 2 + b2
a = 2R sin 60o
1
2
R1 =
( in
∆ABC )
2 3 1
. = 1.
3 2
also a 2 + b 2 ≥ 2ab
B
∴
∴
a + b ≥ 2ab
equality occurs when a = b
2
2
A
C
C
B
Θ A
minimum value of a 2 + b 2 = 2 ab = 120 = 2 30
Alternatively: b = c cos θ;
c 2 |min = 120
Que. 11. (D) In (D) it should be
Que 12. (A) put
1
c2 sin 2θ
∆ = c2 sin θ cos θ =
= 3θ
2
4
a = csin θ
⇒
c2 = 120cos ec2θ
c = 2 30
5 −1
.
8
a = 2R sin A etc.
T1 = 2R sin ( B + C ) cos ( B − C ) = R [sin 2B + sin 2C] etc.
E = R [sin 2B + sin 2C + sin 2C + sin 2A + sin 2A + sin 2B] = 2R ( sin 2A + sin 2B + sin 2C )
= 8R.
a b c
abc
.
.
= 2.
2R 2R 2R R
Que. 13. (D) where n ∈ I.
sin 2 2x = 2 cos 2 x 4sin 2 x cos 2 x = 2 cos 2 x cos 2 x 1 − 2sin 2 x  = 0
cos 2 x = 0
or
sin 2 x =
1
2
∴
b
3 +1 b − c
3 +1− 2
=
;
=
=
Que. 14. (B) c
2
b+c
3 +1+ 2
now using tan
B−C b−c
A
=
cot =
2
b+c
2
(
(
x = nπ ±
3 −1
)
3 +1
3 −1
)
3 +1
.
.
π
2
or
x = nπ ±
π
4
1
3
3
B−C
= 2− 3 ⇒
= 15o
2
3
∴ B − C = 30o
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Question. & Solution. Trigo. Page: A - 16 of A - 34
Que. 15. (C)
Que. 16. (D) sin 2 θ = 1 [sin θ ≠ +1] ⇒ sin θ = −1 ⇒ θ = 2nπ − π / 2 ⇒ infinite roots
Que. 17. (C) In a triangle b + c > a
⇒
b + c − a > 0 ∴ a ( b + c − a ) > 0 |||1y
c (a + b − c) > 0
and
a ( b + c ) + b ( c + a ) + c ( a + b ) > a 2 + b 2 + c2
∴
b (c + a − b) > 0
a2 + b2 + c2
<2
ab + bc + ca
∴
2 ( ab + bc + ca ) > a 2 + b 2 + c2
............(1) also forany a,b,c ∈ R a 2 + b 2 + c 2 ≥ ab + bc + ca
........(2) (equality holds if a = b = c)
Que. 18. (A) Make a quadratic in cos 2θ to get cos 2α + cos 2β =
⇒
2ac
2 cos α + cos β = 2
+ 2;
a + b2
(
2
2
∑a < 2
∑ ab
2
a 2 + b 2 + c2
≥1
ab + bc + ca
)
form (1) and (2)
1≤
2ac
a + b2
2
a 2 + ac + b 2
cos α + cos β =
a 2 + b2
2
2
 b 2  2b 2
a 1 − 2  +
a (1 − tan 2 θ ) 2b.tan θ
a ( a 2 − b 2 ) + 2ab 2 a ( a 2 + b 2 )
a  a

=
=
= a.
Que. 19. (A) 1 + tan 2 θ + 1 + tan 2 θ =
b2
a 2 + b2
a 2 + b2
1+ 2
a
Que. 20. (D) f 4 ( x ) − f 6 ( x ) =
=
1
1
1
1
sin 4 x + cos 2 x ) − ( sin 6 x + cos 6 x ) = (1 − 2sin 2 cos 2 x ) − (1 − 3sin 2 cos 2 x )
(
4
6
4
6
1 1 2  1 3 2  1 1 1
1 − sin 2x  − 1 − sin 2x  = − = .
4  2
 6 4
 4 6 12
Que. 21. (A)
r
5
Que. 22. (D) Given A = 60o ; tan 30o = ⇒ r =
5
B r
5
now tan = =
2 3 3 3
3
1 − tan 2 ( B 2 ) 1 − ( 25 27 )
2
1
cos B =
=
=
=
=
2
1 + tan ( B 2 ) 1 + ( 25 27 ) 521 26
26
(a = ?)
675 15 3
B
1
C
Hence
15 3
sin B =
sin C = sin ( A + B )
26
b
A
30o
30o
5
3 1 1 15 3 1 
 = 4 3 = sin C ⇒ c = a ;
. + .
=
16
3

2 26 2 26
52 
13
sin C sin A
a
I
r
= sin A cos B + cos Asin B =
D B/2
B
c
3
a=
8. 3 13
.
= 13.
2 4 3
Que. 23. (B)
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Question. & Solution. Trigo. Page: A - 17 of A - 34
Que. 24. (A,B,D)
(c)
∑ sin
2
A 1
3 1
= 3 − ( cos A + cos B + cos C )  = − ( cos A + cos B + cos C )
2 2
2 2
but [ cos A + cos B + cos C]max =
3 ∴
2
∑ sin
2
A
3 3 3
= − = ∴

2  min 2 4 4
∑ sin
2
A 3
≥ ⇒ (c) si wrong.
2 4
a,b,c are correct and hold good in an equilateral triangle as their maximumum values.
Que. 25. (A,B,C)
(c)
(d)
(b) RHS =
sin 4α + sin 2α 2sin 3α.cos α
1
=
=
= cos ec α ( using π = 7α ) ⇒ (b).
sin 2α.sin α 4
sin 2α.sin 4α sin α
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Question. & Solution. Trigo. Page: A - 18 of A - 34
1
cos α + cos3α + cos5α sum of a series with constant d = 2α sum = 2 ⇒ (c) is wrong.
continued product = 1 ⇒ (d) is also wrong.
Que. 26. (A,B,C)
(a) x = π 8, ⇒ ( tan x )
n ( sin x )
(b)
x = π 6 , ⇒ 4n 2 < 5n 2 ⇒ True.
(c)
x = π 2, 2n 2 < 3n 2 ⇒ True.
(d)
x = π 4 , 2 0 > 2 − n 2 ⇒ 1 <
1
2 n 2
> 1 and ( cot x )
n ( sin x )
< 1 ⇒ True.
is not correct ⇒ False.
Que. 27. (B) ( cos 2 A − sin 2 B ) − ( sin 2 A − sin 2 B ) = cos 2 A − sin 2 A = cos 2A
(C)
sin 2B;
(D) − sin 2B
Que. 28. (D) (1 − sin 2 5x ) − (1 − sin 2 x ) + sin 4x sin 6x = 0 ⇒ sin 2 x − sin 2 5x + sin 4x.sin 6x = 0
− sin 6x.sin 4x + sin 4x.sin 6x = 0 which is true for all x ∈ [ 0, π] ⇒ it is indentity
 α +β 
 α −β 
 α+β
 α +β 
 α −β 
 α+β
2sin 
 cos 
 + 2 sin 
 cos 
 cos 
 + cos 

 2 
 2 
 2 
 2 =
 2 
 2 
Que. 29. (A)
 α +β 
 α −β 
 α +β 
 α+β
 α −β 
 α+β
2sin 
 cos 
 − 2sin 
 cos 
 cos 
 − cos 

 2 
 2 
 2 
 2 
 2 
 2 
α
β
cos
2
2 = cot α cot β .
=
α
β
2
2
2 sin sin
2
2
2 cos
Que. 30. C. Given tan A.tan B = 2
Let y =
cos A cos B
cos A.cos B
cos A.cos B
1
1
=−
=
=
=
=1
cos C
cos(A + B) sin A sin B − cos A cos B tan A tan B − 1 2 − 1
Que. 31. B. 2sin θ > 1 ⇒ sin θ > 0 ⇒ θ ∈1st or 2nd quadrant , 3sin θ < 1 ⇒ cos θ < 0 ⇒ θ ∈ 2nd or 3rd quadrant hence
θ ∈ 2nd ⇒ possible answer is (B).
A
Que. 32. B.
E = sin A + sin 2B + sin 3C ⇒ E =
3
4 3
15 24
39 − 25 14
+ 2. . − 1 =
+
−1 =
= .
5
5 5
25 25
25
25
5
B
3
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4
C
Que. 33. C.
y = ( sin 2 x + cos 2 x ) + 2 ( sin x cos ecx + cos x sec x ) + sec 2 x + cos ec 2 x
= 5 + 2 + tan 2 x + cot 2 x = 7 + ( tan x − cot x ) + 2 ∴ y min = 9.
2
Que.34. D.
cot 4 x − 2 (1 + cot 2 x ) + a 2 = 0 ⇒ cot 4 x − 2 cot 2 x + a 2 − 2 = 0 ⇒ ( cot 2 x − 1) = 3 − a 2
2
to have
atleast one solution 3 − a 2 ≥ 0 ⇒ a 2 − 3 ≤ 0 ⇒ a ∈  − 3, 3  integral values −1, 0,1 ∴ sum = 0.
Que.35. D.
cos 2θ = t ∴ 8t +
Let
⇒ t = 8 or t =
Que.36. C.
1
1
1
9
3
2
2
(t = 8 is rejected, think !) ∴ cos 2θ = ; 2 cos θ − 1 = ⇒ cos θ = ⇒ cos θ = .
8
8
8
16
4
2sin x + 7 cos px = 9 is possible only if sin x = 1 cos px = 1
x = (4n + 1)
Que.37. A.
8
= 65 ⇒ 8t 2 − 65t + 8 = 0 ⇒ 8t − 64t − t + 8 ⇒ 8t ( t − 8 ) − ( t − 8 ) = 0
t
π
2mπ
π 2mπ
4m
and px = 2mπ ⇒ x =
⇒p=
( m, n,∈ I ) ∴ (4n + 1) =
∴ p ∈ rational.
2
p
2
p
4n + 1
tan θ > 1 ⇒ 0 <
1
<1
tan θ
now,
1
1
1
+
+
+ ........∞ = sin θ + cos θ
2
tan θ tan θ tan 2 θ
1
1
cos θ
⇒ tan θ = sin θ + cos θ ⇒
= sin θ + cos θ ⇒
= sin θ + cos θ
1
tan
θ
−
1
sin
θ
−
cos
θ
1−
tan θ
⇒ cos θ = sin 2 θ − cos 2 θ = 1 − 2 cos 2 θ
cos θ =
Que.38. D.
1
π
or cos θ = −1 (rejected) ⇒ =
⇒ tan θ = 3.
2
3
Let a sin x − cos x = k, k ≥ 0 ............... (1) also sin x + a cos x = b ............... (2) Square and
add (1) and (2)
Que.39. A.
a 2 + 1 = k 2 + b2 ⇒ k 2 = a 2 − b2 + 1 ⇒ k = a 2 − b2 + 1.
cos 2x
3
π
π
π

−
sin 2x = 0 or cos 2x.cos 2x.cos − sin 2x.son ⇒ cos  2x +  = 0
2
2
3
3
3

⇒ 2x +
Que.40. B.
π π
π
π
= ; 2x = ⇒ x = .
3 2
6
12
( sin θ + 2 )( sin θ + 3)( sin θ + 4 ) = 6 LHS
sin θ = −1. ⇒ sin θ = −1 ⇒ θ =
Que.41. A.
⇒ 2 cos 2 θ + cos θ − 1 = 0 ⇒ ( 2 cos θ − 1)( cos θ + 1) = 0
≥ 6 and RHS = 6 ⇒
equality only can hold if
3π 7 π
,
∴ sum = 5π ⇒ 5.
2 2
c
c
c
a sin x + c < 0 ⇒ sin x < − ; − > sin x; − > 1; − c > a ⇒ a + c < 0 ⇒ (A)
a
a
a
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Question. & Solution. Trigo. Page: A - 19 of A - 34
Que.42. A.
In 2nd quadrant sin x < cos x is False (think !)
In 4th quadrant cos x < tan x is False (think !)
 5π 3π 
in 3rd quadrant, i.e.  ,  if tan x < cot x ⇒ tan 2 x < 1 which is not correct hence A can be correct
 4 2 
now
 π
sin x < cos x is true in  0, 4  and tan x < cot x is also true


∴
only the value of x for which cos x < tan x is be determined
∴
now
⇒ sin x =
2
2
cos x = tan x i.e. cos x = sin x or 1 − sin x = sin x ⇒ sin 2 x + sin x − 1 > 0
 5 −1 
−1 ± 5
5 −1
; sin x =
⇒ x = sin −1 

2
2
 2 

5 −1 π 
∴ cos x < tan x in  sin −1
,  and cos x > tan x in
2
4

Que.43. D.

5 −1 
−1
 0,sin

4 

cos θ ( ∑ cos x ) + sin θ ( ∑ sin x ) = 0
1
2 = 3 ∴ cos 32θ = 2 cos 2 θ − 1 = 2. 9 − 1 = 2 = 1
2
4
16
16 8
1+1−
Que.44. D.
cos θ =
7
 1  7.2 7
agin x 2 = 1 + 1 − 2 cos 2θ = 2 (1 − cos 2θ ) = 2 1 −  =
= ⇒x=
4
2
 8 8
A
1
B
2
Que.45. A. sum = sin
0
1
1
2
Que.47. B.
∴
x
1
N
C
20
1
P
π
2π
8π
+ sin 2
+ ........ + sin 2
+ 1 now
18
18
18
Que.46. A. Using ∆ = bc sin A ∴
⇒ tan
M
1/ 2
1
.2
2
(
)
3 − 1 sin A =
sin 2
π
8π
+ sin 2
= 1 etc. ⇒ sum = 5.
18
18
3 −1
1
∴ sin A = ⇒ A = 30o
2
2
B−C b−c
A 3 − 3 3 +1
=
cot =
.
= 3 ⇒ B − C = 120o also B + C = 150o ⇒ C = 15o.
2
b+c
2
3 + 1 3 −1
2
2
sin 2 x + a cos x + a 2 > 1 + cos x put x = 0 ⇒ a + a > 2 ⇒ a + a = 2 > 0 ⇒ ( a + 2 )( a − 1) > 0
−2
1
∴
largest negative integral value of ‘a’ = – 3.
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Question. & Solution. Trigo. Page: A - 20 of A - 34
Que. 48. A.

a 1 n
a 1

sin   .  + ∑ cos(ka)  ⇒ sin   .  + cos a + cos 2a + cos 3a + ...... + cos na 
 2   2 k =1
2 2


 
1
a 1 
3a
a 
5a
3a 
1
1  

sin +  sin − sin  +  sin − sin  + ........ +  sin  n +  a − sin  n −  a  
2
2 2 
2
2 
2
2 
2
2  

 
1
a 1 
1
a 1
1

sin + sin  n +  a − sin  = sin  n +  a.
2
2 2 
2
2 2 
2
Que. 49. D. Since ABC are acute angle
∴ A +B > π/2 ⇒ A >
B>
Again,
π
− B ⇒ sin A − cos B > 0 ⇒ cos B − sin A < 0
2
................. (1)
π
− A ⇒ sin B > cos A ⇒ sin B − cos A > 0
2
................. (2)
Form (1) and (2)
x-coordinate is – ve and y-coordinate is +ve
line in 2nd quadrant only..
⇒
Que. 50. B.
⇒ sin x sin 3 x − cos 2 x + 2sin x + 1 = 0
sin 4 x − cos 2 sin x + 2sin 2 x + sin x = 0
⇒ sin x sin 3 x − 1 + sin 2 x + 2 sin x + 1 = 0
⇒ sin x sin 3 x + sin 2 x + 2 sin x  = 0
⇒ sin 2 x = 0 or sin 2 x + sin x + 2 = 0
⇒ not possible for real x. sin x = 0
⇒ x = 0, π, 2π, 3π,
⇒
4 solution.
Que. 51. D.
(
)
(
)
(
)
(
)
Que. 52. C.
a 2 + b 2 = 4R 2 sin 2 45o − θ + sin 2 135o − θ  = 4R 2 sin 2 45o − θ + cos 2 45o − θ  = 4R 2 .
Que. 53. D.
A
B
C

R = 8r = 8  4R sin sin sin 
2
2
2

∴ 2 sin
A−B
A+B
C 1

⇒  cos
− cos
 sin =
2
2 
2 16

C 1
C 1
sin
− sin + = 0
2 2
2 16
2
cos C = 1 − 2sin 2
⇒
⇒
⇒
− cos θ
cos 2θ
hence
sin
C 1
=
2 4
C
1 7
= 1− = .
2
8 8
− cos θ
1
−2sin 2 θ − cos θ
or
C 1
C 1
.  − sin  = ;
2 2
2  16
2
Que. 54. D. For non trivial solution
0
sin
C
1
 − sin  = 0
2
4
1
2 sin 2 θ
A
B
C 1
sin sin =
2
2
2 16
− cos θ
1
− cos θ = 0 using C1 → C1 → C3
cos 2θ − cos θ
1
cos 2θ
1
− cos θ = 0 ⇒ 2sin θ 0
2
1
− cos θ
cos 2θ
− cos θ = 0 ⇒ sin 2 θ = 0
−1 − cos θ
1
1
1 1 − cos 2 θ  − 1  cos 2 θ − cos 2θ  ⇒ sin 2 θ −  cos 2 θ − (cos 2 θ − sin 2 θ  ⇒ sin 2 θ − sin 2 θ = 0
D = 0 ∀θ∈R ⇒
D.
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Question. & Solution. Trigo. Page: A - 21 of A - 34
9
1 1
+ 
1+

9
−1
4 5 ; α = tan −1   = 19 = 28 = 14 = a ⇒ 14 + 5 = 19.
π


 
tan  + α  when α = tan 
 19  1 − 9 10 5 b
 1− 1 
4

 20 
19
Que. 55. A.
R=
Que. 56. D.
abc
1
θ c
θ
; ∆ = .6.6sin θ = 18sin θ ⇒ a = b = 6 ⇒ sin =
⇒ c = 12 sin .
4∆
2
2 12
2
Alternatively :
OD 2 = 62 + x 2 ⇒
x
= tan ( θ / 2 ) ⇒ OD 2 = 62.sec 2 ( θ / 2 ) ⇒ 4r 2 = 36 sec 2 ( θ / 2 ) ⇒ r = 3sec ( θ / 2 ) .
6
∆ 2 = s ( s − a )( s − b )( s − c ) ⇒ r 2s = ( s − a )( s − b )( s − c )
Que. 57. D.
(∴
r = ∆ / s)
= s3 − s 2 ( a + b + c ) + ( ab + bc + ca ) s − abc ∴ r 2s = s3 − 2s3 + ( ab + bc + ca ) s − abc
using
abc
=∆ ⇒
4R
abc = 4Rrs ⇒
r 2 = ( ab + bc + ca ) − s 2 − 4Rr ∴
∑ ab = r
2
+ s 2 + 4Rr.
E = (1 + tan A)(1 + tan B) = 1 + tan A + tan B + tan A tan B
Que. 58. A.
tan(A + B) =
Now
tan A + tan B
= 1 ⇒ A = 20o and B = 25o ⇒ 1 − tan A tan B = tan A + tan B
1 − tan A tan B
⇒ tan A + tan B + tan A tan B = 1
∴
E = 2. ⇒ A.
Que. 59. B. Expression reduces to 2 corosec 8x
Que. 60. D.
x
1+ x
2
x +1
=
( x + 1)
2
⇒ x 2  (x + 1) 2 + 1 = ( x + 1)  (x 2 + 1)  ⇒ x 2 (x + 1) 2 + x 2 = x 2 (x + 1) 2 + (x + 1) 2
2
+1
x 2 = (x + 1)2 ⇒ x + 1 = x not possible as x → ∞ or x + 1 = − x ⇒ x = −1/ 2 not possible (think!).
Que. 61. D.
cos ( tra −1 ( tan(4 − π) ) ) = cos ( 4 − π ) = cos ( π − 4 ) = − cos 4 > 0
(A)
(
)
(B)
sin cot −1 ( cot ( 4 − π ) ) = sin ( 4 − π ) = − sin 4 > 0
(C)
tan cos −1 ( cos ( 2π − 5 ) ) = tan ( 2π − 5 ) = − tan 5 > 0
(D)
cot sin −1 ( sin ( π − 4 ) ) = cot ( π − 4 ) = − cot 4 < 0
(
(
)
)
(as sin 4 < 0)
( as tan 5 < 0 )
⇒
(D) si correct.
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Question. & Solution. Trigo. Page: A - 22 of A - 34
Que. 62. A.


 2 ( x 2 + 5 | x | +3 ) − 2 
 2
 π
π
π
π
2
2
 = cot cot −1 
− cos −1 cos 
− 2 + ⇒ − 2 + 2
=
−2+
2
2
x + 5 | x | +3
2
x + 5 | x | +3 9 | x |
2


9|x+
 2


0<↓<2


2
⇒
⇒
x − 4 | x | +3 = 0
⇒
x = 1,3
(
x = ±1, ±3.
)
Solution are 3, − 1 , 2 − 3, − 2 + 3 ⇒ Product = 1.
Que. 63. A.
3
Que. 64. A.
o
 75  1 − cos 75
tan 37.5o = tan   =
=
sin 75o
 2 
2
(
o
) (
)=2
6 − 3 + 3 − 2 2 − 3 +1
2
∴
a = 6; b = 4, c = 3, d = 2
y = (3)
Que. 65. C.
3cos x
+ (3)
4sin x
3 −1
2 2 = 2 2 − 3 +1 = 2 2 − 3 +1
2
3 +1
3 +1
2 2
(
1−
2
5/2
3
=
2
3.3. 3
2
=
243
=
)=2
3 −1
6 −4−2 2
6 −4+2 3 −2 2
= 6− 4+ 3− 2
2
ad 12
=
= 1.
bc 12
⇒
1/ 2
33cos x + 34sin x
≥ ( 33cos x.34sin x )
now using AM ≥ GM
2
⇒ 33cos x + 34sin x ≥ 2 33cos x + 4sin x ≥ 2 3−5
=
)(
but
−5 ≤ 3cos x + 4sin x ≤ 5
∴ 33cos x + 34sin x ≥ 2 3−5
4
⇒ a + b = 247.
243
Domain of f is [ −1, 1] ; f ( x ) = sin x + cos x + tan x + sin −1 x + cos −1 x + tan −1 x
Que. 66. A
2
f ' ( x ) = cos x − sin x + sec
x +0+
>1
1
1
+ x2
[1/ 2, 1]
Hence f ' ( x ) > 0
∴
Que. 67. A.
f is increasing ⇒ range is  f ( −1) , f (1) 
f ( x ) |min = f ( −1) = − sin1 + cos1 − tan1 −
M+m π
= + cos1
2
2
⇒
⇒
⇒
( A )]
S = cosθ + cos 2θ + ....... + cos nθ
nθ
2 cos ( n + 1) θ
S=
θ
2
sin
2
sin
θ
2
Hence cos ( n + 1) = 0
π
π π
+ π − = + cos1 − sin1 − tan1
2
4 4
where θ = π / 6 and n = 2009
θ  2010  π
π
⇒ now ( n + 1) 2 =  2  6 = ( 335 ) 2


⇒
S=0
Ans.
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Question. & Solution. Trigo. Page: A - 23 of A - 34
Que. 68. A.
=
(
)
2 +1
⇒ 2  2009 C1

( 2)
2009
2008
−
(
)
2 −1
+ 2009 C3
2009
( 2)
2006
+ 2009 C5
2004
+ ........ + 2009 C2009
( 2 ) 
0
(A)
⇒
= which is an even integer
( 2)
Comprehesion Type
# 1 Paragraph for Q. 1 to Q. 3
1.
C.
2.
(i)
tan
B.
3.
A.
A
∆
r 
∆
A s−a
=
=
 r =  ∴ cot =
2 s (s − a ) s − a 
s
2
r
4s2 ( a + b + c )
A
A s − a + s − b + s − c s s2
 abc 
=
=
.R  ∆
cot
=
cot
=
=
=
in any triangle, ∑
∏
.
4∆
abc
2
2
r
r ∆
 4R 
2
(ii)
A
B
∑ cot 2 cot 2 =
cos
A
B
C
B
C
A
C
A
B
cos sin + cos cos sin + cos cos sin
2
2
2
2
2
2
2
2
2
A
B
C
A
sin sin sin = ∏ sin
2
2
2
2
........(1)
A B C
C
C
A B
 A B
sin  + +  = 1 ⇒ sin  +  cos + cos  +  sin
2
2
2 2
 2 2
 2 2 2
Now consider
A
B
C
B
C
C
A
A
B
C 
A
B
C

 sin cos cos + sin cos cos cos + cos cos cos  −  sin sin sin  = 1
2
2
2
2
2
2
2
2
2
2 
2
2
2

∴ cos
A
B
C
B
C
A
C
A
B
A
cos sin + cos cos sin + cos cos sin = 1 + ∏ sin
2
2
2
2
2
2
2
2
2
2
A
B
∴ ∑ cot cot =
2
2
(iii)
We have
are not cot
A
2
A
∏ sin 2
1 + ∏ sin
A
A

 using r = 4R ∏ sin  =
2

A
s
∑ cot 2 = ∏ cot 2 = r
and
A
B
r
4R = 4R + r .
r
r
4R
1+
∑ cot 2 cot 2 =
4R + r
hence an equation whose roots
r
A 2 
A
B
A

A
B
C
3
, cot and cot is x −  ∑ cot  x +  ∑ cot cot  x − ∏ cot = 0
2
2
2
2
2
2
2


s
s
 4R + r 
π
f (x) = x 3 − x 2 + 
 x − = 0 as A or B C ∈   .
r
r
 r 
2
A
B
C


s 4R + r s
− =0
∴ x = 1 must a root  as cot or cot or cot = 1  ∴ f (1) = 0 ⇒ 1 − +
2
2
2
R
r
r


⇒
r − 2s + 4R + r = 0
⇒
2R + r = s.
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Question. & Solution. Trigo. Page: A - 24 of A - 34
4.
A.
5.
# 2 Paragraph for Q. 4 to Q. 6
6.
C.
B
( AH )( BH )( CH ) = 3
i.e.
3
( 2R cos A )( 2R cos B )( 2R cos C ) = 3 ⇒ ∏ cos A = 8R 3 ........(1)
( HD )( HE )( HF ) = ( 2R cos B cos C )( 2R cos C cos A )( 2R cos A cos B ) = 8R 3 ( cos2 A cos 2 B cos 2 C ) ....(2)
From (1) and (2)
∏ cos A = 3
∑ cos A 8R
2
.
3
9
∏ ( HD ) = 8R . 64R
3
4R 2
3
=
7
14R
= 1 − 2cos A cos Bcos C
∴ 4R 3 − 7R − 3 = 0 ⇒
6
=
9
2
2
2
3 also ( AH ) + ( BH ) + ( CH ) = 7
8R
∴ (1) ÷ ( 3)
Now we know that, in a triangle ABC cos 2 A + cos 2 B + cos 2 C
⇒
7
3
7
3
= 1 − 2. 3 ⇒
= 1−
2
2
4R
8R
4R
4R 3
( R + 1)( 2R + 1)( 2R − 3) = 0
3
∴R= .
2
Assertion & Reason Type
Que. 1. (D) A = π − (B + C) tan A = − tan(B + C) =
tan B + tan C
⇒ S − 2 is True
tan B tan C − 1
hence if A is acute then tan B tan C > 1 if A is obtuse then thanB tan C < 1
⇒ S − 1 is False ⇒ answer is (D)
Que. 2. (C)
Que. 3. (C)
f (x) = ax 2 + bx + c given f (0) + f (1) = 2 ⇒ f (x) > 0∀x ∈ R ⇒ S − 1 is true.
Let f (x) = x 2 − x + 1 ⇒
Que. 4. (A)
a+b =0
⇒
S − 2 is False
cos 2 A + cos 2 B + cos 2 C = 1 − 2 cos A cos B cos C
∴
∑ cos
2
1
1 3
A |min = 1 − 2. = 1 − = .
8
4 4
Que. 5. B.
Que.6. (a) tan θ − tan α − tan β =
tan α + tan β
tan α + tan β
− ( tan α + tan β ) =
( tan α.tan β ) ⇒ tan θ. tan α. tan β
1 − tan α tan β
1 − tan α tan β
Que. 7. C.
Que. 8. B.
Que. 9. D.
Givne tan 2x = 1 ∴ 2x = nπ +
π
(note that tan 4x is not defined)
4
∴ Statement - 1 is false and Statement - 2 is true.
Hence given equation has no solution
Que. 10. A.
Que. 11. C. If it is acute triangle then statement -1 is abviously true
Let A be obtuse say A = 150o
∴
B + C = 30o both angles < 30o and if C = 30o
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Question. & Solution. Trigo. Page: A - 25 of A - 34
Now
cot A and cot ( B + C ) will be of equal magnitke but opposite sign, As cot θ is decreasing
hence, cos B + cos A alone is +ve
Que. 12. B
∴ cos A + cot B + cot C > 0 .
(∴ chord BD subtends equal angle in same segment)
∠BPD = ∠BCD = 60°
∴
∠APB + ∠DPB = 180°
A, P, D are collinear
|||ly
B, P, E and C, P, F are also collinear
hence AD, BE, CF are concurrent at P.
Que. 13. A.
∴
∆BHN and BDN are congruent
HN = ND = 2R cos Bcos C
⇒
HD = 4R cos Bcos C
More than One May Correct Type
Que. 1. (A,B,C,D)
2cos θ + 2 2 = 3sec θ
∴
2cos 2 θ + 2 2 cos θ − 3 = 0
cos θ =
−2 2 ± 32 −2 2 ± 4 2
=
4
4
∴
cos θ =
∴
θ=
⇒
sin θ = −
1
2
π
or
4
or
−
cos θ = −
3
2
( rejected )
π
4
1
; cot θ = −1;
2
tan θ = 1
and
cos θ = −
1
2
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Question. & Solution. Trigo. Page: A - 26 of A - 34
Que. 2 (B,C)
1
1
(A) = sin150o =
2
4
⇒
rational
⇒
irrational
(B)
2 + log 2 7
(C)
log 3 6 = 1 + log 3 2
(D)
8−1/ 3 = 2−1 =
1
⇒
2
⇒
irrational
irrational
Que. 3. (B,C,D)
Opposite angles of a cyclic quadrilateral are supplementary
 11π   5π 
π
 π  1 π 1
sin 
 sin   = sin   cos   = sin   = ∈ Q
 12   12 
 12 
 12  2  6  4
Que. 4. (A,B,C,D) (A)
(B).
1
 9π   4π 
 π  π
cos ec   sec   = − cos ec   sec   =
=−
o
o
 19   5 
 10   5  sin18 cos36
(C).
1
1 3
π
 π
π
 π
π
sin 4   + cos 4   = 1 − 2sin 2   cos 2   = 1 − sin 2   = 1 − = ∈ Q
2
4 4
8
8
8
8
4
(D).
π
2π
4π
1
2cos 2 .2 cos 2 .2cos 2
= 8 ( cos 20o.cos 40o.cos80o ) = ∈ Q
9
9
9
8
1
2
1
2
1
2
Que. 5. (A,C) ra + rb = ab sin C ⇒ r(a + b) = 2∆ ⇒ r =
2∆
a+b
(
16
)(
5 −1
)
5 +1
= −4 ∈ Q
......(1)
C
A
B
R
A
∴ r=
R
I
B
2abc
abc
2ab
C
=
⇒ ( A ) also x =
cos
2
4R ( 2R sin A + 2R sin B ) 4R ( sin A + sin B )
a+b
2
form (1)
1
C
C
C
2. ab sin C 2ab sin cos
2ab cos
2
2 =
2 .sin C = x sin C
r= 2
=
a+b
a+b
a+b
2
2
Que. 6. (A,C) sin x − 3sin 2x + sin 3x = cos x − 3cos 2x + cos 3x or 2sin 2x cos x − 3sin 2x = 2 cos x − 3cos 2x
sin 2x ( 2 cos x ) − 3 = cos 2x [ 2 cos x − 3] ⇒ ( sin 2x − cos 2x )( 2 cos x − 3) = 0 but 2 cos x − 3 ≠ 0
as cos x ≤ 1 hence, sin 2x − cos 2x = 0 ⇒ tan 2x = 1 ⇒ 2x = nπ + π / 4 or x =
nπ π
+ ⇒ a, b, c
2 8
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Question. & Solution. Trigo. Page: A - 27 of A - 34
Que. 7. (A,B,D)
( sin x+ cos x )
2
= 1 − 3sin 2 x cos 2 x >
⇒
2
3
- 3sin 2 xcos 2 x ( sin 2 + cos 2 x ) >
5
8
5
5
⇒ 1 − > 3sin 2 x cos 2 x
8
8
3
π
π
 π π

> 3sin 2 x cos 2 x ⇒ 1 − 2sin 2 2x > 0 ⇒ cos 4x > 0 ⇒ 4x ∈  − ,  ⇒ 4x ∈  2πr − , 2πr + 
8
2
2
 2 2

 nπ π nπ π 
x ∈
− , + n∈I
 2 8 2 8
now verify .
Que. 8. (A,B,D)
Que. 9. (B,C,D)
sin (18o )
sin 24o cos 60o − cos 24o sin 6o
(A) 1 (B) 3 (C) sin 21o cos 39o − sin 39o cos 21o = sin −18o = −1
( )
Que. 10. (A,C)
(A)
Que. 11. B,D.
tan ( α + β ) =
Que. 12. A,B,C.
(A). tan α =
cos 2 α − sin α
= 2 cot 2α
sin α cos α
15
8
π
∴ α + β + γ = ⇒ (B) and (D)
and tan γ =
8
15
2
1 − cos 2θ
⇒ (A) is correct.
sin 2θ
(B).
2 tan θ
1 − tan 2 θ
2 tan θ
sin 2θ =
; cos 2θ =
; tan 2θ =
⇒ (B) is correct.
2
2
1 + tan θ
1 + tan θ
1 − tan 2 θ
(C).
tan 3θ =
(D).
sin θ =
Que. 13. A,B,C,D.
(D) –1
sin θ
⇒
cos 3θ
(C) is correct.
1
which is rational but cos 3θ = cos θ ( 4 cos 2 θ − 3) which is irrational ⇒ (D) is correct.
3
3θ = nπ + ( −1) ( 3α )
n
∴ 3θ = 3α;
3θ = π − 3α; 3θ = −π − 3α or 3θ = 2π + 3α;
3θ = −2π + 3α
Hence θ = α; θ =
π
2π
π

 2π

− α; θ = −  + α  ; θ = 
+ α ; θ = −
+ 3α
3
3
3

 3

π

 2π

⇒ cos θ = cos  ± α  or cos θ = cos 
± α  ⇒ (A), (B), (C) and (D) all are correct.
3

 3

Que. 14. A,B. Given quadratic equaton is an identity ∴ cosec2 θ = 4 and cot θ = − 3 ⇒ cosec θ = 2
or − 2 and tan θ = −
Que. 15. A,B,C.
(a
2
1
⇒
3
θ=
5π
11π
or
6
6
Making quadrtic in sine from a cos θ + b sin θ + c, we get
+ b 2 ) sin 2 θ − 2bc sin + c 2 − a 2 = 0
α
β
................ (1)
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Question. & Solution. Trigo. Page: A - 28 of A - 34
⇒ sin α + sin β =
2bc
c2 − a 2
⇒
(A)
is
correct
⇒
sin
α
+
sin
β
=
⇒ (B) is correct
a 2 + b2
a 2 + b2
Making quadratic equation in cos, we get (changing a and b)
(a
2
α
β
+ b 2 ) cos 2 θ − 2ac cos θ + c 2 − b 2 =
⇒ cos α + cos β =
2bc
c2 − b2
⇒
(C)
is
correct
⇒
cos
α
+
cos
β
=
⇒ (D) is correct
a 2 + b2
a 2 + b2
∑ cos 3A = 1 ⇒ sin
Que. 16. B,D.
alos r = ( s − a ) tan
A
B
C
or ( s − b ) tan or ( s − c ) tan
2
2
2
r = 3 ( s − a ) or
Que. 17. A,D.
3 ( s − b ) or
3 (s − c)
⇒
(D)
3 −1
3 +1
π
π
π

+
= 2 ⇒ sin cos x + cos sin x = sin 2x ⇒ sin 2x = sin  x + 
12
12
2 2 sin x 2 2 cos x
 12 
∴ 2x = x +
Que. 18. A.
3A
3B
3C
2π
2π
2π
.sin
.sin
=0 ∴ A=
or B =
or C =
⇒ (B)
2
2
2
3
3
3
π
π
or 2x = π − x −
12
12
⇒ x=
π
11π
or 3x =
12
12
∴ x=
π
11π
or
⇒ A, C.
12
36
Square and adding 9 + 16 + 24sin ( A + B ) = 37 ⇒ 24sin ( A + B ) = 12 ⇒ sin ( A + B ) = 1
2
1
⇒ sin C = ; C = 30o Or 150o if C = 150 o then even of B = 0 and
2
1
1
3sin A + 4cos B ⇒ 3. + 4 = 5 < 6 hence C = 150o is not possible
2
2
Que. 19. A,B,C,D.
A = 30 o the quantity
⇒ ∠C = 30o only
cos 3θ = cos 3α put n = 0,1 ⇒ 3θ = 2nπ ± 3α
∴ 3θ = 3α or − 3α or 2π + 3α or 2π − 3α
θ = α or − α or
2π
2π
+ α or
−α ⇒
3
3
if n = −1 ⇒ 3θ = −2π ± 3α ⇒ θ = −
(A), (C), (D) are correct.
2π
±α
3
π

 2π

 2π



π

π

sin θ = sin  −
± α  = − sin 
± α  = − sin  π − ± α  = − sin  π −  ± α   = − sin  ± α 
3
 3

 3



3

3


hence (B) is not correct.
Que. 20. B, C, D
i.e.
use power of point B w.r.t. the circle passing through AC1GB1
BC1 × BA = BG × BB1 ⇒
c2 2
c 2 2  2c2 + 2a 2 − b 2 
2
c
2
× ( mb ) ⇒ = 
× c = BB1 × BB1 ⇒

3 3
3 3
4
2
3

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Question. & Solution. Trigo. Page: A - 29 of A - 34
⇒
⇒
2a 2 = b 2 + c 2 Ans.
B, C, D are the answers.
Match Matrix Type
Que. 1. A - R.
A.
B - S.
C- P.
cos 2 2x − sin 2 x = 0 ⇒ cos 3x.cos x = 0 ⇒ cos 3x = 0 or cos x = 0 ⇒ 3x = (2n − 1)
x = ( 2n − 1)
π
⇒ (R)
6
B.
cos x + 3 sin x = 3 ⇒
∴
x−
3 tan 2 x −
D.
⇒
π
2
π
π
π
π
or x = ( 2n − 1) hence general solution is (2n − 1) as ( 2n − 1) is contained in
6
2
6
2
(2n − 1)
C.
D - Q.
(
cos x
3
3
π
π
π
π

+
⇒ cos  x −  = cos ⇒ x − = 2nπ ±
sin x =
2
2
2
3
6
3
6

π
π
π
π
π
= 2nπ + or 2nπ − ⇒ x = 2nπ + or x = 2nπ + ⇒ (S)
3
6
6
2
6
)
3 + 1 tan x + 1 = 0 ⇒ 3 tan x ( tan x − 1) − ( tan x − 1) = 0 ⇒ ( tan x − 1)
∴
tan x − 1
or
tan x =
⇒
1
3
⇒
nπ
nπ
are nπ is contained in
3
3
⇒
nπ
nπ
(rejected) or
2
3
(Q)
Subjective Type ( Up to 4 digit)
Que. 1.
π
= θ; 17θ = π ⇒ 2cos θ.cos9θ + cos 7θ + cos9θ ⇒
17
)
3 tan x − 1 = 0
π
x = nπ + 
4
 ⇒ (P)
π
x = nπ +
6 
tan 3x − tan 2x − tan x = 0 or tan x. tan 2x.tan 3x = 0 ⇒ x = nπ or
general solution
(
cos(100)θ + cos8θ + cos 7θ + cos 9θ = 0
Sum of cosines of supplymentary angles is zero.
Que. 2 tan α
2sin α cos β − 2sin α 2sin α(cos β − 1)
=
= tan α Hence proved.
2cos α cos β − 2cos α 2cos α(cos β − 1)
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Question. & Solution. Trigo. Page: A - 30 of A - 34
Que. 3. Multiple numerator and denominator by 2
cos θ − cos 3θ + cos 3θ − cos 9θ + cos 9θ − cos17θ
sin 3θ − sin θ + sin 9θ − sin 3θ + sin17θ − sin 9θ
∴
=
cos θ − cos17θ 2sin 9θ sin 4θ
=
= tan 9θ = tan kθ ⇒
sin17θ − − sin θ 2sin 4θ cos9θ
k = 9.
3 2 7 3
;
8
16
Que. 4.
8
8
x 
x 
x
x 
x
x
x
x


y =  cos  −  sin  =  cos 4 − sin 4   cos 4 + sin 4  = cos x 1 − 2 sin 2 cos 2 
2 
2 
2
2 
2
2
2
2


 1

= cos x 1 − sin 2 x 
2


(i)
3
3 2
π 1  1 1 1 3
y  =
. =
=
.
1 − .  =
8
2  2 2
2 4 4 2
4
(ii)
3 1 1 7 3
π
y  =
1 − ;  =
 6  2  2 4  16
sin A sin B sin C
52 + 62 − 42 3
=
=
⇒ a = 4k, b = 5k, c = 6k ∴ cos A =
= ;
a
b
c
2.5.6
4
Que. 5. (229)
cos B
cos A cos B cos C
4 2 + 6 2 − 52 9
4 2 + 52 − 6 2 1
=
=
= ;cos C =
= hence
3/ 4
9 /16
1/ 8
2.5.6
16
2.4.5
8
deividing by 16
cos A cos B cos C
=
=
∴ x = 12, y = 9 and z = 2.
12
9
2
Que. 6. A = 3; B = 1;C = −2; D = 2 ⇒ 3 + 1 − (−2) ÷ 2 = 5.
Que.
(
(
)
be independent of x, 1 − 2k 2 = 0 ⇒ k =
2
4
(
Que. 9. L =
) (
6 + 2 + 8+ 4 3
4
)=
1+
)
2
x cos 2 x (1 − 2k 2 ) for this to
1
1
or −
Note :The value of expression for this value of k
2
2
1
.
2
π
1 + cos15o
o
cot
=
cos
7.5
=
=
Que. 8.
24
sin15o
=
(
2
= 1 − 2 − sin 2 x cos2 − k 2  cos2 x + sin 2 x − 4sin 2 x cos2 x  = (1 − k 2
is
)
) − 2 sin
E = cos2 x + sin 2 x − 2sin 2 x cos2 x − k 2 cos2 x − sin 2 x
7.
6+ 2
4+ 6 + 2
4+ 6 + 2
4
=
=
4
6− 2
6− 2
4
(
)(
6+ 2
)
6 + 2 + 2 + 3 = 2 + 3 + 4 + 6 ∴ p + q + r + s = 15.
−2cos 4θ sin 2θ + 2sin 4θ sin 2θ 2sin 2θ ( sin 4θ − cos 4θ )
=
= 2sin 2θ
sin 4θ − cos 4θ
sin 4θ − cos 4θ
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Question. & Solution. Trigo. Page: A - 31 of A - 34
5 +1
2
If θ = 27 o , L = 2sin 54o == 2cos36o ⇒ L =
2 tan x
tan x.
tan x tan 2x
2 tan x
2 tan x
1 − tan 2 x =
M=
=
=
= sin 2x
2
2
2 tan x
tan 2x − tan x
1
+
tan
x
2
−
1
−
tan
x
− tan x
1 − tan 2 x
(
when x = 9o , M = sin18o ⇒ M =
N=
)
5 −1
.
4
1 − cos 4α
1 + cos 4α
2sin 2 2α.cos 2 2α 2cos 2 2α.sin 2 2α
+
=
+
= 2 cos 2 2α + sin 2 2α ⇒ N = 2.
2
2
2
2
sec 2α − 1 cos ec 2α − 1
1 − cos 2α
1 − sin 2α
(
)
(
(
)
)
 5 + 1  5 − 1 
∴ LMN = 

 (2) = 1.

 2  4 
Que. 10. cos ( α + β ) =
4
3
5
5
now 2β = ( α + β ) − ( α − β )
⇒ tan ( α + β ) = ⇒ sin ( α − β ) = ⇒ tan ( α − β ) =
5
4
13
12
3 5 
−
tan ( α + β ) − tan ( α − β )  4 12  16
tan 2β =
=
= .
1 + tan ( α + β ) .tan ( α − β ) 1 + 3 . 5
63
4 12
Que. 11. 2 (1 − sin 2 x ) − sin 4 x + k = 0 put sin 2 x = t, t ∈ [ 0,1] ⇒ 2 (1 − t ) − t 2 + k = 0 ⇒ t 2 − 4t + k + 2 = 0
2
2
since sum of the roots is 4 ⇒ one root in (0,1) and other greater then 1 as shown
y
x
O
now
f (0) ≥ 0 and f (1) ≤ 0 ⇒ k + 2 ≤ 0 and k − 1 ≥ 0 ⇒ k ∈ [ −2,1] .
Alternatively : 2 cos 4 x − sin 4 x + k = 0 ⇒ cos 4 x + ( cos 4 x − sin 4 x ) + k = 0 ⇒ cos 4 x + cos 2x + k = 0
2
 1 + cos 2x 
2
2

 + cos 2x + k = 0 ⇒ 1 + cos 2x + 6 cos 2x + 4k = 0 .....(A) ⇒ cos 2x = t ⇒ t + 6t + 1 + 4k = 0
2


⇒ ( t + 3) = 8 − 4k ⇒ ( t + 3) max. = 16 ⇒ ( t + 3)min. = 4 ∴ 4 ≤ 8 − 4k ≤ 16 ⇒ −4 ≤ −4k ≤ 18 ⇒ 1 ≥ k ≥ −2
2
2
Alternatively : After step (A)
2
cos 2x =
−6 ± 36 − 16k − 4 −6 ± 32 − 16k
=
2
2
cos 2x = −3 + 2 2 − k or − 3 − 2 2 − k (rejected, think !)
⇒ −1 ≤ −3 + 2 2 − k ≤ 1 ⇒ 2 ≤ 2 2 − k ≤ 4 ⇒ 1 ≤ 2 − k ≤ 2 ⇒ 1 ≤ (2 − k) ≤ 4 ⇒ −1 ≤ − k ≤ 2 ⇒ 1 ≥ k ≥ −2


Que. 12. Let a = sin θ; b = sin  θ +
2π 
4π 
2π 
4π 



 ; c = sin  θ +
 Hence, a + b + c = sin θ + sin  θ +
 + sin  θ +

3 
3 
3 
3 



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Question. & Solution. Trigo. Page: A - 32 of A - 34
 
2π 

2π  
π

π


π

= sin θ + sin  − θ  − sin  + θ  use : sin  π +  = sin  π −  θ +
  = sin  − θ   (using C-D)
3 
3 
3

3


3


 
π
= sin θ − 2 cos sin θ = sin θ − sin θ = 0 since a + b + c = 0 hence a 3 + b3 + c3 = abc
3
2π 
4π 
π


π
 π



∴ sin 3 θ + sin 3  θ +  + sin 3  θ +  = −3sin θ sin  − θ  sin  + θ  = −3sin θ  sin 2 − sin 2 θ 
3 
3 
3


3
 3



=−
3
3
3sin θ − 4sin 3 θ ) = − sin 3θ. H.P.
(
4
4
sin ( 2 r − 2r −1 ) sin 2 r cos r −1 − cos 2r sin 2r −1
sin 2r −1
Que. 13. Tr =
=
=
= tan 2 r − tan 2r −1
r −1
r
r −1
r
r −1
r
cos 2 .cos 2 cos 2 .cos 2
cos 2 .cos 2
n
∴ Sum = ∑ ( tan 2r − tan 2r −1 ) = tan 2 − tan1 + tan 22 − tan 2 + tan 23 − tan 2 2 + ......... + tan 2 n − tan 2n −1
r =1
Sum = tan 2n − tan (1)
Que. 14. (0250.00) Dividing
by
15 tan 4 α + 10 = 6sec 4 α ⇒ 15 tan 4 α + 10 = 6(1 + tan 2 α ) 2
cos 4 α
⇒ 9 tan 4 α − 12 tan 2 α + 4 = 0 ⇒ ( 3 tan 2 α − 2 ) = 0 ⇒ tan 2 α =
2
2
3
3
 3
 2
8 cosec α + 27 sec α ⇒ 8 (1 + cot α ) + 27 (1 + tan α ) ⇒ 8 1 +  + 27 1 +  ⇒ 125 + 125 = 250.
 2
 3
Now
6
6
2
3
2
3
x + sin y = 2008
Que. 15. (2008).
∴ y=
Que. 16. (92)
Subtract
x + 2008cos y = 2007
⇒ sin y = 1 + 2008cos y This is possible only if cos y = 0
sin y − 2008cos y = 1
π
π
and x = 2007 ⇒ x + y = 2007 + ⇒ [ x + y ] = 2008.
2
2
2sin x sin1 = cos ( x − 1) − cos ( x + 1)
44
∴ S∑ cos ( x − = 1) − cos ( x + 1)  1 + sec ( x − 1) .sec ( x + 1) 
x =2


1
1


= ∑ cos ( x − 1) +
− cos(x − 1) −
cos(x + 1)
cos ( x − 1) 
x =2 


44
 1 − cos 2 ( x + 1) 1 − cos 2 (x − 1) 
= ∑
−

 cos ( x + 1)
cos(x − 1) 
x =2 
44
2
44 
sin 2 (x + 1) sin ( x − 1) 
sin 2 3 sin 2 1 sin 2 4 sin 2 2 sin 2 4 sin 2 3
sin 2 44 sin 2 42
= ∑ 
−
∴
S
=
−
+
−
+
−
........
+
−

cos ( x − 1) 
cos 3 cos1 cos 4 cos 2 cos 5 cos 3
cos 44 cos 42
x = 2  cos ( x + 1)
+
sin 2 44 sin 2 42 sin 2 45 sin 2 43
sin 2 44 sin 2 45 sin 2 1 sin 2 2
−
+
−
⇒ S=
+
−
−
cos 44 cos 42 cos 45 cos 43
cos 44 cos 45 cos1 cos 2
sin 2 1 sin 2 44 sin 2 2 sin 2 45
S=−
+
−
+
which resembles 4 term of
cos1 cos 44 cos 2 cos 45
∴
4
∑ ( −1)
n =1
n
φ2 ( θn )
ψ ( θn )
θ1 + θ2 + θ3 + θ4 = 1 + 2 + 44 + 45 = 92.
Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
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Question. & Solution. Trigo. Page: A - 33 of A - 34
Que. 17. (3388)
f (x) = cos −1 ( sin x ) =
g(x) = sin −1 ( cos x ) =
π
− sin −1 ( sin x ) ...........(1)
2
π
− cos −1 (cos x)
2
and
............... (2) both f(x) and g(x) are periodic with period
−1
−1
2π. The graphs of sin ( sin x ) and cos ( cos x ) as follows
π
 2 − x if x ∈ [ 0, π / 2]

π
π
3π

<x≤
hence f (x) = x −
if
;

2
2
2
 5π
 − x if 3π < x ≤ 2π
 2
2
π
 2 − x if x ∈ [ 0, π]
g(x) 
 x − 3π if x ∈ [ π, 2π]

2
Now
Area enclosed between the two curves is the area of the rectangle ABCD in one pepriod.
now
Now
and
and
π2 π2
π2
π 
AD =
+
=
=

4
4
2
2⇒

DC = 2π

DC = 2π ∴ A = 7.π2 = 7 π2
π
. 2 [= π2
2
( in [ −7π, 7π])
49A = 49.7 π2 = 7.22.22 = 7.484 = 3388.
Address : Plot No. 27, III- Floor, Near Patidar Studio, Above Bond Classes, Zone-2, M.P. NAGAR, Bhopal
“IITJEE MATHS BY SUHAG SIR” ON WWW.YOUTUBE .COM
Teko Classes, Maths : Suhag R. Kariya (S. R. K. Sir), Bhopal Phone : 0 903 903 7779, 0 98930 58881.
Also Available online www.MathsBySuhag.com DOWNLOAD FREE STUDY PACKAGE FROM WEBSITE WWW.TEKOCLASSES.COM
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Question. & Solution. Trigo. Page: A - 34 of A - 34