Extra Practice Questions
Trigonometry (R.D. Sharma Level)
1. If 5 tan α = 4, show that 5 sin α − 3 cos α5 sin α + 2 cos α = 16.
Given: tan α = 4/5.
Divide the numerator and denominator of the expression by cos α:
LHS = (5 sin α/cos α) − (3 cos α/cos α)(5 sin α/cos α) + (2 cos α/cos α) = 5 tan α − 35 tan α + 2
Substitute the value of tan α:
LHS = 5(4/5) − 35(4/5) + 2 = 4 − 34 + 2 = 16 = RHS. (Proved)
2. If θ is an acute angle and tan θ = ab, find the value of a sin θ − b cos θa sin θ + b cos θ.
Divide numerator and denominator by cos θ:
Expression = a tan θ − ba tan θ + b
Substitute tan θ = a/b:
= a(a/b) − ba(a/b) + b = a2/b − ba2/b + b = (a2 − b2)/b(a2 + b2)/b = a2 − b2a2 + b2.
3. Find the value of x if 2 sin(3x) = √3.
Given: 2 sin(3x) = √3 => sin(3x) = √32.
We know that sin 60° = √32.
Therefore, 3x = 60°.
x = 60°3 = 20°.
4. Evaluate: 4(sin4 30° + cos4 60°) − 3(cos2 45° − sin2 90°).
Substitute the known values: sin 30° = 1/2, cos 60° = 1/2, cos 45° = 1/√2, sin 90° = 1.
= 4[ (1/2)4 + (1/2)4 ] – 3[ (1/√2)2 – (1)2 ]
= 4[ 1/16 + 1/16 ] – 3[ 1/2 – 1 ]
= 4[ 2/16 ] – 3[ -1/2 ]
= 4(1/8) + 3/2 = 1/2 + 3/2 = 4/2 = 2.
5. If A = 60° and B = 30°, verify that sin(A – B) = sin A cos B − cos A sin B.
LHS: sin(A – B) = sin(60° – 30°) = sin(30°) = 12.
RHS: sin A cos B – cos A sin B = sin 60° cos 30° – cos 60° sin 30°
= (√32)(√32) – (12)(12)
= 34 – 14 = 24 = 12.
Since LHS = RHS, the identity is verified.
… (Questions 6-20 follow)
6. Prove that (sin θ + cos θ)(tan θ + cot θ) = sec θ + cosec θ.
LHS = (sin θ + cos θ)(sin θcos θ + cos θsin θ)
= (sin θ + cos θ)(sin2θ + cos2θsin θ cos θ)
= (sin θ + cos θ)(1sin θ cos θ) = sin θ + cos θsin θ cos θ
= sin θsin θ cos θ + cos θsin θ cos θ = 1cos θ + 1sin θ = sec θ + cosec θ = RHS. (Proved)
7. Prove that tan A + sec A – 1tan A – sec A + 1 = 1 + sin Acos A.
In the numerator, replace 1 with (sec2A – tan2A):
LHS = (tan A + sec A) – (sec2A – tan2A)tan A – sec A + 1
= (sec A + tan A) – (sec A – tan A)(sec A + tan A)tan A – sec A + 1
Factor out (sec A + tan A):
= (sec A + tan A) [1 – (sec A – tan A)]tan A – sec A + 1 = (sec A + tan A) [1 – sec A + tan A]tan A – sec A + 1
= sec A + tan A = 1cos A + sin Acos A = 1 + sin Acos A = RHS. (Proved)
8. If cos θ + sin θ = √2 cos θ, show that cos θ – sin θ = √2 sin θ.
Given: sin θ = √2 cos θ – cos θ = cos θ(√2 – 1).
sin θ = cos θ(√2 – 1) × √2 + 1√2 + 1 = cos θ(2 – 1)√2 + 1
sin θ = cos θ√2 + 1
(√2 + 1)sin θ = cos θ => √2 sin θ + sin θ = cos θ
cos θ – sin θ = √2 sin θ. (Proved)
9. If tan A + sin A = m and tan A – sin A = n, prove that m2 – n2 = 4√(mn).
LHS: m2 – n2 = (m-n)(m+n)
= [(tan A + sin A) – (tan A – sin A)][(tan A + sin A) + (tan A – sin A)]
= (2 sin A)(2 tan A) = 4 tan A sin A.
RHS: 4√(mn) = 4√[(tan A + sin A)(tan A – sin A)]
= 4√(tan2A – sin2A) = 4√(sin2Acos2A – sin2A)
= 4√[sin2A(1cos2A – 1)] = 4√[sin2A(sec2A – 1)]
= 4√(sin2A tan2A) = 4 sin A tan A.
Since LHS = RHS, it is proved.
10. Prove that sin6θ + cos6θ = 1 – 3sin2θcos2θ.
LHS = (sin2θ)3 + (cos2θ)3
Using a3 + b3 = (a+b)(a2 – ab + b2):
= (sin2θ + cos2θ)(sin4θ – sin2θcos2θ + cos4θ)
= (1)[(sin2θ + cos2θ)2 – 2sin2θcos2θ – sin2θcos2θ]
= (1)2 – 3sin2θcos2θ = 1 – 3sin2θcos2θ = RHS. (Proved)
11. Prove: √(sec2θ + cosec2θ) = tan θ + cot θ.
LHS = √[(1 + tan2θ) + (1 + cot2θ)]
= √(tan2θ + cot2θ + 2)
Since 2 = 2 × tan θ × cot θ (as tanθcotθ=1):
= √(tan2θ + cot2θ + 2tanθcotθ)
= √[(tan θ + cot θ)2] = tan θ + cot θ = RHS. (Proved)
12. If x = a sec θ and y = b tan θ, eliminate θ.
We have sec θ = x/a and tan θ = y/b.
Using the identity sec2θ – tan2θ = 1:
(x/a)2 – (y/b)2 = 1
x2a2 – y2b2 = 1. This is the required relation without θ.
13. Find acute angles A and B, if sin(A + 2B) = √3/2 and cos(A + 4B) = 0.
sin(A + 2B) = √3/2 => A + 2B = 60° —(1)
cos(A + 4B) = 0 => A + 4B = 90° —(2)
Subtracting (1) from (2): (A + 4B) – (A + 2B) = 90° – 60°
2B = 30° => B = 15°.
Substitute B = 15° in (1): A + 2(15°) = 60° => A + 30° = 60° => A = 30°.
So, A = 30° and B = 15°.
14. Prove that 1 + cos θ – sin2θsin θ(1 + cos θ) = cot θ.
LHS = (1 – sin2θ) + cos θsin θ(1 + cos θ)
= cos2θ + cos θsin θ(1 + cos θ)
= cos θ(cos θ + 1)sin θ(1 + cos θ)
= cos θsin θ = cot θ = RHS. (Proved)
15. If x = h + a cos θ and y = k + b sin θ, eliminate θ.
From the equations, we get: cos θ = x – ha and sin θ = y – kb.
Using the identity sin2θ + cos2θ = 1:
(y – kb)2 + (x – ha)2 = 1.
(x – h)2a2 + (y – k)2b2 = 1.
16. If sec θ = x + 14x, prove that sec θ + tan θ = 2x or 12x.
We know tan2θ = sec2θ – 1.
tan2θ = (x + 14x)2 – 1 = x2 + 2(x)(14x) + 116x2 – 1
= x2 + 12 + 116x2 – 1 = x2 – 12 + 116x2 = (x – 14x)2.
So, tan θ = ± (x – 14x).
Case 1: tan θ = x – 14x. Then sec θ + tan θ = (x + 14x) + (x – 14x) = 2x.
Case 2: tan θ = -(x – 14x). Then sec θ + tan θ = (x + 14x) – (x – 14x) = 24x = 12x. (Proved)
17. Prove: cot A + cosec A – 1cot A – cosec A + 1 = 1 + cos Asin A.
This is a variation of Q7. We use 1 = cosec2A – cot2A in the numerator.
LHS = (cot A + cosec A) – (cosec2A – cot2A)cot A – cosec A + 1
= (cosec A + cot A) [1 – (cosec A – cot A)]cot A – cosec A + 1
= (cosec A + cot A) [1 – cosec A + cot A]1 – cosec A + cot A = cosec A + cot A.
= 1sin A + cos Asin A = 1 + cos Asin A = RHS. (Proved)
18. If cot θ = 1√3, find the value of 1 – cos2θ2 – sin2θ.
Given cot θ = 1/√3, so θ = 60°.
The expression is sin2θ2 – sin2θ.
Substitute θ = 60°. sin 60° = √3/2.
= (√3/2)22 – (√3/2)2 = 3/42 – 3/4 = 3/4(8-3)/4 = 3/45/4 = 35.
19. Prove that sin θcot θ + cosec θ = 2 + sin θcot θ – cosec θ.
Rearrange the equation: sin θcot θ + cosec θ – sin θcot θ – cosec θ = 2.
LHS = sin θ [1cot θ + cosec θ – 1cot θ – cosec θ]
= sin θ [(cot θ – cosec θ) – (cot θ + cosec θ)cot2θ – cosec2θ]
= sin θ [-2 cosec θ-1] = sin θ [2 cosec θ]
= sin θ [2 (1sin θ)] = 2 = RHS. (Proved)
20. If cosec θ – sin θ = a and sec θ – cos θ = b, prove that a2b2(a2 + b2 + 3) = 1.
a = 1sin θ – sin θ = 1-sin2θsin θ = cos2θsin θ.
b = 1cos θ – cos θ = 1-cos2θcos θ = sin2θcos θ.
ab = (cos2θsin θ)(sin2θcos θ) = sin θ cos θ.
LHS = (ab)2(a2 + b2 + 3) = (sinθcosθ)2[cos4θsin2θ + sin4θcos2θ + 3]
= (sinθcosθ)2[cos6θ + sin6θsin2θcos2θ + 3]
= cos6θ + sin6θ + 3sin2θcos2θ
= (cos2θ)3 + (sin2θ)3 + 3sin2θcos2θ(1)
= (cos2θ)3 + (sin2θ)3 + 3sin2θcos2θ(sin2θ+cos2θ)
This is the expansion of (a+b)3 = a3+b3+3ab(a+b). So it equals (sin2θ + cos2θ)3 = (1)3 = 1 = RHS. (Proved)
