GROUP (A) – CLASS WORK PROBLEMS
π
 7π 

cot  π +  = cot 

6
 6 

Q-1) Find the principal solution of the following
equations.
i)
sin x =
3
2
ii)
sin x =
2
3
iii)
cot x =
3
7π
π
< 2π
π and θ <
< 2π
π
6
6
Such that θ <
Thus, required principal solutions are
x=
7π
π
and x =
6
6
Q-2) Find the principal solution of the following
equations.
Ans. i)
sin x
3
2
=
π
= sin  
3
∴ sin x
π

 2π 
= sin  π –  = sin  
3


 3 
and x
such that θ <
3 cosec x + 2 = 0
iii)
3 sec x + 2 = 0
Ans. i)
tan x
= – 3
tan x
π
= – tan  
3
∴
tan x
π

= tan  π – 
3


2π
π
π
and x =
3
3
sec x =
ii)
ii)
π
2π
< 2π and θ <
< 2π
3
3
Thus, required principal solutions are
x=
tan x = – 3
i)
 2π 
= tan   and
 3 
2
tan x
3
π
∴ sec x = sec   and
6
such that θ <
x=
Such that θ <
11π
π
< 2π
π and θ <
< 2π
π
6
6
Thus, required principal solutions are
π
x = and x = 11π
6
6
iii)
cot x =
3
π
∴ cot x = cot   and
6
2π
5π
< 2π
π and θ <
< 2π
π
3
3
Thus, required principal solutions are
π

sec x = sec  2π – 
6


 11π 
= sec 

 6 
π

 5π 
= tan  2π –  = tan 

3


 3 
ii)
5π
2π
and x =
3
3
3 cosec x + 2 = 0
∴ cosec x =
–
2
3
π
cosec x = – cos ec  
3
π

∴ cosec x = cosec  π + 
3

4π  and
= cosec 

 3 
Trignomatric Functions
Mahesh Tutorials Science
2
π

∴ cosec x = cosec  2π – 
3


 5π 
= cosec 

 3 
5π
4π
< 2π
π and θ <
< 2π
π
3
3
such that θ <
iii)
viii) cosec 3x
=
–2
ix)
4 cos x = 1
x)
4 sin2 x – 3 = 0
xi)
sin mx – sin nx = 0
xii)
cos 3x = sin 2x
Thus, required principal solutions are
xiii) 4sin x cos x + 2 sin x + 2cos x + 1 = 0
5π
4π
and x =
x=
3
3
xiv)
sec2 2x = 1 – tan 2x
xv)
sinx tanx = tan x – sin x + 1
3 sec x + 2 = 0
xvi)
∴ sec x =
∴ sec x =
3
– sec
3 cos x – sin x = 1
xvii) cot x + tan x = 2cosec x
2
–
π

sin  x +  = 0
5

Ans. i)
π
6
∴
π

 5π 
∴ sec x = sec  π –  = sec 
 and
6

 6 
π

sin  x +  = sin θ
5

Now, sin θ = sin α,
n
θ = nπ + ( –1) α, n ∈ z.
π

sec x = sec  π + 
6


∴
 7π 
= sec 

 6 
Such that θ <
x=
x+
π
n
= nπ ± ( –1) . 0 = nπ
5
Thus required general solution is
5π
7π
< 2π
π and θ <
< 2π
π
6
6
Thus, required principal solutions are
π
, where n ∈ z
5
x = nπ –
ii)
7π
5π
and x =
6
6
∴
cos
5x
= 0
2
cos
π
5x

= cos  n π +  .
2
2

Q-3) Find the principal solution of each of the
n = ∀ 1,3,5, ...
following equations
i)
ii)
∴
π

sin  x +  = 0
5

cos
5x
=0
2
π

cos  x –
 =0
10


iv)
sec x =
v)
tan x = –1
vii)
cos x =
2
–1
2
 2x 
tan 
 =
 3 
Trignometric Functions
5x
2
iii)
3,
π
2
π
5
π

cos  x –
 =0
10 

∴
x–
∴
x
iv)
= nπ +
x = (2n + 1)
∴
iii)
vi)
3
2
π
10
= (2n + 1)
= (2n + 1)
sec x =
sec x =
2
π
4
π
2
π
π
–
2 10
Mahesh Tutorials Science
3
Now, secθ = secα,θ =2πn ± α
∴
v)
Now, cosec θ = cosec α,
n
π
4
x = 2 πn ±
θ = nπ + ( –1) α, n ∈ z.
3x = nπ + ( –1)
tan x = –1
tan x = tan
3π
4
x =
Thus, required general solution is
ix)
vi)
cos x =
3π
where n ∈ z
4
nπ
n 4π
+ ( –1)
, where n ∈ z.
3
9
4 cos2 x = 1
cos 2 x
–1
2
2
cos x
 π
cos x = –cos  
3
1
4
=
π

=  cos 
3


2
2 π
= cos  
3
π

= cos  π – 
3

2
Now, cos x
=
cos2 α ,
x = nπ ± α, n ∈ z.
 2π 
= cos 

 3 
Thus, required general solution is
Now, cos θ = cos α,θ = 2nπ ± α, n ∈ z.
x = nπ ±
Thus, required general solution is
2π
x = 2nπ ±
, where n ∈ z
3
vii)
4π
3
Thus, required general solution is
Now, tanθ
θ = tanα
α,θ
θ = nπ ± α, n ∈ z.
x = nπ +
n
 2x 
tan 

 3 
=
 2x 
tan 

 3 
 π
= tan  
3
3,
x)
2π
, where n ∈ z.
3
4 sin2 x – 3 = 0
∴ 4sin2 x
= 3
2
∴ sin x
=
3
4
 3

= 
 2 
2
Now, tan θ = tan α,θ = nπ + α, n ∈ z.
2x
3
π
3
π

=  sin 
3

2
Thus, required general solution is

2 π
=  sin

3

3n π π
+ , where n ∈ z
2
2
Now, sin2 x = α, x = nπ ± α, n ∈ z.
x =
viii)
= nπ +
2
∴ sin x
cosec 3x
cosec 3x
=
Thus, required general solution is
–2
3
cosec 3x
π
= – cosec  
3
π

= cosec  π + 
3

= cos ec
4π
3
π
= – cosec  
3
π

= cosec  π + 
3


= cos ec
4π
3
Now, cosec θ = cosec α,
Trignometric Functions
Mahesh Tutorials Science
4
n
θ = nπ + ( –1) α, n ∈ z.
3x = nπ + ( –1)
n
or x
4π
3
where n ∈ z.
Thus, required general solution is
xiii)
∴ 2sinx (2cos x – 1) –1(2cos x + 1) = 0
∴ (2cos x + 1)(2sinx + 1) = 0
4 cos x = 1
ix)
2
1
4
∴
cos x
=
cos 2 x
π

=  cos 
3

cos x =
–1
or
2
sin x =
–1
2
π

cos x = cos  π – 
3

2
π

or sin x = sin  π + 
6

2 π
= cos  
3
 2π 
cos x = cos 

 3 
2
= cos α ,
2
Now, cos x
4sin x cos x + 2 sin x + 2cos x + 1 = 0
4sin x cos x + 2 sin x + 2cos x + 1 = 0
nπ
n 4π
+ ( –1)
, where n ∈ z.
3
9
x =
1

=  – 2n  π
2


x = nπ ± α, n ∈ z.
 7π 
or sin x = sin 

 6 
Thus, required general solution is
x = nπ ±
2π
, where n ∈ z.
3
Thus required general solution are
x = 2nπ ±
4sin2 x – 3 = 0
x)
∴ 4sin2 x
= 3
2
∴ sin x
=
or x = mπ + ( –1)
3
4
2
π

=  sin 
3

2
∴ sin x
xii)
2
Now, sin2 x = α, x = nπ ± α, n ∈ z.
x
or x
=
2
∴ tan 2x + tan 2x = 0
∴ tan 2x (tan 2x + 1) = 0
∴ tan 2x = 0 or tan 2x + 1 = 0
∴ 2x = nπ or tan 2x = –1
∴ 2x = nπ
π

π
or tan 2x = – tan   = tan  π – 
4
4


 
π
+ 3x
2
π
– 2n π
2
Thus, required solution are
x
=
 3π 
= tan 

 4 
2n π π
+
5 10
= x=
2n π π
+
5 10
Trignometric Functions
sec 2 2x = 1 – tan 2x
2
∴ 1 + tan 2x = 1 – tan 2x
Thus, required general solution is
∴
7π
,
6
sec 2 2x = 1 – tan 2x

2 π
=  sin 
3

= 2n π –
m
where n, m ∈ z
 3

= 
 2 
or –x
2π
3
Thus required general solution are
x =
xv)
n π mπ 3π
or
±
, where n, m ∈ z
2
2
8
sinx tanx = tan x – sin x + 1
sinx tanx = tan x – sin x + 1
∴ sinx tanx – tanx + sinx – 1 = 0
Mahesh Tutorials Science
5
∴ tanx (sinx –1) + 1(sin x – 1) = 0
∴ 2cosx = 1
∴ (sinx – 1) (tanx + 1) = 0
sin x = sin
∴ cosx =
 3π 
π
or tan x = tan 

 4 
2
π
∴ cosx = cos  
3
Thus required general solution are
x = nπ + ( –1)
x = mπ +
n
π
2
1
2
Now, cosθ = cosα,θ = 2nπ ± α,n ∈ z
or
Thus required general solution are
3π
, where n, m ∈ z
4
x = 2nπ ±
π
, where n ∈ z
3
3 cos x – sin x = 1
xvi)
3 cos x – sin x = 1
Dividing throughout by 2, we get
1
1
3
cos x –
sin x =
2
2
2
π
π
cos   cos x – sin   sin x
6
6
π
= cos  
3
GROUP (A)–HOME WORK PROBLEMS
Q-1) Find the principal solution of the following
equations.
1
2
i)
cos x =
ii)
cosec x = 2
iii)
tanx =
Ans. i) cos x =
3
1
2
π

π
cos  x +  = cos  
6

3
cos x
π
= cos   and
3
Now, cosθ = cosα,θ = 2nπ ± α, n ∈ z
cos x
π

= cos  2π – 
3


x+
π
6
= 2n π ±
π
3
x+
π
6
= 2n π +
π
or
3
x+
π
6
= 2n π –
π
3
 5π 
= cos 

 3 
such that θ <
Thus, required principal solution are
x=
Thus required general solution are
π
∴ cosec x = cos ec   and
6
π
, where n ∈ z
2
xvii) cot x + tan x = 2cosec x
cot x + tan x
π

cosec x = cos ec  π – 
6

 5π 
= cos ec 

 6 
= 2cosec x
i.e 2cosec x = cotx + tanx
∴
2
sin x
=
=
∴ cos2 x + sin2 x
5π
π
and x =
3
3
ii) cosec x = 2
π
x = 2nπ +
6
or x = 2nπ +
5π
π
< 2π
< 2π and θ <
3
3
cos x sin x
+
sin x cos x
cos2 x + sin2 x
sin x .cos x
such that θ <
π
5π
< 2π
< 2π and θ <
6
6
Thus, required principal solution are
x=
π
5π
and x =
6
6
= 2cosx
Trignometric Functions
Mahesh Tutorials Science
6
iii)
3
tanx =
tan x =
π
∴ tanx = tan   and
3
tan x =
π

tan  2π – 
4


 7π 
= tan 

 4 
π

tan  π + 
3

such that θ <
 4π 
= tan 

 3 
such that θ <
x=
π
4π
and x =
3
3
Q-2) Find the principal solution of the following
equations.
i)
ii)
iii)
Thus, required principal solution are
4π
π
< 2π
< 2π and θ <
3
3
Thus, required principal solution are
1
sin x =
2
tan x = –1
x=
iii)
7π
3π
and x =
4
4
2 cos x +1 = 0
=
∴ cos x
=
∴ cos x
π

= cos  π – 
4

cos x
Thus, required principal solution are
x=
7π
11π
and x =
3
6
5π
3π
< 2π
< 2π and θ <
4
4
Thus, required principal solution are
x=
5π
3π
and x =
4
4
Q-3) Find the general solution of each of the
following equations.
i) cos 2x = θ
ii) sin 3x = θ
3
1
iv) sin2x =
2
2
v) sec 3x = –2
vi) sin x = tan x
vii) cos x + sin x = 1
viii) 2tan x – cot x + 1 = 0
Ans. i) cos 2x = θ
iii)
sinx =
ii) tan x = –1
∴ tan x =
π
– tan  
4
π

tan x = tan  π – 
4


 3π 
= tan 
 and
 4 
Trignometric Functions
π

= cos  π + 
4

such that θ <
 7π 
= sin 
 and
 6 
11π
7π
< 2π
such that θ <
< 2π and θ <
3
6
π
– cos  
 4
 5π 
= cos 

 4 
π
∴ sin x = sin  
6
 11π 
= sin 

 6 
2
 3π 
= cos 
 and
 4 
1
Ans. i) sin x =
2
π

sin x = sin  2π – 
6

–1
∴ cos x
2cos x +1 = 0
π

sin x = sin  π + 
6

7π
3π
< 2π
< 2π and θ <
6
4
cos(2n – 1)
π
= θ
2
2x = (2n – 1)
π
2
∴ x = (2n – 1)
π
4
Mahesh Tutorials Science
ii) sin 3x
7
=θ
n
Now, sin θ = sin α, θ = nπ + ( –1) α,n ∈ z.
n
∴ 3x = nπ + n ( –1) = nπ
x=
∴ sinx = sin 0
x = nπ or x = 2nπ,
where n ∈ z.
vii) cos x + sin x = 1
π
3
Dividing throughout by 2 , we get
n
1
2
Thus, required general solution is
n
sin 2x
cos x +
1
2
sin x =
Now, cos θ = cos α, θ = 2nπ ± α,n ∈ z.
= sin
π
6
π
4
= 2n π ±
π
4
x
= 2n π ±
π π
+
4 4
x
= 2n π ±
π π
+
4 4
x
= 2n π –
π π
+
4 4
∴ x–
n
Now, sin θ = sin α, θ = nπ + ( –1) α,n ∈ z.
2x = nπ + ( –1)
n
π
6
Thus, required general solution is
nπ
n π
x=
– ( –1)
12
2
x = 2n π +
v) sec 3x = –2
=
π
– sec  
3
π

= sec  π – 
3

= sec
2π
3
viii)
2π
= 2nπ ±
3
Thus, required general solution is
2n π 2π
±
3
9
where n ∈ z.
vi) sin x = tan x
∴ sin x = tan x
∴ sinx =
sin x
cos x
π
or
2
x = 2nπ, where n ∈ z.
2tan x – cot x + 1 = 0
2tan x –
1
+1
tan x
= 0
2 tan2 x + tan x – 1 = 0
2tan x (tan x + 1) – 1(tan x + 1) = 0
(2tan x – 1)(tan x + 1) = 0
Now, sec θ = sec α, θ = 2nπ ± α,n ∈ z.
x=
or
Thus, requied general solutions are
where n ∈ z.
∴ 3x
2
π

π
∴ cos  π –  = cos  
4


 4
1
2
∴ sec 3x
1
π
π
π
∴ cos   cos x + sin   sin x = cos  
 4
 4
4
π
3
where n ∈ z.
iv) sin2x =
x = 2nπ ± 0
Thus, required general solution are
Now, sin θ = sin α, θ = nπ + ( –1) α,n ∈ z.
x = nπ + ( –1)
cos x = cos 0
0
3
2
sinx = sin
or
∴ x = nπ ± ( –1) (0) or
nπ
where n ∈ z
3
iii) sin x =
∴ sinx cos x – sin x = 0
∴ sinx (cos x – 1) = 0
or
cos x = 1
∴ sinx = 0
tan x =
1
2
or tan x = –1

2  1 
tan x = tan  tan    or
 2 

tan x = tan
3π
4
Now, tan θ = tan α, θ = nπ ± α,n ∈ z.
Thus, requied general solutions are
Trignometric Functions
Mahesh Tutorials Science
8
3π
–1  1 
x = n π + tan   or x = m π +
4
2
 
Q.2) Find polar co-ordinates of points whose
Cartesian oc-ordinates are
where n, m ∈ z.
GROUP (B) – CLASS WORK PROBLEMS
Q-1) Find Cartesian co–ordinates of points whose
i)
 1
(x, y) =  0, 
 2
ii)
 –1 –1 
,
(x, y) = 

 2 2
iii)
(x, y) =
(–
polar co–ordinates are
 π
i) Polar co–ordinates are  4 
 2
3

ii) Polar co-ordinates are  ,1350 
4

x = r cos θ and y = r sin θ
π
π
∴ x = 4cos   and y = 4 sin  
2
 2
∴ x = 4(0) = 0 and y = 4(1) = 4
Thus, the Cartesiam co-ordinates are
(0,4).
ii)
 1
Ans. i) (x, y) =  0, 
 2
=
x2 + y2
=
0+
1
4
1
2
=
and tanθ
=
r sin θ
r cos θ
=
2
= ∞
0
θ =
=
3
,
4
θ = 135 0
θ =
x =
3
cos 1350 and
4
y =
3
sin 1350
4
x =
3  –1 

 =
4 2
y
(
3 1 
=

 =
4 2
)
)
–3
4 2
 –1 –1 
,

ii) (x,y) = 
 2 2
and
3
.... (given)
Using x = r cos θ and y = r sin θ,
where (r,θ) are the required co-ordinates.
4 2
–1
2
r
= r cos θ,
=
and tanθ
∴ θ =
Trignometric Functions
π
2
1 π 
Polar co-ordinates are  , 
2 2
Thus, the Cartesian co-ordinates are
 –3
3 
,


4 2 4 2
3π
π
or
2
2
But the point lies on the side of Y-axis.
x = r cos θ and y = r sin θ
(
.... (given)
where (r,θ) are the required co-ordinates.
3

Polar co-ordinates are  ,1350 
4

∴ r
)
Using x = r cos θ and y = r sin θ,
r
π
Ans. i) ∴ x = 4, θ =
2
2, 2
x 2 + y2
=
π
5π
or
4
4
–1
2
= r sin θ
=
r sin θ
r cos θ
1 1
+
2 2 = 1
= 1
Mahesh Tutorials Science
9
But, point lies in 3rd quadrant.
∴θ =
5π
4
 5π 
∴ Polar co-ordinates are 1,

 4 
(–
(x,y) =
iii)
2, 2
)
r sin θ
r cos θ
tanθ =
∴θ =
2 = r sin θ
x 2 + y2
=
0
= 45
∴ ∠A
= 1800 – 600 – 450 = 750
2+2
=
=
2
– 2
Q-4) In ∆ ABC prove that
A
B – C b –c 
sin 
=
 cos
2
 2   a 
co-ordinates.
r
∴ ∠C
2
.... (given)
where (r,θ) are the required
2 = r cos θ,
=
Thus the angles of ∠ABC are ∠A = 750 ,
Using x = r cos θ and y = r sin θ,
–
1
∴ sin C
A
 B – C b – c 
Ans. sin 
=
 cos
2
 2   a 
= 2
Consider R.H.S.
A
b – c 
= 
 cos
2
 a 
= –1
3π
7π
or
4
4
=
k sin B – k sin C
A
.cos
2
k sin A
But, point lies in 2nd quadrant.
∴θ =
... (by Sine rule)
3π
4
A
 sin B – sin C 
= 
 cos
sin A
2


 3π 
∴ Polar co-ordinates are  2,

 4 
=
Q-3) The angles of ∆ ABC are in A.P. and
But
b : c = 3 : 2 , then find ∠A, B, C.
Ans. ∵ angles are in A.P.
Let the angles of ∆ ABC be a – d, a, a + d
respectively
As measures of all angles of a triangle is 1800
∴
B – C
B+C
2sin 
 cos 

 2 
 2  cos A
sin A
2
A+B+C
= π
B+C
= π–A
B+C
2
=
A
π
–
2
2
π A
 B +C 
∴ cos 
 = cos  2 – 2 


 2 
a + a + d + a – d – 180 0
= 60
∴ ∠B
= 60 0
But
sin B
sin b
=
∴
sin B
sin C
=
∴
sin60
sin C
=
∴
3
=
= sin
0
∴ a
... (i)
A
2
... (ii)
Puting (ii) in (i) we get
sin C
c
R.H.S.
=
b
c
=
3
A
A
B – C
sin 
  2sin cos 
2
2
 2 
sin A
B – C
sin 
 sin A
 2 
sin A
2
3
2
.sin C
B – C
= sin 

 2 
= L.H.S.
Trignometric Functions
Mahesh Tutorials Science
10
∴ a 3 sin ( C – A )
A
B – C b – c
∴ sin 
=
 cos
2
a
2

 

+ b 3 sin ( C – A ) + c 3 sin ( A + B )
Q-5) With the usual notations prove that
C
A

2 a sin2 + c sin2  = (a + c – b).
2
2

2
Ans. sin θ
=
1 – cos 2θ
2
)
+ kc
2
(
sin (a
2
–b
2
)
)
 a 2b 2 – a 2c 2 + b 2c 2 – a 2b 2 
= k 

+ c 2a 2 – b 2c 2 

A

2 C
+ c sin2 
= 2 a sin
2
2

L.H.S.
(
2
2
2
2
2
2
= ka b – c + kb sin c – a
= k(0)
= 0
a (1 – cos C ) c (1 – cos A ) 
+
= 2

2
2


∴ a 3 sin (B – C) + b 3 sin (C – A) c 3
sin (A – B) = 0
= a – acos C + c – c cos A
= (a + c) – (a cos C + c cos A)
= a+c+b
= R.H.S.
Q-7) In any ∆ ABC if a 2 , b 2 , c 2 are in A.P. then
prove that cot A, cot B, cot C are in A.P.
Ans. a 2 , b 2 , c 2 are in A.P.
A

2 C
+ c sin2  = a + c + b
∴ 2 a sin
2
2

3
Q-6) In ∆ ABC prove that a sin (B – C) +
3
b3 sin (C – A) + c sin (A – B) = 0
Ans. a 3 sin (B – C)
2
= 2b
∴ a2 + c2
2
2
2
∴ a + c – b2 = b
 a 2 + c 2 – b2 
∴ 
 =
2ac


cos B
∴
=
3
= a (sin B cos C – cos B sin C)
3
= a (kb cos C – kc cos B)
  a2 + b2 – c 2 
 a 2 + c 2 – b 2 
2
= a k ab 
 – ac 

2ab
2ac


 
 
2
2
2
a2 + c 2 – b2 
2 a + b – c
–
= a k

2
2


=
k 2
a a 2 + b2 – c 2 – a 2 – c 2 – b2
2
=
k 2
a 2b 2 – 2c 2
2
( )(
(
k 2 sin2 B
(
2k sin A k 2 sin C
∴
cos B
=
k 2 sin2 B
2k 2 sin A sin C
∴
cos B
=
k 2 sin2 B
2k 2 sin A sin C
cos B
=
sin B
sin B
sin A sin C
∴
)
b2
2ac
)
... (by Cosine and Sine rule)
... (by Sine rule)
2
= ka (ab cos C – ac cos B)
.... (given)
2
∴ 2cot B
=
sin  π – ( A + C ) 
sin A sin C
.... (∵ A + B + C = π )
)
=
2
2
2
= ka (b – c )
Similarly, we can prove that b 3 sin ( C – A )
2
2
2
= kb (c – a ) and
∴ 2cot B
(
2
2
2
c 3 sin ( A + B ) = kc a – b
)
sin ( A + C )
sin A sin C
=
sin A cos C – cos A sin C
sin A sin C
=
sin A cos C cos A sin C
+
sin A sin C sin A sin C
= cot C + cot A
Hence, cot A, cot B, cot C are in A.P.
Trignometric Functions
Mahesh Tutorials Science
11
... (by sine rule)
Q.8) In ∆ ABC, if acos A = bcos B, then prove
that the triangle is either a right-angied
or an isosceles triangle.
Ans. acos A = b cos B
.... (given)
∴
3.4
sin 700
=
a
sin 250
∴
∴
k sin A cos A = k sin B cos B
sin A cos A
= sin B cos B
∴
3.4
0.9397
=
a
0.4226
∴
∴
2sin A cos A
sin 2A
∴
a
=
3.4 × 0.4226
0.9397
∴
2A = π – 2B or 2A – 2B
= 2sin B cos B
= sin 2B
= 1.5290 = 1.53
c
sin C
=
b
sin B
∴
3.4
sin 700
=
b
sin 850
and hence ∆ ABC is a right-angled triangle
∴
3.4
0.9397
=
b
0.9962
....(i)
A = B and hence ∆ ABC is an isoscles
∴
b
=
3.4 × 0.9962
0.9397
∴
∴
A + B=
π
2
C
π
2
=
Also
triangle.
....(ii)
From (i) and (ii), ∆ ABC is a right angled
= 3.6044
= 3.6
triangle or it is an isosceles triangle.
Q.9) With the usual notations show that
2(bc cos A + ac cos B + ab cos C) =
a 2 + b2 + c2
Ans. Consider R.H.S.
= 2(bc cos A + ac cos B + ab cos C)
= 2bc cos A + 2ac cos B + 2ab cos C
 b2 + c 2 – a 2 
 a 2 + c 2 – b2 
= 2bc 
 + 2ac 

2bc
2ac




 a 2 + b2 – c 2 
– 2ab 

2ac


.... (by Cosine rule)
2
2
2
2
2
2
2
2
2
= b +c – a +a + c – b +a + b – c
2
2
2
= a +b + c
L.H.S.
2(bc cos A + ac cos B + ab cos C)
2
2
2
= a +b + c
Q.10) Solve the triangle in which c = 3.4,
Thus, in ∆ ABC, we have a = 1.53 units,
b = 3.6 units ∠C = 700
Q-11) Solve the triangle in which a =
b=
(
)
3 +1 ,
)
3 + 1 and ∠C = 600 .
Ans. a = ( 3
0
+1) and ∠C = 60
+1) , b = ( 3
∴
a=b
It is an isosceles triangle such that
∠A = ∠B = x
0
But ∠A + ∠B + ∠C = 180
∴
x + x + 60 = 180
2x = 120
∴
∴
0
x = 60
0
∴
∠A = ∠B = ∠C = 60
Thus, it is an equilateral triangle.
∴
c=
∠A = 250 and ∠B = 850
3
+1
A
0
0
Ans. In ∆ABC, c = 180 = 3.4, ∠A = 25 and
∠B = 850
x
.... (given)
a= 3+1
Now, ∠A + ∠B + ∠C = 180 0 , we have
∠C
(
0
0
0
= 180 – ( 85 + 25 )
= 70 0
we have
c
sin C
0
x
=
a
sin A
B
60
b= 3+1
C
Trignometric Functions
Mahesh Tutorials Science
12
Q-12) In any ∆ ABC prove the following
(
2
2
= sin A (b – c )
)
i) a2 sin( B– C) = b2 – c2 sinA
= R.H.S.
 A–B
tan 

(a + b )
 2 
=
ii)
( a + b ) tan  A + B 


 2 
iii)
iv)
ii)
cos A
(b 2 – c 2 ) ( sin A )
a 2 sin ( B – C ) =
(a + b )
(a + b )
cos 2A cos 2B  1
1 
=
= 2 – 2
2
2
a
b
b 
a
=
tan 

tan 

Consider L.H.S
cos B
=
a –b
a +b
cos C
a 2 + b2 + c2
=
2abc
( a cos B + b cos A )
=
k sin A – k sin B
k sin A + k sin B
=
sin A – sin B
sin A + sin B
+
a 2 sin ( B – C )
+
sin B + sin C
+
b 2 sin ( C – A )
sin C + sin A
c 2 sin ( A – B )
sin A + sin B
2sin 

2sin 

=0
2
2
2
Ans. i) a sin ( B – C ) = (b – c ) sin A
Consider L.H.S
=
2
= a sin ( B – C )
= k 2 sin A sin ( B – C )
2
2
= k sin
.... 

= k 

a
sin A
( π – ( B+C ) sin ( B – C )
=
2
= k sin ( B+C ) [sin ( B+ C ) sin ( B – C )]
k2
2
k2
2
=
=
sin ( B + C ) [2sin ( B + C ) sin ( B – C )]
k2
2
=
A–B 

 cos 
2


A+B 

 sin 
2


k2
2
∴
tan 


tan 

A–B
2
A+B
2
k
2

sin ( B + C ) 

k2
k2
b
k2
2
–
b
sin B
sin ( π – A ) (b 2 – c 2 )
Trignometric Functions
c
k2
=
a –b
a +b
cos 2A
a2



c
sin C






A+B
2
A–B
2



A+B
2






=
tan 


tan 







A–B
2
A+B
2
iii)
sin ( B + C ) × 2 ( sin2 B – sin2 C )
2


 cot 


A–B
2






A+B
2
A–B
2
= R.H.S.
sin ( B + C ) [cos 2C – cos 2B]
... ∵

=
A–B 

 cos 
2


A+B 

 cos 
2


sin ( B + C ) 1 – 2sin2 C – 1 + 2 sin2 B 
2
=
sin 

cos 


= tan 

= k 2 sin2 ( B+C ) sin ( B – C )
=






A–B
2
A+B
2
( c cos B + b cos C ) (a cos C + c cos A )
+
v)
∴
=
cos 2B
b2
= 

1
a2
–
1
b2
Consider L.H.S



=
cos 2A
a2
=
1 – 2sin2 A
a2
–
cos 2B
b2
–
1 – 2sin2 B
b2



Mahesh Tutorials Science
13
(
.... ∵ cos 2 θ =1 – 2sin2θ
=
1
a2
–2
sin2 A
a2
=
1
a2
1
b2
–
+2
1
b2
2
– 2k –
)
sin2 B
b2
=
a 2 + b2 + c 2
2abc
v)
a 2 sin ( B – C )
sin B + sin C
+
+ 2k
2
a 2 sin ( B – C )
sin B + sin C
... (by Sine rule)
=
1
a2
1
b2
–
cos 2A
a2

= 

1
a2



1
b2
=
a 2 (bk cos C – ck cos B )
bk + ck
=
ak [ab cos C – ac cos B]
k (b + c )
iv)
cos A
(c cos B + b cos C )
+
+
cos B
(a cos C + c cos A )
cos C
(a cos B + b cos A )
=
=
=
=
cos B
a
cos
C + cos A )
(
+
  a 2 + b2 – c 2 

ab 
2ab

 
2
2

 c + a – b2
– ac . 

2ac


a
b +c
a 2 + b2 + c 2
2abc
Consider L.H.S
cos A
c
cos
B
+ b cos C )
(
a 2 sin ( B – C )
sin B + sin C
Consider
cos 2B
b2
–
–
b 2 sin ( C – A )
sin C + sin A
2
c sin ( A – B )
=0
sin A + sin B
+
+
= R.H.S.
=
b 2 sin ( C – A )
sin C + sin A
2
c sin ( A – B )
=0
sin A + sin B
+
+
cos C
(a cos B + b cos A )
cos B
b
cos C
c
a
b
( +c)
1
2






(a 2 + b 2 – c 2 – c 2 – a 2 + b 2 )
=
a
b +c
1
2
=
a
(b + c )
(b + c ) (b – c )
( 2b 2 – 2c 2 )
= a(b – a)
=
cos A
a
+
+
...(by projection rule)
=
b2 + c 2 – a 2 + a 2 + c 2 – b2 + a2 + b2 – c 2
2abc
similarly,
b 2 sin ( C – A )
sin C + sin A
= b(c – a) and
c 2 sin ( A – B )
sin A + sin B
= c( a – b)
∴ L.H.S
=
a 2 + b2 + c 2
2abc
=
= R.H.S.
∴
+
+
cos A
(c cos B + b cos C )
cos B
(a cos C + c cos A )
+
a 2 sin ( B – C )
sin B + sin C
cos C
(a cos B + b cos A )
b 2 sin ( C – A )
sin C + sin A
2
c sin ( A – B )
sin A + sin B
+
= a(b – c) + b(c – a) + c(a – b)
= ab – ac + bc – ab + ac – bc
= 0
= R.H.S.
Trignometric Functions
Mahesh Tutorials Science
14
cos A cos B
=
then show that it
a
b
is an isosecles triangle
sin 

B–C 
A
 sin
2
2

 π – (B + C ) 
cos 

2


Q-13) In ∆ ABC if
cos A
a
Ans.
=
cos B
b
∴
cos A
k sin A
=
cos B
k sin B
∴
cos A
sin A
=
cos B
sin B
=
... (given)
∴
sin A cos B
= cos A sin B
∴
sin A cos B
= cos A sin B = 0
∴
sin (A – B)
= 0
∴
A–B
= 0
∴
A=B
=
B–C 
sin 

2


(B + C)
 π
cos 
–
2
 2
=
sin 

sin 

=
sin
Hence, it is isosceles triangle
Q-14) In ∆ ABC prove that
(b – c )
a






B–C
2
B+C
2
B
2
B
2
sin
C
2
C
2
cos
cos
+ cos
=
b –c
a
=
k sin B – k sin C
k sin A
=
sin B – sin C
sin A
=
tan
b –c
a
=
sin
B
C
cos , we get
2
2
=
tan
2sin 

2sin 

2sin 

B
2
B
2
C
2
C
2
– tan
+ tan
Q-15) Show with that usual noations that
a sin ( B – C )
B–C
2
B–C
2
(b
Ans. Let


 cos 


sin A


 cos 


sin A
B+C
2
π–A
2



B–C
2
2sin 



 cos 


sin A
π
2
B–C 
A
 sin
2
2

A
A
2 sin
cos
2
2
Trignometric Functions
–
–c
2
)
sin A
a



=
b sin ( C – A )
=
c2 – a 2
sin B
b
=
c sin ( A – B )
=
(a
2
– b2
)
sin C
C
a sin ( B – C )
b2 – c 2
=
a ( sin B cos C – sin C cos B )
b2 – c 2
=
k (ab cos C – ac cos B )
b2 – c 2



A
2
2
Consider
∵ A + B + C = π

∴ B + C = π – A or A = – B + C 
(
)

=
C
2
C
2
sin
tan
Ans. Consider L.H.S.
=
B
2
B
2
– cos
Dividing Nr and Dr by cos
B
C
– tan
2
2
=
B
C
tan + tan
2
2



  a 2 + b2 – c 2 
 ab 

2ab

 
2
2

 a + c – b2
– ac 

2ac








=
k
b2 – c 2
=
k
2 (b 2 – c 2 )
(a 2 + b 2 – c 2 – a 2 – c 2 + b 2 )
=
k
2 (b 2 – c 2 )
× 2 (b 2 – c 2 )
= k
Mahesh Tutorials Science
15
Similarly, we can prove that
Q-17) Show with the usual notations that
(c
2
b sin ( C – A )
c 2 – a2
= k and
)
(
=
c sin ( A – B )
a 2 – b2
=k
=
cos A =
b sin ( C – A )
c 2 – a2
=
∴
2bc cos A
cos2 B – cos 2 A
b+a
cos 2 B – cos 2 C
b+c
=
=
cos 2 C – cos 2 A
c +a
+
1 – sin2 B – (1 – sin2 C )
= 4∆
And
b +c
c +a
sin2 C – sin2 B
b +c
2
2
k 2 (c – b ) (c + b )
b +c
2
2 2
k a –k c
c +a
+
+
=
c 2 + a 2 – b2
2ac
c2 + a2 – b2 = 2ac cos B
Consider (a2 – b2 + c2) tan B
= 2ac cos B tan B
= 2ac sin B
= 4∆
...(ii)
Similarly, we can prove (b2 – c2 + a2)
sin2 A – sin2 C
c +a
+
cos B
...(i)
1

= 4  ac sin B 
4


(1 – sin2 C ) – (1 – sin2 A )
k c –k b
b +c
= (b2 + c2 – a2)
1

= 4 ×  bc sin A 
2


∴
Ans. L.H.S.
2
) tan C
= 2 bc sin A
=
2 2
2
= 2 bc cos A. tan A
cos2 B – cos2 C cos 2 C – cos 2 A
+
b+c
c+a
=
– c2 + a
Consider L.H.S. = (c2 – a2 + b2) tan A
c sin ( A – B )
(a 2 – b 2 )
+
2
b2 + c 2 – a 2
2bc
Q-16) In ∆ ABC prove that
=
(b
Ans. By projection rule, we have
Hence,
a sin ( B – C )
(b 2 – c 2 )
)
– a2 + b2 tanA = a2 – b2 + c2 tan B
tan C = 4∆
...(iii)
From (i), (ii) and (iii), we get
(c2 – a2 + b2)tan A = (a2 – b2 + c2)
tan B = (b2 – c2 + a2)tan C
k 2 (a – c ) (a + c )
c +a
= k 2 [c – b + a – c ]
= k 2 [a – b ]
=
=
k 2 a 2 – b 2 
a +b
(ka )2 – (kb )2
b +a
=
sin2 A – sin2 B
b +a
=
cos2 B – cos2 A
b +a
= R.H.S.
Trignometric Functions
Mahesh Tutorials Science
16
Q-18) Show that
A
B
 C   A∆ ( ABC ) 
sin   sin   sin   = 
abcs
2
2
2
2
∴
(a + c) + b = 3b
∴
a + c = 2b
.... (by projection rule)
Hene. a, b, c are in A.P.
for ∆ ABC
Q-20) With the usual notations prove that
Ans. Consider L.H.S.
A
2
= sin
B
2
sin
sin
(s – b ) (s – c )
(s – a ) ( s – c )
bc
ac
=
A
A ( ∆ABC ) = s ( s – a ) tan  
2
B
C
 
= s ( s – b ) tan   = s ( s – c ) tan  
2
2
C
2
( s – a ) (s – c )
Ans. We know that A(∆ ABC) = sin
ab
=
( s – b )2 ( s – c )2 ( s – a )2
=
a 2b 2c 2
s ( s – a )( s – b ) ( s – c )
Consider s ( s – a ) tan
(s – a ) (s – b ) ( s – c )
=
abc
=
=
[ A ( ∆ABC )]2
=
= R.H.S.
.... (i)
Also s ( s – b ) tan
A
2
B
2
sin
C
2
sin
[ A ( ∆ABC )]
B
2
( s – a ) (s – c )
s (s – b )
= s (s – b )
2
=
s ( s – a )( s – b ) ( s – c )
= A (∆ABC)
abcs
= sin
A
2
( s – b ) (s – c )
s (s – a )
= s (s – a )
s ( s – a ) ( s – b ) (s – c )
abcs
abcs
=
s ( s – a )( s – b ) ( s – c )
= A (∆ABC)
Q-19) If in ∆ ABC,
s ( s – c ) tan
then prove that a, b, c, are in A.P.
2 
Ans. a cos 

C
2

2 
 + c cos 


A
2
.... (ii)
Similarly, we can prove that
C
 A  3b
a cos2   + c cos2   =
2
2
 
2

 =

C
= A (∆ABC)
2
a 

1 + cos C
2
 + c  1 + cos A
2



2
....  cos θ =

∴
∴
a + a cos C + c + cos A
2
=

=

A (∆ABC)
3b
2
1+ cos 2θ
2
3b
2
(a + c) + (a cos C + c cos A) = 3b
.... (iii)
From (i),(ii) and (iii), we get
3b
2
= s ( s – a ) tan
A
2
= s ( s – b ) tan
B
2
= s ( s – c ) tan
C
2
...(given)
∴
A
B
C
sin sin
2
2
2



Q-21) Show that in ∆ ABC
A 
B a +b – c

 tan   tan  =
2
2  a +b +c


Ans. Cosider L.H.S.
Trignometric Functions
Trignomatric Functions
Mahesh Tutorials Science
17
Q-2) Find the polar co-ordinates of the points
A 
B

=  tan   tan 
2
2


(s – b )(s – c ) (s – a )(s – c )
s (s – a )
s (s – b )
=
(s – a )(s – b )( s – c )( s – c )
s 2 ( s – a )( s – b )
=
=
s –c
s
=
a +b +c
–c
2
a +b +c
2
=
∴
whose cartesian co-ordinates are
i)
(
2, 2
(
∴
(x,y) =
2, 2
r
∴
= r cos θ ,
=
x2 + y2
2 = r sin θ ,
2+2
=
r sin θ
r cos θ
2
=
2
=
4 =2
= 1
π
4
2
ii) (–3,0)
(x,y) = (–3,0)
....(given)
using x = r cos θ and y = r sin θ , where
(r,θ
θ) are the required co-ordinates,
–3
= r cos θ , 0 = r sin θ
=
GROUP (B)–HOME WORK PROBLEMS
Q-1) Find Cartesian co-ordinates of points whose
polar co-ordinates are
tan θ =
π

Ans. Polar co-ordinates are  2, 
4

π
4
x2 + y2
r sin θ
r cos θ
=
=
9+0
0
–3
= 3
= 0
∴
θ = 0 or π
But the point lies on negative X-axis
θ = π
∴
Polar co-ordinates are (3,π
π)
π

 2, 
4

2,θ=
.... (given),
 π
Polar co-ordinates are  2, 
 4
r
x =
)
π 3π
or
4
4
But the point lies in the 1st quadrant.
A 
B

=  tan   tan 
2 
2

∴
2, 2
θ =
θ =
a +b – c
a +b +c
(
and tan θ =
R.H.S.
=
)
using x = r cos θ and y = r sin θ , where
(r,θ ) are the required co-ordinates,
a +b – c
a +b +c
=
ii) (–3,0)
Ans. i)
2
 A ( ∆ABC ) 
abcs
)
Q-3) In ∆ ABC if ∠A = 450 , ∠B = 600 and
x = r cos θ and y = r sinθ
θ
∴
y =
π
2 cos   and y =
4
π
2 sin  
4
∴
x =
 1 
2

 2
 1 
2

 2
∴
x = 1 and y = 1
and y =
Thus, the cartesian co-ordinates are (1,1)
∠C = 750 then find the ratio of its sides.
Ans. By sine rule, we have
we have
∴
a
sin A
=
b
sin B
=
c
sin C
a = k sin A, b = k sin B and c = k sin C
a
Sin B
Sin A
b
and
=
=
b
Sin C
Sin B
c
a : b : c = sin A : sin B : sin C
a : b : c = sin 450 : sin 600 : sin 750
Trignometric Functions
Mahesh Tutorials Science
18
1
Now, sin 450 =
2
(
, sin 60 0 =
0
and sin 750 = sin 30 + 45
0
A –B 
= cos 
 = L.H.S.
 2 
1
2
)
∴
0
0
0
0
= sin 30 cos 45 + cos 30 sin 45
=
=
1 1
3
1
×
+
×
2
2
2
2
C
 A –B 
a –b 
cos 
 = cos 
 sin
2
 2 
 2 
Q-5) With the usual notations prove that
sin ( A – B )
sin ( A + B )
3 +1
2 2
i.e. a : b : c =
1
2
:
=
3
3 +1
:
2
2 2
a : b : c = 2 : 6 : 3 +1
Q-4) In ∆ ABC prove that
C
 A – B   a +b 
cos 
=
 sin
2
c
2

 

C
 A – B   a +b 
Ans. cos 
=
 sin
2
 2   c 
sin A cos B – cos A sin B
sin A cos B + cos A sin B
=
ak cos B – bk cos A
ak cos B + bk cos A
=
a cos B – b cos A
a cos B + b cos A
=
 b2 + c 2 – a 2 
1   a 2 + c 2 – b2 
a. 
 –b

c  
2ac
2bc


 
...(by Sine rule)
C
 sin A + sin B 
= 
 sin
sin C
2


 A +B 
 A +B 
2sin 
 cos 

 2 
 2  sin C
... (i)
=
sin C
2
A+B+C=π
A+B=π–C
... (ii)
Putting (ii) in (i), we get
=
=
C
C

 A –B 
 2sin cos  cos 

2
2

 2 
sin C
1 a 2 + c 2 – b 2 b 2 + c 2 – a 2 
–


c
2c
2c

=
1
a 2 + c 2 – b2 – b2 – c 2 + a 2
2c 2
=
1
2a 2 – 2b 2
2c 2
=
a 2 – b2
c2
(
(
)
)
sin ( A – B )
sin ( A + B )
=
a 2 – b2
c2
Q-6) In ∆ ABC, if cos A = sin B – cos C then
show that it is a right angled triangle.
Ans. cos A = sin B – cos C
∴
cos A + cos C = sin B
∴
 A +C 
A –C 
2cos 
 cos 
 = sin B
 2 
 2 
∴
 A +C 
A –C 
B
B
2cos 
 cos 
 = 2sin cos
2
2
2
2




A –B 
sin C cos 

 2 
sin C
Trignometric Functions
... (by projection rule)
= L.H.S.
∴
π C
–
2 2
C
 A +B 
π C
sin 
 = sin  –  = cos
2
 2 
2 2
R.H.S
... (by Sine rule)
=
C
 k sin A + k sin B 
= 
 sin
k sin C
2


∴
sin ( A + B )
... (by Cosine rule)
C
 a +b 
RHS = 
 sin
2
 c 
 A +B 

 =
 2 
sin ( A – B )
=
Consider
But,
a 2 – b2
c2
Ans. Consider R. H.S
Thus, required ratio of the sides of ∆ ABC
∴
=
.... (i)
∴
But A + B + C = π
A+C–π=–B
Mahesh Tutorials Science
19
∴
 A +C 


 2 
π –B
= 

 2 
∴
 A +C 
cos 

 2 
π –B 
= cos 

 2 
∴
sin C =
π B 
= cos  – 
2 2 
∴
∠C
= sin
Also,
B
2
∴
∴
∴
∴
A–C=B
∴
2A = π
∴
A=
= cos
=
π
π
, ∠B =
2
6
π
3
a = 72cm, ∠B = 1080 and ∠A = 250
Ans. In ∆ ABC , a = 72cm, ∠B = 1080 and
∠A = 250
∴
250 +1080 + ∠C = 1800
∴
∠C
= 1800 – (1330 )
Hence, ∆ ABC is a right angled triangle.
= 470
We have
Q-7) Solve the triangle in which
a = 2, b = 1and c = 3
Ans. In a triangle a = 2, b = 1 and c =
72
b
=
0
sin 25
sin1800
∴
72
b
=
0.4226 0.9511
∴
b=
2
( 3)
= 1+3=4
∴
b2 + c 2 = a 2
Hence, it is a right-angled triangle at A.
∴
A = 900
=
π
2
sin A sin B sin C
=
=
a
b
c
∴
sin 90 sin B sin C
=
=
2
1
3
∴
1
= sin B =
2
∴
sin B =
1
2
∴
∠B
π
6
sin c
3
a
b
=
sin A sin B
∴
3
.... (given)
Now, b 2 + c 2 = 12 +
.... (given)
Now, ∠A + ∠B + ∠C = 1800
π
2
=
π
3
Q-8) Solve the triangle in which
B
2
A=B+C
A+A=A+B=C
But
1
2
3
2
and ∠C =
B
B
B
A –C 
cos 
 = 2sin cos
2
2
2
 2 
A –C 
cos 

 2 
3
=
Thus, in ∆ ABC,we have ∠A =
.... (ii)
Putting (ii) in (i), we get
2sin
sin C
72 × 0.9511
= 162.04 = 162
0.4226
Also,
a
c
=
sin A sin C
∴
72
c
=
sin 250 sin 470
∴
72
c
=
0.4226 0.7314
∴
c=
72 × 0.7314
= 124.61
0.4226
Thus in ∆ ABC, we have b = 162cm,
c = 124.6cm and ∠C = 470
Trignometric Functions
Mahesh Tutorials Science
20
Q-9) In any ∆ ABC prove the following
iii)
i)
c – b cos A cos B
=
b – cos A
cos C
ii)
cos 2 A cos 2B  1
1 
–
= 2 – 2
a2
b2
b 
a
a 2 sin ( B – C )
iii)
sin A
+
a 2 sin ( B – C )
sin A
+
b 2 sin (C – A )
+
L.H.S. =
b 2 sin ( C – A )
sin B
c 2 sin ( A – B )
sin C
a 2 sin ( B – C )
sin A
+
sin B
Ans. i)
sin C
+
=
=
=
+
c – b cos A
b – cos A
+
 b2 + c 2 – a 2 
c–

2c


=
 b2 + c 2 – a 2 
+ c 2 – a2 
 b–

2bc
2b



a 2 + c 2 – b2 a 2
2c
=
b2 + c 2 – b2 b2
2b
=
ii)
+ c 2 – b2
2ac
+ c 2 – b2
2ab
+
ck
k
b (ck cos A – ak cos C )
k
c (ak cos B – bk cos A )
k
= ab cos C – ac cos B + bc cos A – ab
cos C + ac cos B – bc cos A
= 0
= R.H.S.
= R.H.S.
a 2 sin ( B – C )
sin A
+
+
sin B
c sin ( A – B )
sin C
=0
cos 2A cos 2B
–
a2
b2
1 – 2sin2 A 1 – 2sin2 B
–
=
a2
b2
2
2
2
Q-10) In ∆ ABC if sin A + sin B = sin C
then show that the triangle is a right
angled triangle.
...( ∵ cos 2θ = 1 – 2sin2 θ )
=
1
sin2 A 1
sin2 B
–
2
–
+
2
a2
a2
b2
b2
=
1
1
– 2k 2 – 2 + 2k 2
2
a
b
...(by Sine rule)
=
b 2 sin (C – A )
2
cos 2A cos 2B  1
1 
–
= 2 – 2 
a2
b2
b 
a
Consider L.H.S =
1
1
– 2
2
a
b
= R.H.S
∴
bk
c ( sin A cos B – cos A sin B )
a (bk cos C – ck cos B )
+
c – b cos A cos B
=
b – cos A
cos C
∴
ak
b 2 ( sin C cos A – cos C sin A )
2
+ c 2 – a2 

2bc

cos B
cos C
sin C
a 2 ( sin B cos C – cos B sin C )
Consider L.H.S.
 b2
c –b

=
 b2
b –c

sin B
c sin ( A – B )
=0
c – b cos A cos B
=
b – cos A
cos C
=
b 2 sin (C – A )
2
2
c sin ( A – B )
+
=0
cos 2A cos 2B  1
1 
–
= 2 – 2 
2
2
a
b
b 
a
Trignometric Functions
2
2
2
Ans. sin A + sin B = sin C
∴
ka 2 + k 2b 2 = k 2c 2
∴
k 2 a 2 + b 2 = k 2c 2
(
)
a 2 + b2 = c 2
Hence, ∆ ABC is right-angled triangle
Mahesh Tutorials Science
21
Q-11) In ∆ ABC show that
(
)
(
a 2 cos 2 B – cos 2 C + b 2 cos 2 C – cos 2 A
(
)
)
+ c 2 cos 2 A – cos 2 B = 0
vii) tan A
Ans. a = 18, b = 24, c = 30,
a +b +c
2
s =
Ans. Consider L.H.S.
(
a 2 cos2 B – cos2 C
=
sin A
vi)
(
(cos
)
i)
cos A
)
B)
=
b2 + c 2 – a 2
2bc
=
576 + 900 – 324
2 ( 30 )( 24 )
=
1152
2 × 30 × 24
=
576
30 × 24
=
4
5
+ b 2 cos2 C – cos2 A
+c2
2
A – cos2
2
2
2
2
= a cos B – a cos C
2
2
2
2
+ b cos C – b cos A
+ c 2 cos 2 A – c 2 cos 2 B
=
(a
2
cos2 B – b 2 cos2 A
(
+ (c
=
2
2
2
)
)
C)
+ b cos C – c 2 cos2 B
=
2
2
cos A – a cos
2
(a cos B – b cos A )(a cos B + b cos A )
+ (b cos C – c cos B ) (b cos C + c cos B )
+ (c cos A – a cos C )(c cos A + a cos C )
ii)
sin
A
2
=
....(by projection rule)
= ac cos B – bc cos A + ab cos C – ac cos B
+ bc cos A – ab cos C
(s – b )( s – c )
=
bc
12 × 6
24 × 30
=
(a cos B – b cos A ) c + (b cos C – c cos B ) a
+ (c cos A – a cos C ) b
iii)
cos
A
2
1
10
s (s – a )
=
bc
= 0
= R.H.S.
(
+b
36 × 18
24 × 30
=
a 2 cos 2 B – cos 2 C
∴
2
+c2
(cos
( cos
)
2
=
)
B) = 0
2
C – cos A
2
A – cos 2
iv)
Q-12) In ∆ ABC if a = 18, b = 24 and c = 30 then
find the values of
i)
cos A
ii)
A
sin  
2
iii)
A
cos  
2
iv)
A
tan  
2
v)
A∆ (ABC)
tan
A
2
3
10
=
(s – b )( s – c )
s (s – a )
=
12 × 6
36 × 18
=
v)
= 36
A(∆
∆ ABC) =
1
3
s (s – a ) (s – b ) (s – c )
36 × 18 × 12 × 6
=
= 6×6×6
= 216 sq. units
Trignometric Functions
Mahesh Tutorials Science
22
vi)
A=
1
bc sin A
2
216 =
tan x = – tan
1
× 24 × 30sin A
2
∴
2 × 216
= sin A
24 × 30
∴
sin A =
vii)
tan A =
=
=
As
–π  π π 
∈ – ,
range for principal
6  2 2 
value of tan–1 x
Hence, the required principal value
3
5
of x is
sin A
cos A
–π
4
 1
cos –1  – 
 2
iii)
3
5
4
5
–1  1 
Let x = cos  – 
 2
–
∴ cos x =
– cos
3
4
=
Q-1) Find the principal valus of
i)
1
sin –1  
 2
ii)
tan –1 ( –1)
1
2
∴ cos x =
GROUP (C) – CLASS WORK PROBLEMS
iii)
 π
= tan  – 
 4
π
4
As
π
π

= cos  π – 
3
3


 2π 
cos 

 3 
2π
∈ [0, π ] range for principal
3
–1
value of cos x
Hence, the required principal value
of x is
 1
cos –1  – 
 2
2π
3
Q-2) Find the values of
Ans.
i)
1
Let sin–1   = x
 2
1
= sin x
2
π
sin = sin x
6
and
π
 π π
∈  – ,  range for sin–1 x
6
 2 2
Hence, the required principal value
of x is
ii)
tan
–1
π
6
( –1)
–1
Let tan ( –1)
tan x = –1 = x
Trignometric Functions
–1
–1  1 
–1  1 
i) tan (1) + cos   + sin  
 2
2
–1
ii) tan
Ans. i)
( 3 ) – sec
–1
( –2 )
1
1
tan –1 (1) + cos –1   + sin –1  
2
2
=
π π π
+ +
4 3 6
=
3π
4
ii) tan–1
( 3 ) – sec –1 ( –2)
–1
= tan
( 3 ) – sec –1 ( –2)
–1
But, sec –1 ( x ) = π – sec x
Mahesh Tutorials Science
23
∴ tan –1 3 – sec –1 ( –2 )
∴
(
–1
3 – π – sec –1 ( –2)
= tan
)
3
tan α =  
4
π 
π
– π – 
=
3 
3
–1  3 
α = tan  
4
∴
π 2π
–
=
3 3
... (iv)
From (iii) and (iv),
–π
=
3
4
3
cos –1   = tan–1  
5
 
4
Q-3) Show that
sin –1
4
cos α =  
5
∴
12
4
63
+ cos –1 + tan –1
=0
13
5
16
Ans.
 12 
4
 63 
sin–1   + cos –1   + sin–1 

 13 
5
 16 
–1  12 
–1  3 
–1  63 
= tan   + tan   + tan 

5
4
 
 
 16 
A
13
12
θ
C
  12   3  
  +  
 63 
–1   5   4  
tan
+ tan–1 
=

 12   3  

 16 
1
–





 5   4  


–1
–1
–1  x + y  
∵ tan x + tan y = tan 

 1 – xy  

B
5
–1  12 
Let θ = sin  
 13 
∴
... (i)
= tan
12
sin θ =
13
tan θ =
–1  –63 
–1  63 
= tan 
 + tan 

 16 
 16 
12
5
–1  12 
θ = tan  
 5 
∴
=
.... (ii)
 63 
–1  63 
– tan–1 
 + tan 

 16 
 16 
(∵ tan–1 ( –x ) = – tan–1 ( x ))
–1  12 
From (i) and (ii), sin  
 13 
= tan
= 0
–1  12 
 
 5 
–1  12 
–1  4 
–1  63 
∴ sin   + cos   + tan 
=0
 13 
5
 16 
–1  4 
Also, let α = cos  
5
... (iii)
7π 

Q-4) Find the values of tan –1  tan

6 

P
5
3
Ans. As
7π  π π 
∉ – ,
6  2 2 
–1
which is the range of tan x
α
R
 63 
20  + tan–1  63 




 16 
 –16 
 20 
–1 
4
Q
Trignometric Functions
Mahesh Tutorials Science
24
7π 

tan–1  tan

6 

∴
 5 
sin y =  
 13 
π 

–1 
= tan  tan  π +  
6



–1  5 
∴ y = sin x  
 13 
But, tan ( π + θ ) = tan θ
∴ cos (x + y)
7π 

tan–1  tan

6 

∴
 4   12   3   5 
=  ×  –  × 
 5   13   5   13 
π
–1 
= tan  tan 
6


And
∴
= cos x cos y – sin x sin y
 48 – 15 
= 

 65 
π  π π
∈ – ,
6  2 2 
 33 
= 

 65 
7π 
π

tan–1  tan
 = 6
6 

–1  33 
∴ x + y = cos 

 65 
–1  33 
–1  4 
–1  12 
∴ cos   + cos   = cos 

 66 
5
 13 
Q-5) Show that
4
 12 
–1  33 
cos –1   + cos –1 
 = cos 

 5
 13 
 65 
Q-6) Prove that tan –1 x =
Ans.
1
1 – x 
cos –1 

2
1+ x 
if x ∈ [0,1]
5
3
2
x = tan θ
Ans. Let
∴
x
= tanθ and θ =
x
4
Now, R.H.S.
–1  4 
4
= cos   , then cos –1 –  
5
5
Let x
4
sin –1 x =  
5
–1  3 
= sin x  
5
∴ x
tan –1 x
=
1
1 – x 
cos –1 

2
 1+ x 
=
 1 – tan2 θ 
1
cos –1 

 1+ tan2 θ 
2


=
1
cos –1 ( cos 2 θ )
2
=
1
(2 θ)
2
–1
x = L.H.S.
= θ = tan
∴ tan–1 x
13
=
1
1 – x 
cos –1 

2
 1+ x 
5
 a cos x – b sin x 
Q-7) Simplify tan –1 

 b cos x + a sin x 
y
12
if
 12 
–1  12 
Let x = cos   , then cos y =  
 13 
 13 
Trignometric Functions
a
tan x > –1
b
–1  a cos x – b sin x 
Ans. Let y = tan 

 b cos x + a sin x 
Mahesh Tutorials Science
25
Dividing Nr and Dr by b cos x, we get
2 2
3
cos x =
 a cos x – b sin x 
tan–1 

 b cos x + a sin x 
–1  2 2 
x = cos 

 3 
=
 a–

tan x 
–1  b
tan 

1+ a tan x 
b


1
∴ sin–1  
3
=
a 
tan–1   – tan–1 ( tan x )
b 
2 2
1
∴ sin –1   + sin–1 

3
 3 
=
a 
tan–1   – x
b 
–1  2 2 
–1  2 2 
 + sin 

= cos 
 3 
 3 
9π 9
1 9
2 2
– sin –1 = sin –1
Q-8) Show that
8 4
3 4
3
9
9π 9
1
–1  2 2 
– sin–1   = sin 

4
8 4
3
 3 
=
2 2
9
sin–1 

4
 3 
2 2
π
–1  1 
– sin–1 
 + sin  
2
3
3


 1+ x – 1 – x
tan–1 
 1+ x + 1 – x



=
 1+ cos θ – 1 – cos θ 
tan–1 

 1+ cos θ + 1 – cos θ 
=

θ
2 θ
– 2sin2 
 2cos
2
2
tan–1 

θ
θ
2
+ 2sin2 
 2cos
2
2


1
3



π 1
1
– cos –1 x for
≤ x ≤ 1.
2 2
2
x
∴ sin x =
π
2
–1
θ = cos x
1
–1  1 
Let x = sin  
3
=
Ans. Let x = cosθ
3
22
π
2
Q-9) Show that
=
–1  2 2 
π
1
i.e. – sin–1   = sin 

2
3
 3 
=
2 2
1
sin –1   + sin–1 

3
 3 
 1+ x – 1 – x
tan –1 
 1+ x + 1 – x

9π
–1  1  
 – sin   
42
 3 
π
1
–1  2 2 
i.e. – sin–1   = sin 

2
3
 3 
i.e.
–1
–1
But cos x + sin x
∴
Ans. We want to prove
i.e.
–1  2 2 
= cos 

 3 
=
tan
–1 



θ
θ
– 2 sin 
2
2

θ
θ
2 cos + 2 sin 
2
2
2 cos
Dividing Nr and Dr by
 1+ x – 1 – x
tan–1 
 1+ x + 1 – x
2 cos
θ
, we get
2



Trignometric Functions
Mahesh Tutorials Science
26
θ 

1 – tan

2 
tan–1 

 1+ tan θ (1) 
2


=
6 x + 5x – 1 = 0
∴
6 x 2 – 6x – x – 1 = 0
6x (x + 1) – (x + 1) = 0
=
θ

tan–1 (1) – tan–1  tan 
2

=
π 1
– θ
2 2
tan
2
∴
∴
(x + 1) (6x – 1) = 0
∴
x = – 1 ot x = 1 6
Q-12) Write the expression
 1 – cos x
tan –1 
 1 + cos x

–1 
1+ x – 1 – x 


 1+ x + 1 – x 

 , x < π

in the simplest expression
π 1
– cos –1 x
2 2
=
 1 – cos x 
Ans. tan–1 

 1 + cos x 
1


Q-10) If sin  sin –1 + cos –1 x  = 1 , then find the
5


∴


–1 1
+ cos –1 x 
 sin
5


–1
= sin 1
∴
1
sin–1   + cos –1 ( x )
5
=
π
2
And we know that sin –1 x + cos –1 x =
∴
1
sin–1  
5
∴
x
=
x
2
2 x
2sin
1
2sin2
=
tan
=
x

tan–1  tan 
2

=
x
2
value of ‘x’.
1


Ans. sin  sin–1 + cos –1 x  =1
5


–1
Q-13) Find the value of the expression
π
2
tan
–1
= sin ( x )
2
1
2x
–1
–1 1 – y 
sin
+
cos


2
1+ x2
1+ y2 
where x < 0 and y > 0 such that xy < 1
1
5
Ans. tan
1  –1 2x
1 – y2 
+ cos –1
sin

2 
1+ x 2
1+ y 2 
Put x = tanα and y – tanβ
π
Q-11) If tan 2x + tan 3x =
then find the
4
–1
–1
∴
tan
value of ‘x’.
Ans. tan–1 2x + tan–1 3x =
∴
∴
 2x + 3x 
tan–1 
 1 – 2x ( 3x ) 


5x
1 – 6x
2
= tan
π
4
π
4
=
π
4
= 1
=
 –1  2 tan α 

sin 

2 
 1 + tan α 
1

tan 
2
 1 – tan β  
2

+ cos –1 

 1+ tan2 β  




=
tan
1
sin–1 ( sin 2α ) + cos –1 ( cos 2β ) 

2
=
tan
1
( 2α + 2β )
2
2
5x = 1 – 6 x
Trignometric Functions
1  –1 2x
1 – y2 
+ cos –1
sin

2 
1+ x 2
1+ y 2 
Mahesh Tutorials Science
=
tan ( α + β )
=
tan tan–1 x + tan–1 y
=

 x + y 
tan  tan–1 
 
 1 – xy  

(
tan
∴
27
)
1
1
Q-2) Find the values of cos –1   + 2 sin –1  
2
 
 2
1
1
Ans. cos –1   + 2sin–1  
2
2
1  –1 2x
1 – y2 
+ cos –1
sin

2 
1+ x 2
1+ y 2 
x +y
1 – xy
=
=
π
π
+ 2 
3
6
=
π π
+
3 3
=
2π
3
GROUP (C)–HOME WORK PROBLEMS
Q-1) Find the principal values of
i)
cosec –1 ( 2 )
ii)
 1 
sin 

 2
3π 
–1 
Q-3) Find the values of sin  sin

5 

Ans. We know that sin–1
–1
As
Ans. i) cosec –1 ( 2)
Let x
–1
= cosec 2
3π 
–1 
Let y = sin  sin

5 

1
2
sin x
=
sin x
= sin
 5π – 2π  
–1 
= sin sin 

5



π
6
2π  

–1 
= sin sin  π –

5  


π  π π
∈ – ,
range for principal value of
6  2 2 
2π 
–1 
= sin  sin

5 

cosec –1x
Hence, the required principal value of x
... (∵ sin ( π – θ ) = sin θ )
π
6
∴
–1  1 
ii) sin 

 2
Let x
3π  π π 
∉ – , 
5
 2 2
–1
which is the range of sin x
cosec x = 2
is
–1  1 
= sin 

 2
=
∴ sin x
= sin
3π 

sin–1  sin
 =
5 

 8 
–1  3 
–1  77 
sin –1 
 + sin   = sin 

 17 
 5
 85 
A
Ans.
2
π
4
17
π  π π
As ∈  – ,  range for principal value
4  2 2
–1
of sin x
Hence, the required principal value of x
π
4
2π
5
Q-4) Show that
1
∴ sin x
is
π  –π π 
∈
,
6  2 2 
x
C
15
8
B
–1  8 
Let x = sin 

 17  Trignometric Functions
Mahesh Tutorials Science
28
∴
8
15
and hence cos x =
17
17
sin x =
P
5
3
 5 
tan–1   + cos–1   =
5
 13 
4
= tan–1  
3
 5 
tan–1  
 13 
 5  4
  13  +  3  
   
= tan–1 
5 4 

 1 – 13  3  
 

4
5
cos y =
∴
sin (x + y ) = sin x cos y + cos x sin y
=
8 4 15 3
× +
×
17 5 17 5
=
32 + 45
85


–1
–1
–1  x + y 
∵ tan x + tan y = tan 
 
1
–
xy



  67  
  39  


= tan–1 
19


  39  


77
85
sin (x + y ) =
sin–1
∴
3
3
and hence sin y =
5
5
∴
∴
3
cos–1  
5
R
4
–1
Let y = sin
x+y
∴
3
y
∴
4
x = tan–1  
3
4
+ tan–1  
3
Q
∴
∴
∴
–1  77 
= sin 

 85 
8
3
 77 
+ sin–1 = sin–1 

17
5
 85 
 5 
3
 67 
tan–1   + cos –1   = tan–1 

 13 
5
 19 
π
–1  x – 1 
–1  x +1 
Q -6) If tan 
 + tan 
=
 x – 2
 x +2 4
find x
Q-5) Show that tan
–1
63
5
3
= tan –1
+ cos –1
16
13
5
π
 x –1
–1  x +1 
Ans. tan–1 
 + tan 
=
x
–
2
x
+
2
4




∴

x – 1 x +1 
+


x
– 2 x +2  = π
tan 
 x – 1   x +1   2

1 –  x – 2   x + 2  


∴
( x – 1)( x + 2) + ( x +1)( x – 2)
π
= tan
2
( x – 2) ( x + 2) – ( x – 1)( x +1)
Ans.
5
4
x
3
–1  3 
Let x = cos  
5
∴
∴
–1
x2 + x – 2 + x2 – x – 2
(
∴
2x 2 – 4
=1
–3
3
cos x =  
5
∴
2x2 – 4 = –3
4
tan x =  
5
∴
2x2 = 1
∴
x = ±
Trignometric Functions
)
x2 – 4 – x2 – 1
∴ x2 =
1
2
1
2
=1
then
Mahesh Tutorials Science
29
Q-7) Solve the equation
2 tan
–1
–1  1 
–1  1 
= tan   + tan  
2
 
2
–1
( cos x ) = tan ( 2cosecx )
Ans. 2 tan–1 ( cos x ) = tan–1 ( 2cosecx )
∴
tan
–1
1 1
 +2
 
–1  2 
= tan 
11 

 1– 22 
 

( cos x ) + tan ( cos x )
–1
–1
= tan ( 2 cosec x )
∴
 cos x + cos x 
tan–1 

 1 – cos x ( cos x ) 


 1 
= tan 

 1 – 1 

4
–1
–1
= tan ( 2cosec x )
∴
2 cos x
sin2 x
∴
cos x
= 1
sin x
∴
cot x = 1
∴
x=
=
2
sin x
–1  4 
= tan  
3
∴
4 1
 +7
 
–1  3 
= tan 
41 

 1– 37 
 

π
4
–1
–1
Q-8) If sin (1 – x ) – 2 sin x =
π
2
–1  28 + 3 
= tan 

 21 – 4 
then find the value of ‘x’.
Ans. sin –1 (1 – x ) – 2sin –1 x =
–1  31 
= tan 

 17 
π
2
Let x = sin θ
∴
∴
sin –1 (1 – sin θ ) – 2sin –1 ( sin θ ) =
∴
sin –1 (1 – sin θ ) – 2 θ =
∴
π

1 – sin θ = sin  + 2 θ 
2


∴
1 – sin θ = cos 2θ
∴
1 – sin θ = 1 – 2sin2 θ
∴
2sin2 θ – sin θ = 0
sin θ (2 sin θ – 1) = 0
∴
∴
∴
 4
1
x = tan–1   + tan–1  
3
7
π
2
1
1
 31 
2 tan–1   + tan–1   = tan–1  
2
7
 17 
π
2
sin θ = 0
x=0
Q-9) Prove that 2 tan –1
Ans. Let x = 2 tan –1
1
1
31
+ tan –1 = tan –1
2
7
17
1
1
+ tan –1
2
7
1
But 2 tan–1  
2
Trignometric Functions
Mahesh Tutorials Science
30
BASIC ASSIGNMENT (BA) :
∴
BA–1
 3π 
cot 4x = cot 

 4 
Now, cot θ = cot α ; θ = nπ ± α,n∈Z.
Q-1) Find the general solutions of the following
equations.
3x
=0
2
i)
sin
iii)
cosec x = – 2
1
ii)
cos x = 1
iv)
cot 4x = –1
v)
cos 3x =
vii)
cos 4x = cos 2x
2
vi)
∴
Thus, required general solution is
x=
sin 2x + sin 4x + sin 6x = 0
x)
cos x – sin x = 1
∴
∴
∴
π
cos 3x = cos  
4
∴
3x
=0
2
π
4
Thus, required general solution is
x=
2n π
π
, where n∈Z.
±
3
12
vi)
tan2 x = 1
Now, sin θ = sin α, θ = nπ
π + (–1)n α, n∈
∈Z.
∴
tan2 x = (1)2
3x
= nπ
π + n(–1)n .(0) = nπ
π
2
∴
π

tan2 x =  tan  = tan2
4

2
Now, tan2 x = tan2 α, x = nπ ± α,
α n∈Z.
2n π
, where n∈
∈Z.
3
ii)
cos x = 1
∴
cos x = cos 0
Thus, required general solution is
x = nπ ±
Now, cos θ = cos α, θ = 2nπ
π ± α,n∈
∈Z.
iii)
cosec x = – 2
∴
cosec x = – cosec
π
π

= cosec  π + 
4
4

cos 4x
∴
cos 4x – cos 2x = 0
∴
 4x – 2x 
 4x + 2x 
– 2 sin 
 =0
 . sin 
2
2




∴
–2 sin 3x . sin x = 0
∴
sin 3x = 0 or sin x = 0
∴
3x = nπ or x = mπ, where n,m∈Z.
= cos 2x
Thus, required general solution is
 5π 
= cosec 

 4 
 5π 
x = nπ + (–1)n 
 , where n∈Z.
 4 
π
, where n∈Z.
4
vii)
x = 2nπ
π±0
Thus required general solution is
x = 2nπ
∈Z.
π, where n∈
iv)
3x = 2nπ ±
 3x  = sin 0
sin 

 2 
x=
∴
2
Now, cos θ = cos α, θ = 2nπ ± α,n∈Z.
Thus, required general solution is
∴
1
cos 3x =
tan2 x = 1
ix)
sin
n π 3π
+
, where n∈Z.
4 16
v)
viii) tan3 x – 3 tan x = 0
Ans. i)
3π
4
4x = nπ +
x=
viii)
nπ
or x = mπ , where n, m∈Z.
3
tan3 x – 3 tan x = 0
tan x (tan2 x – 3) = 0
cot 4x = –1
2
∴
Trigonometric Functions
tan x = 0 or tan2 x = 3 =
( 3)
Mahesh Tutorials Science
∴
∴
31
π

x = nπ or tan2 x =  tan 
3

x = nπ or x = mπ ±
Now, cos θ = cos α, θ = 2nπ
π ± α,n∈
∈Z.
2
π
3
∴
x +
π
π
= 2nπ
π±
4
4
∴
x +
π
π
π
π
= 2nπ
π+
or x +
= 2nπ
π–
4
4
4
4
Thus, required general solution is
Thus, required general solution is
π
x = nπ or x = mπ ±
, where n, m∈Z.
3
ix)
sin 2x + sin 4x + sin 6x = 0
∴
(sin 2x + sin 6x) + sin 4x = 0
∴
 6x + 2x  . sin  6x – 2x 
2 sin 



2


2


+ sin 4x = 0
∴
2 sin 4x cos 2x + sin 4x = 0
∴
sin 4x (2 cos 2x +1) = 0
∴
sin 4x = 0 or cos 2x =
∴
2π
π

sin 4x = 0 or cos 2x = cos  π –  = cos
3
3

–1
2
x = 2nπ
π or x = 2nπ
π–
BA–2
Q-1) Find Cartesian co-ordinates of points
whose polar co-ordinates are
i)
1
0
 , 210 
2

ii)
3 3 3
 ,

2 
2
iii)
In ∆ ABC prove that
(a – b)2 cos2
Now, sin 4x = sin 0
∴
4x = nπ + (–1)n (0) = 2π
π
Ans. i)
∴
x=
nπ
4
∴
 2π 
Also, cos 2x = cos 

 3 
∴
2x = 2mπ
π±
∴
2x = mπ
π±
r =
2π
3
π
3
∴
x=
1
1
cos (2100) and y = sin(2100)
2
2
∴
x=
1– 3
– 3

 =
and
2  2 
4
y=
1  –1  –1
 =
2 2  4
Thus the Cartesian co-ordinates are
cos x – sin x = 1
cos x –
1
2
sin x =
 – 3 –1 
, 

4 
 4
2 , we get
Dividing throughout by
2
1
, θ = 2100
2
x = r cos θ and y = r sinθ
θ
nπ
π
x=
or x = mπ
π±
, where n, m∈
∈Z.
4
3
1
C
C
+ (a + b)2 sin2
= c2
2
2
1

Polar co-ordinates are  ,2100 
2

Thus, required general solution is
x)
π
, where n∈
∈Z.
2
1
2
∴
π
π
π
cos   cos x – sin   sin x = cos  
4
 4
 4
∴
π
π

cos  x +  = cos  
4
 4


ii)
3 3 3
(x, y) ≡  ,
 ...given
2
2


  1   3 
(x, y) ≡  3   ,3 
 
 2
 2 

Using x = r cos θ and y = r sin θ, where
(r, θ) are the required co-ordinates, we get
Trigonometric Functions
Mahesh Tutorials Science
32
r = 3, cos θ =
1
3
, sin θ =
2
2
∴
r = 3 and θ =
π
6
∴
 π
Polar co-ordinates are  3, 
 6
iii)
Consider R.H.S.

A –B
 A + B 
= k . 2sin 
 cos 

 2 
 2 


A +B 
 A – B 
2sin  2  cos  2  





=

A –B
 A – B 
2sin  2  cos  2  






A+B 
 A + B 
2sin  2  cos  2  





C
C
=(a – b)2 cos2
+ (a + b)2 sin2
2
2
= k . sin (A – B) sin (A + B)
C
=(a2 + b2 – 2ab)2 cos2 + (a2 + b2 + 2ab)2
2
...( ∵ 2 sin θ cos θ = sin 2θ)
= k sin (A – B) sin (π – C)
C
sin
2
2
...( ∵ A + B = π – C)
= k sin (A – B)sin C
C
C
=(a2 + b2) cos2 – 2ab cos2
+
2
2
= (k sin C) sin (A – B)
= c sin (A – B)
C
C
+ 2ab sin2
(a2 + b2) sin2
2
2
C
C

=(a2 + b2)  cos2 + sin2 
2
2

= R.H.S.
∴
a sin A – b sin B = c sin (A – B)
ii)
Consider L.H.S.
= ac cos B – bc cos A = (a2 – b2)
C

2 C
– sin2 
– 2ab  cos
2
2

(
 a 2 + c 2 – b2 
 b2 + c 2 – a 2 
= ac 
 – bc 

2ac
2bc




)
2
2
= a + b (1) – 2ab cos
(
2
2
..... ∵ cos θ – sin θ = cos 2θ
)
=
1 2
(a + c2 – b2 – b2 – c2 + a2)
2
2
2
 2

= a2 + b2 – 2ab  a + b – c 
2ab


=
1
× 2(a2 – b2)
2
...(by Cosine rule)
= a + b – a – b + c = c2
= (a2 – b2)
R.H.S.
2
2
2
2
2
= L.H.S.
∴
(a – b)2 cos2
C
C 2
+ (a + b)2 sin2
+c
2
2
∴
ac cos B – bc cos A = (a2 – b2)
iii)
Consider L.H.S.
= (b + c) cos A + (c + a) cos B
+ (a + b) cos C
Q-2) In any ∆ ABC prove the following
i)
a sin A – b sin B = c sin (A – B)
ii)
ac cos B – bc cos A = (a2 – b2)
iii)
(b + c) cos A + (c + a) cos B + (a + b)
cos C = (a + b + c)
Ans. i)
= b cos A + c cos A + c cos B + a cos B
+ a cos C + b cos C
= (b cos A + a cos B)
= c+b+a
Consider L.H.S.
= ( a + b + c)
= a sin A – b sin B
= (k sin A). sin A – (k sin B). sin B
= k (sin A – sin B) (sin A + sin B)
Trigonometric Functions
+ (c cos A + a cos C)
...(by projection rule)
= R.H.S.
∴
(b + c) cos A + (c + a) cos B + (a + b)
cos C = (a + b + c)
Mahesh Tutorials Science
π
then prove that
2
Q-3) In ∆ ABC if ∠C =
sin (A – B) =
33
∴
(c + a – b) tan
B
∆
=2
...(ii)
2
s
similary, we can prove that
a2 – b2
a2 + b2
(a + b – c) tan
Ans. In ∆ ABC , ∠C =
∴
π
2
C 2∆
=
...(iii)
2
s
from (i), (ii) and (iii), we get
sin c = 1
(b + c – a) tan
and A + B =
π
2
tan
π
π
– B and B = – A
2
2
∴
A=
∴
a 2 – b2
k 2 sin2 A – k 2 sin2 B
L.H.S. = 2
2 =
a +b
k 2 sin2 A + k 2 sin2 B
=
=
=
=
=
A
= ( c + a – b)
2
C
B
= (a + b – c) tan
2
2
Q-5) Show that ∆ ABC b cos2
Ans. Consider L.H.S.
sin2 A – sin2 B
sin2 A + sin2 B
=
b cos2
sin A. ( sin A ) – sin B . ( sin B )
sin2 A + sin2 B
C
B
+ c cos2
2
2
b (1 + cos C )
=
2
+
c (1 + cos B )
sin A cos B – sin B cos A
sin2 A + cos 2 A
1
b + b cos C + c + c cos B
2
=
sin (A – B) = R.H.S.
B
A
= (c + a – b) tan =
2
2
=
A
2
2
(b + c ) + a
=
C
2
Ans. Consider (b + c – a) tan
(b + c ) + (b cos C + c cos B )
=
Q-4) In ∆ ABC prove that
(a + b – c) tan
2
...  cos 2θ = 1 + cos θ 
2


sin ( A – B )
(b + c – a) tan
C
B
+ c cos2
= s.
2
2
∴
...(by projection rule)
2
s = R.H.S.
b cos2
C
B
+ c cos2
= s
2
2
BA–3
A
= [(b + c – a) – 2a] tan
2
Q-1) Find the principal value of
= (2s – 2a)
= 2(s – a)
=2
=2
( s – b )(s – c )
s (s – a )
( s – b )(s – c )
s (s – a )
(
)
i)
tan –1 – 3
ii)
cosec –1 – 2 + cot –1
iii)
24
Show that 2 sin–1 3 = tan–1
7
5
iv)
Show that tan–1
(
)
(s – a ) (s – b ) (s – c )
s
( s – a )(s – b ) (s – c )
s2
( 3)
1
1
+ tan–1 1 + tan–1
3
5
7
+ tan–1
π
1
=
4
8
Trigonometric Functions
Mahesh Tutorials Science
34
v)
Solve the equation
3
3
= tan–1   + tan–1  
4
 
4
1 – x  1
tan–1 
(tan–1x) for x > 0
=
1+ x  2
Ans. i)
Let x = tan
–1
3 3
4+4
  
–1  
= tan
33 

 1– 44 
 

(– 3)
∴
tan x = – 3
∴
tan x = – tan
π
= tan  – π 
4
 4
 6 


= tan  4 
 7 
 16 
–1
π  π π
∈ – , range for principal
4  2 2 
value of tan–1 x.
Hence, the required principal value of
As –
π
x is – .
3
ii)
(
)
cosec –1 – 2 + cot –1
( 3)
= tan–1  6 × 16 
4 7 
∴
3
2sin–1   = tan–1  24 
5
 
 7 
iv)
Consider L.H.S
But, cosec–1(–x) = –cosec–1x
∴
(
=
( 3) =
–cosec –1
( 2 ) + cot ( 3 )
–π π
–π
+
=
4 6
12
5
3
iii)
x
–1
1
1
1
1
+ tan–1 + tan–1 + tan–1
3
8
7
5
 1 1 
 5+7 
  
–1  
= tan
1 1

1 – 5 ×  7  
 

 1 1 
 3+8 
   
–1 
+ tan
1 1

1 – 3 ×  8  
 

3
Let x = sin–1  
5
  11  
  12  


  35  
  + tan–1   24  
= tan–1  
  23  
  34  
  24  
  35  




3
sin x =  
5
 11 
 6 
–1
= tan–1 
 + tan  23 


 17 
3
tan x =  
4
  6   11  
  17  +  23  

 

= tan–1 
6  11  

 1 – 17 ×  23  



4
∴
= tan–1
)
cosec –1 – 2 + cot –1
∴
3
x = tan–1  
 4
∴
3
3
sin–1   = tan–1  
5
 
 4
∴
3
3
2sin–1   = 2 tan–1  
5
 4
Trigonometric Functions
 138 +187 
 17 × 23 
= tan–1 

 391 – 66 
 17 × 23 
Mahesh Tutorials Science
35
 325 
= tan–1 

 325 
= tan–1(1) =
∴ tan–1
π
4
1 π
1
1
1
+ tan–1 + tan–1 + tan–1 =
8 4
3
7
5
v)
1 – x  1
–1
tan–1 
 = tan x
 1+ x  2
∴
tan–1(1) – tan–1 x =
∴
1
π
– tan–1 x = tan–1x
2
4
∴
π
– 2 tan–1x = tan–1x
2
1
tan–1x
2
π
= 3 tan–1x
2
∴
tan–1x =
∴
x = tan
∴
x =
π
6
π
6
1
3
Trigonometric Functions