Class: X
Subject: Math’s
Topic: Trigonometry
No. of Questions: 20
If sin x= (a2 – b2)/(a2 + b2), find tan x.
Sol:
sin x= (a2 – b2)/(a2 + b2) = P/H
H2= P2 + B2
(a2 + b2)= (a2 – b2) + B2
a4 + b4 + 2a2b2= a4 + b4 – 2a2b2 + B2
B= 2ab
tan x= P/B= (a2 – b2)/2ab
Q.2
If cosec x=2, find the value of 1/tan x + sin x/(1+ cos x)
Sol:
cosec x= H/P = 2
H2= B2 + P2
4= B2 + 1
4 – 1= B2
B= 3
Q.3
Sol:
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tan x= P/B= 1/ 3
sin x= P/H= 1/2
cos x= B/H= 3 /2
1/ tan x + sin x/ (1+ cos x)= 2
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Q.1
In ∆ABC, right angled at C, if tan A= 1/ 3 , find the value of sin A cos B+ cos A sin B.
(CBSE-2008)
tan A= 1/ 3 =P/B
P2 + B2= H2
1 + ( 3 )2= H2
sin A= 1/2 , cos A= 3 /2 , cos B= 1/2, sin B= 3 /2
sin A cos B + cos A sin B= 1
Q.4
If sin B=1/2, find the value of 3 cos B – 4 cos2 B
Sol:
sin B=1/2 = P/H
H2 = P2 + B2
4= 1 + B2
B= 3
cos B= 3 /2= B/H
3 cos B – 4 cos2B= 0
4 tan x  5 cos x
sec x  4 cot x
If sin x= 4/5, find the value of
Sol:
sin x = 4/5
H2= P2 + B2
25= 16 + B2
B=3
cos x=B/H= 3/5, tan x= P/B= 4/3, cot x=B/P=3/4, sec x= H/B=5/3
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Q.5
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4 tan x  5 cos x
= 1/2
sec x  4 cot x
Given 16 cot A=12, find the value of
Sol:
sin A  cos A
sin A  cos A
sin A  cos A
sin A  cos A
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Q.6
=
1  cot A
1  cot A
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Dividing the numerator and denominator by sin A,
Now, 16 cot A=12
cot A= 12/16= 3/4
sin A  cos A
=7
sin A  cos A
Q.7
If tan x= 12/13, evaluate (2sinx cos x)/ (cos2x – sin2x)
Sol:
tan x= 12/13
(2sinx cos x)/ (cos2x – sin2x)
Dividing the numerator and denominator by cos2x
= 2tanx/(1 – tan2x)= 24/13 * 169/25= 312/25
Q.8
Find the value of 2(cos2450 + tan2600) – 6(sin2450 – tan2300)
Sol:
cos 450= 1/ 2 , tan 600=
3 , sin 450= 1/ 2 , tan 300= 1/ 3
2(cos2450 + tan2600) – 6(sin2450 – tan2300)= 2[(1/ 2 )2 + ( 3 )2] – 6[(1/ 2 )2 – (1/ 3 )2]
=6
Q.9
Find the value of x in tan3x = sin 450 cos 450 + sin 300
Sol:
tan3x = sin 450 cos 450 + sin 300
ns
tan3x= 1/ 2 * 1/ 2 + 1/2=1
tan3x= tan 450
3x=450
x=150
Q.10
Find the acute angle x. when (cos x- sin x)/(cos x + sin x)=(1 -
Sol:
(cos x- sin x)/(cos x + sin x)= (1 - 3 )/(1 +
Dividing LHS by cos x
(1 – tan x)/ (1 + tan x)= 1 - 3 )/(1 + 3 )
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tan x=
x= 600
If sin(A + B)=1 and cos(A – B)=
: sin(A + B)= 1
sin(A + B)= sin 450
cos(A – B)= 3 /2
cos(A – B)= cos 300
A + B=900
A – B=450
Add both the above equations
2A= 1200
A=600
600 + B= 900
B=300
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Q.11
Sol:
3)
3 )/(1 + 3 )
3 /2, 00 < A+B≤900, A>B then find A and B.
Q.12
If x is an acute angle and sin x= cos x, find the value of 2tan2x + sin2x -1
Sol:
2tan2x + sin2x -1
sin x= cos x (given)
sin x/cos x= cos x/ cos x
2tan2x + sin2x -1= 2 + 1/2 – 1= 3/2
Q.13
Given that sin (A+B)= sin A cos b + cos A sin B, find the value of sin 750.
Sol:
750 = 450 + 300
sin(300 + 450)= sin 300 cos 450 + cos 300 sin 450
sin 750= 1/2 * 1/ 2 +
3 /2 * 1/ 2 = ( 3 + 1)/ 2 2
If each of α, β and γ is appositive acute angle such that sin(α + β + γ)=1/2 , cos(β + γ – α)=
1/2 and tan(γ + α – β)=1, find the values of α, β and γ.
Sol:
sin(α + β + γ)=1/2
sin(α + β + γ)=sin 300
α + β + γ= 300
cos(β + γ – α)= 1/2
cos(β + γ – α)= cos600
β + γ – α= 600
tan(γ + α – β)=1
tan(γ + α – β)=tan 450
γ + α – β=450
add the 3 equations
2β= 900 and 2α=750
β=450 and α=37.50
γ= 52.50
Q.15
Evaluate cos(400 – x) – sin(500 + x)+ (cos2400 + cos2500)/(sin2400 + sin2500)
Sol:
cos(400 – x) – sin(500 + x)+ (cos2400 + cos2500)/(sin2400 + sin2500)
cos(400 – x) – sin(90-(400 – x))+ [cos2400 + cos2(900 - 400)]/[sin2400 + sin2(900 - 400)]
cos(400 – x) – cos (400 - x)+ (cos2400 + sin2400)/(sin2400 + cos2400)
= 0 + 1=1
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Q.14
(CBSE-2002)
Q.16
Evaluate cot 120 cot 380 cot 520 cot 600 cot 780
Sol:
cot 120 cot 380 cot 520 cot 600 cot 780
= (cot 120 cot 780) (cot380 cot 520) (cot600)
=(cot 120 tan120) (cot380 tan 380) (cot600)
(CBSE-2001)
= 1 * 1/ 3 = 1/ 3
Evaluate tan 10 tan 20 tan 30 ……. tan 890.
Sol:
tan 10 tan 20 tan 30 ……. tan 890
= tan(900 – 890) tan(900 – 880)…… tan 880 tan 890
=cot 890 cot 880 ……. tan880 tan 890.
= (cot 890 tan 890) ……. (cot 440 tan 440)tan450
= 1 * 1 * 1 * 1…… * 1= 1
Q.18
If sec 4A= cosec (A – 200), where 4A is an acute angle, find the value of A.
(CBSE-2008)
Sol:
sec 4A= cosec(A-200)
= sec[900 – (A – 200)]
4A= 1100 – A
5A= 1100
A=220.
Q.19
Evaluate cot2 x- 1/sin2x
Sol:
cot2x – 1/sin2x= cot2x – cosec2x= -1
Q.20
(1 + tan2x)(1 + sin x)(1 – sin x), find the value.
Sol:
(1 + tan2x)(1 + sin x)(1 – sin x)
=(1 + tan2x)(1 - sin2 x) =sec2 x * cos2 x=1
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Q.17