NCERT Exemplar: Trigonometric Functions

Q.1. Prove that

Ans.

[Rationalizing the denominator]

R.H.S. Hence proved.

Q.2. If  then prove that

is also equal to y.

Ans.
Given that:

Hence proved.

Q.3. If m sin θ = n sin (θ + 2α), then prove that tan (θ + α) cot α =
[Hint: Express  and apply componendo and dividendo]
Ans.
Given that: m sin θ = n sin (θ + 2α)

Using componendo and dividendo theorem we get

Hence proved.

Q.4. If cos (α + β) =  and sin (α – β) = where α lie between 0 and  find the value of tan2α.
[Hint: Express tan 2 α as tan (α + β + α – β]
Ans.
Given that:

Now    tan 2α  = tan [α + β + α – β]
= tan [(α  + β) + (α  – β)]

Hence,

Q.5. If tan x =   then find the value of
Ans.
Given that: tan x =

Hence,

Q.6. Prove that cosθ cos  sin 7θ sin 8θ.
[Hint: Express L.H.S. =
Ans.
L.H.S.

= – sin 8θ sin (– 7θ) = sin 7θ sin 8θ     [∵ sin (-θ) = -sin θ]
L.H.S. = R.H.S. Hence proved.

Q.7. If  a cos θ + b sin θ = m and a sin θ – b cos θ = n, then show that a2 + b2 = m2 + n2
Ans.
Given that: a cos θ + b sin θ
= m and a sin θ  b cos θ = n
R.H.S. = m2 + n2 = (a cos θ + b sin θ)2 + (a sin θ – b cos θ)2
= a2 cos2 θ + b2 sin2 θ + 2ab sin θ cos θ + a2 sin2 θ + b2 cos2 θ  2ab sin θ cos θ
= a2 cos2 θ + b2  sin2 θ + a2  sin2 θ + b2  cos2 θ
= a2(cos2 θ + sin2 θ) + b2(sin2 θ + cos2 θ)
= a2.1 + b2.1 = a2 + b2  L.H.S.
L.H.S = R.H.S.
Hence    proved.

Q.8. Find the value of tan 22°30 ′ .
[Hint: Let θ = 45°, use
Ans.
Let 22°30’  ∴ θ =45°
tan 22°30’
Put θ = 45°

Hence, tan 22°30’ = √2 - 1

Q.9. Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A.
Ans.
L.H.S. sin 4A = sin (A + 3A)
= sin A cos 3A + cos A sin 3A
= sin A(4 cosA – 3 cos A) + cos A(3 sin A – 4 sin3 A)
= 4 sin A cos3 A – 3 sin A cos A + 3 sin A cos A – 4 cos A sin3 A
= 4 sin A cos3 A – 4 cos A sin3 A. R.H.S.
L.H.S. = R.H.S.
Hence proved.

Q.10. If tanθ + sinθ = m and tanθ – sinθ = n, then prove that m2 – n2 = 4sinθ tanθ
[Hint: m + n = 2tanθ, m – n = 2 sinθ, then use m2 – n2 = (m + n) (m – n)]
Ans.
Given that: tan θ  + sin θ = m and tan θ – sin θ = n
L.H.S. m2 – n2 = (m + n)(m – n)
= [(tan θ + sin θ) + (tan θ – sin θ)]. [(tan θ + sin θ) – (tan θ – sin θ)]
= (tan θ + sin θ + tan θ – sin θ).(tan θ + sin θ – tan θ + sin θ)
= 2 tan θ.2 sin θ= 4 sin θ tan θ. R.H.S.
L.H.S. = R.H.S. Hence proved.

Q.11. If tan (A + B) = p, tan (A – B) = q, then show that tan 2 A =
[Hint: Use 2A = (A + B) + (A – B)]
Ans.
Given that: tan (A + B) = p, tan (A – B) = q
tan 2A = tan (A + B + A – B) = tan [(A + B) + (A – B)]

Q.12. If cosα + cosβ = 0 = sinα + sinβ, then prove that cos 2α + cos 2β = – 2cos (α + β).
[Hint: (cosα + cosβ)2 – (sinα + sinβ)2 = 0]
Ans.
Given that: cos α + cos β = 0 ................(i)
and sin α + sin α = 0 ...............(ii)
From  (i) and (ii) we have
(cos α + cos β)2  (sinα + sin β)2 = 0
⇒(cos2 α + cos2 β  + 2 cos α cos  β) – (sin2 α + sin2  β + 2 sin α sin  β) = 0
⇒cos2 α + cos2  β + 2 cos α cos  β – sin2 α – sin2  β – 2 sin α sin  β = 0
⇒(cos2 α – sin2 α) + (cos2  β  – sin2  β) + 2(cos α cos  β – sin α sin  β) = 0
⇒cos2 α + cos2 β + 2 cos (α +  β) = 0
Hence, cos 2α + cos 2 β = – 2 cos (α +  β).
Hence proved.

Q.13. If   then show that

[Hint: Use Componendo and Dividendo].
Ans.
Given that:

(Using componendo and dividendo theorem)

Q.14. If tanθ =   then show that sinα + cosα = √2 cosθ.
[Hint: Express tanθ = tan
Ans.
Given that:
⇒

⇒   √2 cos θ = cos α + sin α
⇒ sin α + cos α = √2 cos θ.
Hence proved.

Q.15. If sinθ + cosθ = 1, then find the general value of θ.
Ans.
Given that: sin θ + cos θ  = 1
Dividing both sides by  we get
...(1)

Hence, the general values of θ are 2nπ,n∈Z
Alternate method:
From eqn. (i) we get

Hence, the general value of

Q.16. Find the most general value of θ satisfying the equation tanθ = –1 and

Ans.
Given that: tan θ = – 1 and cos θ =
tan θ = – 1

Hence, the general solution is

Q.17. If cotθ + tanθ = 2 cosecθ, then find the general value of θ.
Ans.
Given that: cot θ + tan θ = 2 cosec θ

⇒  2sin θ cos θ = sin θ
⇒ 2 sin θ cos θ  sin θ = 0
⇒ sin θ (2 cos θ  1) = 0
⇒ sin θ = 0 or 2 cos θ  1 = 0 or
⇒
⇒
Hence, the general values of θ is

Q.18. If 2sin2θ = 3cosθ, where 0 ≤ θ ≤ 2π, then find the value of θ.
Ans.
Given that: 2 sin2 θ = 3 cos θ
⇒ 2(1 – cos2 θ) = 3 cos θ
⇒ 2 – 2 cos2 θ – 3 cos θ = 0
⇒ 2 cos2 θ + 3 cos θ – 2 = 0
⇒ 2 cos2 θ+ 4 cos θ – cos θ – 2 = 0
⇒2 cos θ (cos θ + 2) – 1(cos θ + 2) = 0
⇒ (cos θ + 2) (2 cos θ – 1) = 0
⇒ cos θ + 2 = 0 or 2 cos θ – 1 = 0
⇒ cos θ  - 2 [- 1  cos θ  1]
2 cos θ  1 = 0

∴
and
Hence, the value of θ are

Q.19. If secx cos5x + 1 = 0, where 0 < x ≤ then find the value of x.
Ans.
Given that: sec x cos 5x + 1 = 0

⇒ cos 5x + cos x = 0

⇒ cos 3x . cos 2x = 0
⇒ cos 3x = 0 or cos 2x = 0

Hence, the values of x are

Q.20. If sin (θ + α) = a and sin (θ + β) = b, then prove that cos 2(α – β) – 4ab cos (α – β) = 1 – 2a2 – 2b2
[Hint: Express cos (α – β) = cos ((θ + α) – (θ + β))]
Ans.
Given that:
sin (θ + α) = a and sin  + β) = b  .... (i)
cos  - β) = cos  + α - θ  β) = cos [(θ + α)   + β)]
∴ cos  - β)= cos  + α) cos  + β) + sin  + α) sin  + β)

Now cos 2(α – β) – 4ab cos (α – β)
= 2 cos2 (α – β) – 1 – 4ab cos (α – β) [ cos 2θ = 2 cos2 θ – 1]

= 2a2 b2 + 2 - 2a2 - 2b2 + 2a2 b2 +

= 1  2a2  2b2
Hence, cos  2  - β)  4ab cos  - β) = 1  2a2  2b2.
Hence proved.

Q.21. If cos (θ + φ) = m cos (θ – φ), then prove that
[Hint: Express  and apply Componendo and Dividendo]
Ans.
Given that:
cos (θ + φ) = m cos (θ – φ)

Using componendo and dividendo theorem, we get

Hence proved.

Q.22. Find the value of the expression

Ans.
Given that:

= 3[ cos4 α + sin4  + α)]  2[cos6 α + sin6  - α)] = 3[cos4 α + sin4 α]  2[cos6 α + sin6 α]
= 3[cos4 α + sin4 α + 2 sin2 α cos2 α - 2 sin2 α cos2 α]  2[(cos2 α + sin2 α)3  3 cos2 α sin2 α (cos2 α + sin2 α)]
=  3[(cos2 α + sin2 α)2  2 sin2 α cos2 α]  2[1  3 cos2 α sin2 α]
= 3[1  2 sin2 α cos2 α]  2[1  3 cos2 α sin2 α]
= 3 – 6  sin2 α cos2 α - 2 + 6 cos2 α sin2 α
= 3  2 = 1
Hence, the value of the given expression is 1.

Q.23. If a cos 2θ + b sin 2θ = c has α and β as its roots, then prove that tanα + tan β =
[Hint: Use the identities cos 2θ =
Ans.
Given that: a cos  + b sin  = c..... (i)

a – a tan2θ + 2b tanθ = c(1 +  tan2 θ)
⇒ a  a tan2θ + 2b tan θ = c + c tan2 θ
⇒ a  a tan2θ + 2b tan θ  c tan2 θ  c = 0
⇒ - (a + c) tan2 θ + 2b tan θ + (a  c) = 0
(a + c) tan2 θ  2b tan θ + (c  a) = 0..... (ii)
Since α and β are the roots of this equation

Hence proved.

Q.24.  If x = sec φ – tan φ and y = cosec φ + cot φ then show that xy + x – y + 1 = 0
[Hint: Find xy + 1 and then show that x – y = – (xy + 1)]
Ans.
Given that: x = sec φ – tan φ
and y = cosec φ + cot φ
xy + x – y + 1 = 0
L.H.S.   xy + x – y + 1 = (sec φ – tan φ)
(cosec φ  + cot φ) + (sec φ – tan φ) – (cosec  φ + cot φ) + 1

R.H.S.
L.H.S. = R.H.S.
Hence proved.

Q.25. If θ lies in the first quadrant and cosθ =  then find the value of
cos (30° + θ) + cos (45° – θ) + cos (120° – θ).
Ans.
Given that:

But θ lies in I quadrant.
∴
Now cos (30° + θ) + cos (45° – θ) + cos (120° – θ)
= cos 30° cos θ – sin 30° sin θ + cos 45° cos θ + sin 45° sin θ + cos 120° cos θ + sin 120° sin θ

Hence, the required solution =

Q.26. Find the value of the expression
[Hint: Simplify the expression to

Ans.

Hence, the required value of the expression =

Q.27.Find the general solution of the equation
5cos2θ + 7sin2θ – 6 = 0
Ans.
5 cos2 θ + 7 sin2 θ  6 = 0
⇒ 5 cos2 θ + 7(1  cos2 θ)  6 = 0
⇒ 5 cos2 θ + 7– 7 cos2 θ  6 = 0
- 2 cos2 θ + 1 = 0
⇒ 2 cos2 θ = 1

∴
Hence, the general solution of

Q.28. Find the general solution of the equation
sinx – 3sin2x + sin3x = cosx – 3cos2x + cos3x
Ans.
Given that: sin x – 3 sin 2x + sin 3x = cos x – 3 cos 2x + cos 3x
⇒ (sin 3x + sin x) – 3 sin 2x = (cos 3x + cos x) – 3 cos 2x

⇒ 2 sin 2x.cos x – 3 sin 2x = 2 cos 2x .cos x – 3 cos 2x
⇒ 2 sin 2x cos x – 2 cos 2x.cos x = 3 sin 2x – 3 cos 2x
⇒ 2 cos x(sin 2x – cos 2x) = 3(sin 2x – cos 2x)
⇒ 2 cos x (sin 2x – cos 2x) – 3(sin 2x – cos 2x) = 0
⇒ (sin 2x – cos 2x) (2 cos x – 3) = 0
⇒ sin 2x – cos 2x = 0 and 2 cos x – 3 ≠0 [∵  -1 ≤ cos x ≤ 1]

Hence, the general solution of the equation is

Q.29. Find the general solution of the equation ( √3 – 1) cosθ + ( √3 + 1) sinθ = 2
[Hint: Put √3 – 1=  r sinα, √3 + 1 = r cosα which gives tanα = tan

Ans.
Given that: ( √3 - 1) cos θ + ( √3 + 1) sin θ = 2
Put √3 - 1 = r sin α, √3 + 1 = r cos α
r2 = √3 + 1 - 2 √3 + √3 + 1 + 2 √3
r2 = 8 r = ± 2 √2
Now the given equation can be written as
r sin α cos θ + r cos α sin θ = 2
⇒ r (sin α cos θ + cos α sin θ = 2
⇒ r (sin α cos θ + cos α sin θ) = 2
⇒2 √2 sin ( α + θ) = 2
⇒ sin (α + θ) =
sin (α+ θ) =

Now

Putting the value of α in equation (i) we get

Hence, the general solution of the given equation is

Objective Type Questions
Q.30. If sin θ + cosec θ = 2, then sin2 θ + cosec2 θ is equal to
(a) 1
(b) 4
(c) 2
(d) None of these
Ans. (c)
Solution.
Given that: sin θ + cosec θ = 2
Squaring both sides, we get
(sin θ +  cosec θ)2 = (2)2
⇒ sin2  θ +  cosec2 θ + 2 sin θ cosec θ = 4
⇒ sin2 θ + cosec2 θ + 2
⇒ sin2 θ + cosec2 θ + 2 = 4
⇒ sin2 θ + cosec2 θ = 2
Hence, the correct option is (c).

Q.31. If f (x) = cos2 x + sec2 x, then
(a) f (x) < 1
(b) f (x) = 1
(c) 2 < f (x) < 1
(d) f(x) ≥ 2
[Hint: A.M ≥ G.M.]
Ans. (d)
Solution.
Given that: f(x) = cos2 x + sec2 x
We know that AM ≥ GM

⇒ f(x)  2
Hence, the correct option is (d)

Q.32. If tan θ =  and tan φ = then the value of θ + φ is
(a) π/6
(b) π
(c) 0
(d) π/4
Ans. (d)
Solution.
We know that

Hence the correct option is (d).

Q.33. Which of the following is not correct?
(a)
(b) cos θ = 1
(c)
(d) tan θ = 20
Ans. (c)
Solution.
sin θ = is correct. ∵1  sin θ  1
So (a) is correct.
cos θ = 1 is correct. ∵ cos  = 1
So (b) is correct.

⇒ cos θ = 2 is not correct.
- 1  cos θ  1
Hence, (c) is not correct.

Q.34. The value of tan 1° tan 2° tan 3° ... tan 89° is
(a) 0
(b) 1
(c) 1/2
(d) Not defined
Ans. (b)
Solution.
Given that: tan 1° tan 2° tan 3° .... tan 89°
= tan 1° tan 2° tan 3° .... tan 45°.tan (90 – 44°).tan (90 – 43°) ...tan (90 – 1°)
= tan 1° cot 1°.tan 2°.cot 2°.tan 3°.cot 3° ... tan 89°.cot 89°
= 1.1.1.1 ... 1.1 = 1
Hence, the correct option is (b).

Q.35. The value of
(a) 1
(b) √3
(c) √3/2
(d) 2
Ans. (c)
Solution.
Given that:
Let θ = 15°   = 30°

Hence, the correct option is (c).

Q.36. The value of cos 1° cos 2° cos 3° ... cos 179° is
(a) 1/√2
(b) 0
(c) 1
(d) –1

Ans. (b)
Solution.
Given expression is cos 1°.cos 2°.cos 3° ... cos 179°
⇒cos 1°.cos 2°.cos 3° ... cos 90°.cos 91° ... cos 179°
⇒ 0 [∵ cos 90° = 0]
Hence, the correct option is (b).

Q.37. If tan θ = 3 and θ lies in third quadrant, then the value of sin θ is
(a)
(b)
(c)
(d)
Ans. (c)
Solution.
tan θ =  3,  θ lies in third quadrant, it is positive.
where θ lies in third quadrant
Hence the correct option is (c).

Q.38. The value of tan 75° – cot 75° is equal to
(a) 2√3
(b) 2 + √3
(c) 2 − √3
(d) 1
Ans. (a)
Solution.
The given expression is tan 75° – cot 75°
tan 75° – cot 75° = tan 75° – cot (90 – 15°)
= tan 75° – tan 15° =

Hence, the correct option is (a).

Q.39. Which of the following is correct?
(a) sin1° > sin 1
(b) sin 1° < sin 1
(c) sin 1° = sin 1
(d)
Ans. (b)
Solution.
We know that if  θ  increases  then the value of  sin θ  also increases
So, sin  < sin 1
Hence the correct option is (b).

Q.40. If tan α =  then α + β is equal to
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
Given that

⇒ tan  + β)
Hence, the correct option is (d).

Q.41. The minimum value of 3 cosx + 4 sinx + 8 is

(a) 5
(b) 9
(c) 7
(d) 3
Ans. (d)
Solution.
The given expression is 3 cos x + 4 sin x + 8
Let y = 3 cos x + 4 sin x + 8
⇒ y  8 = 3 cos x  + 4 sin x
Minimum value of y  8
⇒ y – 8 =
⇒ y = 8 – 5 = 3
So, the minimum value of the given expression is 3.
Hence, the correct option is (d).

Q.42. The value of tan 3A – tan 2A – tan A is equal to
(a) tan 3A tan 2A tan A
(b) – tan 3A tan 2A tan A
(c) tan A tan 2A – tan 2A tan 3A – tan 3A tan A
(d) None of these
Ans. (a)
Solution.
The given expression is tan 3A – tan 2A – tan A
tan 3A = tan (2A + A)

⇒ tan 3A(1 – tan 2A tan A) = tan 2A + tan A
⇒ tan 3A – tan 3A tan 2A tan A = tan 2A + tan A
⇒ tan 3A – tan 2A – tan A = tan 3A tan 2A tan A
Hence, the correct option is (a).

Q.43.  The value of sin (45° + θ) – cos (45° – θ) is
(a) 2 cosθ
(b) 2 sinθ
(c) 1
(d) 0
Ans. (d)
Solution.
Given expression is (sin 45° + θ)  cos  (45°- θ)
Sin (45° + θ) = sin 45° cos θ + cos 45° sin θ

Cos (45° - θ) = cos 45° cos θ + sin 45° sin θ

Sin (45°+ θ - cos (45°- θ)

= 0.
Hence, the correct option is (d).

Q.44. The value of
(a) 2 cosθ
(b) 2 sinθ
(c) 1
(d) 0
Ans. (c)
Solution.

Hence, the correct option is (c).

Q.45. cos 2θ cos 2φ + sin2 (θ – φ) – sin2 (θ + φ) is equal to
(a) sin 2(θ + φ)
(b) cos 2(θ + φ)
(c) sin 2(θ – φ)
(d) cos 2(θ – φ)
[Hint: Use sin2 A – sin2 B = sin (A + B) sin (A – B)]
Ans. (b)
Solution.
Given that: cos 2θ.cos 2 φ + sin2  - φ)  sin2 + φ)
Cos  cos 2φ + sin2  - φ)  sin2  + φ)
= cos  cos 2φ + sin  - φ + θ + φ).sin  - φ - θ - φ)
[∵ sin2 A   sin2 B =  sin (A +  B).sin  (A  B)]
= cos  cos 2φ + sin 2θ.sin ( - 2φ)
= cos  cos 2φ - sin  sin 2φ   [  sin(φ - θ) = - sin θ]
=    cos 2(θ + φ)
Hence, the correct option is (b).

Q.46. The value of cos 12° + cos 84° + cos 156° + cos 132° is
(a)
(b) 1
(c)
(d)
Ans. (c)
Solution.
The given expression is cos 12° + cos 84° + cos 156° + cos 132° (cos 132° + cos 12°) + (cos 156° + cos 84°)

= 2 cos 72°.cos 60° + 2 cos 120°.cos 36°

= cos 72° – cos 36°
= cos (90° – 18°) – cos 36° = sin 18° – cos 36°

Hence, the correct option is (c).

Q.47. If tan A =  tan B = then tan (2A + B) is equal to
(a) 1
(b) 2
(c) 3
(d) 4
Ans. (c)
Solution.
Given that: tan A =  and tan B =
tan 2A =

So, tan 2A =  and tan B =
tan (2A + B) =

Hence, the correct option is (c).

Q.48. The value of  is
(a)
(b)
(c)
(d) 1
Ans. (c)
Solution.

= – sin 18°.sin 54°
= – sin 18°.sin (90° – 36°) = – sin 18°.cos 36°

Hence, the correct option is (c).

Q.49. The value of sin 50° – sin 70° + sin 10° is equal to
(a) 1
(b) 0
(c) 1/2
(d) 2
Ans. (b)
Solution.
Given expression is sin 50° – sin 70° + sin 10°
(sin 50° – sin 70°) + sin 10°=
= 2 cos 60°.(– sin 10°) + sin 10°

= – sin 10° + sin 10°
= 0
Hence, the correct option is (b).

Q.50. If sin θ + cos θ = 1, then the value of sin 2θ is equal to

(a) 1
(b) 1/2
(c) 0
(d) –1
Ans. (c)
Solution.
Given that:
sin θ + cos θ = 1
Squaring both sides, we get,
⇒ (sin θ + cos θ)2 = (1)2
⇒ sin2 θ + cos2 θ + 2 sin θ cos θ = 1
⇒ 1 + sinθ = 1
⇒ sinθ = 1  1 = 0
Hence, the correct option is (c).

Q.51. If α + β = then the value of (1 + tan α) (1 + tan β) is
(a) 1
(b) 2
(c) – 2
(d) Not defined
Ans. (b)
Solution.
Given that:

⇒ tan α + tan β = 1– tan α tan β
⇒ tan α + tan β + tan α tan β = 1
On adding 1 both sides, we get,
⇒ 1 + tan α + tan β + tan α tan β = 1 + 1
⇒ 1(1 + tan α) + tan β (1 + tan α) = 2
⇒ (1 + tan α )(1 + tan β) = 2
Hence, the correct option is (b)

Q.52. If sin θ =  and θ lies in third quadrant then the value of

is
(a)
(b)
(c)
(d)
Ans. (c)
Solution.
Given that:  lies in third quadrant

Hence, the correct option is (c).

Q.53. Number of solutions of the equation tan x + sec x = 2 cosx lying in the interval [0, 2π] is
(a) 0
(b) 1
(c) 2
(d) 3

Ans. (c)
Solution.
Given equation is tan x + sec x = 2 cos x

⇒ 1 + sin x = 2 cos2 x
⇒ 2 cos2 x – sin x – 1 = 0
2(1 – sin2 x) – sin x – 1 = 0
⇒  2 – 2 sin2 x – sin x – 1 = 0
⇒ – 2 sin2 x – sin x + 1 = 0
⇒  sin2 x + sin x – 1 = 0
Since, the equation is a quadratic equation in sin x.
So it will have 2 solutions.
Hence, the correct option is (c).

Q.54. The value of is given by
(a)
(b) 1
(c)
(d)
Ans. (a)
Solution.
The given expression is

Hence, the correct option is (a).

Q.55. If A lies in the second quadrant and 3 tan A + 4 = 0, then the value of 2 cot A  –  5 cos A + sin A is equal to
(a)
(b)
(c)
(d)
Ans. (b)
Solution.
Given that: 3 tan A + 4 = 0, A lies in second quadrant

and
∴ 2 cot A – 5 cos A + sin A

Hence, the correct option is (b).

Q.56. The value of cos2 48° – sin2 12° is
(a)
(b)
(c)
(d)
[Hint:  Use cos2 A – sin2 B = cos (A + B) cos (A – B)]
Ans. (a)
Solution.
Given expression is cos48° – sin2 12°
cos2 48° – sin2 12° = cos (48° + 12°).cos (48° – 12°)
[∵ cos2 A – sin2 B = cos (A + B).cos (A – B)]
= cos 60°.cos 36°

Hence, the correct option is (a).

Q.57. If tan α =  tan β = then cos 2α is equal to
(a) sin 2β
(b) sin 4β
(c) sin 3β
(d) cos 2β
Ans. (b)
Solution.
Given that:

Now

cos 2α = sin 4β =
Hence, the correct option is (b).

Q.58. If tan θ =   then b cos 2θ  + a sin 2θ  is equal to
(a) a
(b) b
(c) a/b

(d) None
Ans. (b)
Solution.
Given that:
b cos  +  a sin

Hence, the correct option is (b).

Q.59. If for real values of x, cos θ = then
(a) θ is an acute angle
(b) θ is right angle
(c) θ is an obtuse angle
(d) No value of θ is possible
Ans. (d)
Solution.
Given that:

⇒ x2 + 1 =  x cos θ
x2  x cos θ + 1 = 0
For real value of x, b2  4ac  0
⇒ (- cos θ)2  4 × 1 × 1  0
⇒ cos2 θ  4  0
cos2 θ  4
⇒ cos θ  2 [- 1  cos θ  1]
So, the value of θ is not possible.
Hence, the correct options (d).

Fill in the blanks

Q.60. The value of is _______ .
Ans.

Hence, the value of filler is 1.

Q.61. If k = then the numerical value of k is _______.
Ans.
Given that: k =
⇒ k = sin 10°. Sin 50°. Sin 70°
⇒ k = sin 10° sin (90° -  40°) sin (90° - 20°)
⇒ k = sin 10° cos 40° cos 20°

Hence, the value of the filler is

Q.62. If tan A =, then tan 2A = _______.
Ans.
Given that:  tan A =

So, tan 2A = tan B
Hence, the value of the filler is tan B.

Q.63. If sin x + cos x = a, then
(i) sin6 x + cos6 x = _______
(ii) | sin x – cos x | = _______.
Ans.
Given that: sin x + cos x = a
Squaring both sides, we get,
(sin x + cos x)2 = a2
⇒ sin2 x + cos2 x + 2 sin x cos x = a2
⇒ 1 + 2 sin x cos x = a2
⇒ sin x cos x =  ...(i)
(i) sin6 x + cos6 x = (sin2 x)3 + (cos2 x)3
= (sin2 x + cos2 x)3 – 3 sin2 x cos2 x(sin2 x + cos2 x)

Hence, the value of the filler is
(ii) |sin x  cos x|2 = sin2 x + cos2 x   2 sin x cos x

= 1 – (a2 – 1)
= 1 – a2 + 1
= 2  a2
|sin x  cos x| =

Hence, the value of the filler is

Q.64. In a triangle ABC with ∠C = 90° the equation whose roots are tan A and tan B is _______.
[Hint: A + B = 90° ⇒ tan A tan B = 1 and tan A + tan B =
Ans.
Given a ΔABC with ∠C = 90°
x2    (tan A + tan B)x + tan A.tan B = 0
A + B = 90°  [∵  ∠C = 90°]
⇒ tan (A + B) = tan 90°

1 – tan A tan B = 0
tan A tan B = 1……(i)
Now  tan A + tan B =

∴ tan A + tan B =……(ii)
From (i) and (ii) we get

Hence, the value of the filler is

Q.65. 3 (sin x – cos x)4 + 6 (sin x + cos x)2 + 4 (sin6 x + cos6 x) = ______.
Ans.
Given expression is 3(sin x – cos x)4 + 6(sin x + cos x)2 + 4(sin6 x + cos6 x)
= 3[sin2 x + cos2 x – 2 sin x cos x]2 + 6(sin2 x + cos2 x + 2 sin x cos x) + 4 [(sin2 x)3 + (cos2 x)3]
= 3[1 – 2 sin x cos x]2 + 6(1 + 2 sin x cos x) + 4[(sin2 x + cos2 x)3 – 3 sinx cos2 x (sin2 x + cos2 x)]
= 3[1 + 4 sinx cos2 x – 4 sin x cos x] + 6(1 + 2 sin x cos x) + 4[1 – 3 sin2 x cos2 x]
= 3 + 12 sin2 x cos2 x – 12 sin x cos x + 6 + 12 sin x cos x + 4 – 12 sin2 x cos2 x
= 3 + 6 + 4 = 13
Hence, the value of the filler is 13.

Q.66. Given x > 0, the values of f (x) = – 3 cos  lie in the interval _______.
Ans.
Given that: f(x) =
Put
f(x) = - 3 cos y
- 1  cos y  1
3  - 3 cos y  -3
⇒ - 3  - 3 cos y  3
- 3  - 3 cos ≤ , x > 0
Hence, the value of the filler is [– 3, 3].

Q.67. The maximum distance of a point on the graph of the function y = √3 sin x + cos x from x-axis is _______.
Ans.
Given that y = √3 sin x + cos x ...(i)
∴ The maximum distance from a point on the graph of eqn. (i) from x-axis

Hence, the value of the filler is 2.

state whether the statements is True or False? Also give justification.
Q.68. If tan A = then tan 2A = tan B
Ans.
Given that: tan A
tan 2A =
∴ tan 2A = tan B
Hence, the statement is ‘True’.

Q.69. The equality sin A + sin 2A + sin 3A = 3 holds for some real value of A.
Ans.
Given that: sin A + sin 2A + sin 3A = 3
Since the maximum value of sin A is 1 but for sin 2A
and sin 3A it is not equal to 1. So it is not possible.
Hence, the given statement is ‘False’.

Q.70. sin 10° is greater than cos 10°.

Ans.
If sin 10° > cos 10°
⇒ sin 10° > cos (90° – 80°)
⇒sin 10° > sin 80°
which is not possible.
Hence, the statement is ‘False’.

Q.71.
Ans.
L.H.S.
= cos 24°.cos 48°.cos 96°.cos 192°
[(2 sin 24 cos 24 )(2 cos 48 )(2 cos 96 )(2 cos 192 )]
[sin 48 .2 cos 48 (2 cos 96 )(2 cos 192 )]
[2 sin 48 cos 48 (2 cos 96 )(2 cos 192 )]
[sin 96 (2 cos 96 )(2 cos 192 )]
[2 sin 96 . cos 96 (2 cos 192 )]
[sin 192 .(2 cos 192 )]
2 sin 192 cos 192

[∵ sin (360° + θ) = sin θ]
R.H.S.
Hence, the given statement is ‘True’.

Q.72. One value of θ which satisfies the equation sin4 θ – 2sin2 θ – 1 lies between 0 and 2π.
Ans. Given equation is
sin4 θ – 2 sin2 θ – 1 = 0

= 1 ± √2
∴ sin2 θ = (1 + √2 ) or (1 - √2 )
⇒ - 1 ≤ sin q ≤ 1
⇒ sin2 θ ≤  1 but sin2 θ = (1 + √2 ) or (1 - √2 )
Which is not possible.
Hence, the given statement is ‘False’.

Q.73. If cosec x = 1 + cot x then x = 2nπ, 2nπ +
Ans.
Given that: cosec x = 1 + cot x

⇒ sin x + cos x = 1

or
Hence, the given statement is ‘True’.

Q.74. If tan θ + tan 2θ + √3 tan θ tan 2θ = √3 , then
Ans.
Given that:tan θ + tan 2θ + √3 tan θ tan 2θ = √3
⇒ tan θ + tan 2θ = √3 - √3 tan θ tan 2θ
⇒ tan θ + tan 2θ = √3 (1 - tan θ tan 2θ)

⇒ tan (θ + 2θ) = √3

So
Hence, the given statement is ‘True’.

Q.75. If tan (π cosθ) = cot (π sinθ), then
Ans.
Given that: tan (π cos θ) = cot (π sin θ)

Hence, the given statement is ‘True’.

Q.76. In the following match each item given under the column C1 to its correct answer given under the column C2 :

 (a) sin (x + y) sin (x – y) (i) cos2 x – sin2 y (b) cos (x + y) cos (x – y) (iv) sin2 x – sin2 y

Ans.
(a) sin (x + y) sin (x  y) = sin2 x  sin2 y
(a)↔(iv)
(b) cos (x + y) cos (x  y) = cos2 x  cos2 y
(b)  (i)
(c)

(c)  (ii)
(d)
(d)  (iii)
Hence,
(a)  (iv)
(b) ↔ (i)
(c) ↔ (ii)
(d) ↔ (iii).

The document NCERT Exemplar: Trigonometric Functions | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

## FAQs on NCERT Exemplar: Trigonometric Functions - Mathematics (Maths) for JEE Main & Advanced

 1. What are trigonometric functions?
Ans. Trigonometric functions are mathematical functions that relate the angles of a triangle to the lengths of its sides. They include sine, cosine, tangent, cosecant, secant, and cotangent.
 2. How are trigonometric functions used in real life?
Ans. Trigonometric functions are used in various real-life applications such as engineering, architecture, navigation, physics, and computer graphics. They help in calculating distances, angles, heights, and solving problems related to waves, vibrations, and periodic phenomena.
 3. What is the unit circle in trigonometry?
Ans. The unit circle is a circle with a radius of 1 unit, centered at the origin of a coordinate system. It is widely used in trigonometry to define the values of trigonometric functions for any angle. The coordinates on the unit circle correspond to the cosine and sine values of the angles.
 4. How do trigonometric functions relate to right triangles?
Ans. Trigonometric functions are primarily defined and used in right triangles. In a right triangle, the sine of an angle is the ratio of the length of the side opposite the angle to the hypotenuse. The cosine is the ratio of the length of the adjacent side to the hypotenuse, and the tangent is the ratio of the opposite side to the adjacent side.
 5. How are trigonometric functions related to periodicity?
Ans. Trigonometric functions are periodic functions, meaning they repeat their values after a certain interval. For example, the sine and cosine functions have a period of 2π radians or 360 degrees. This periodicity is essential in analyzing and modeling periodic phenomena like waves, sound, and oscillations.

## Mathematics (Maths) for JEE Main & Advanced

209 videos|443 docs|143 tests

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