NCERT Exemplar: Trigonometric Functions | Mathematics (Maths) Class 11 - Commerce PDF Download

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 cos³ A – 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

Long Answer Type
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 (θ + β) = b  .... (i)
cos (α - β) = cos [θ + α - θ – β) = cos [(θ + α) – (θ + β)] 
∴ cos (α - β)= cos (θ + α) cos (θ + β) + sin (θ + α) sin (θ + 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 (α - β) – 4ab cos (α - 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 (π + α)] – 2[cos⁶ α + sin⁶ (π - 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 2θ + b sin 2θ = 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
Squaring and adding, we get
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 0° = 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° ∴ 2θ = 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 1° < 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 (α + 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 (θ - φ) – sin² (θ + φ)
Cos 2θ cos 2φ + sin² (θ - φ) – sin² (θ + φ)
= cos 2θ cos 2φ + sin (θ - φ + θ + φ).sin (θ - φ - θ - φ)
[∵ sin² A –  sin² B =  sin (A +  B).sin  (A – B)]
= cos 2θ cos 2φ + sin 2θ.sin ( - 2φ)
= cos 2θ cos 2φ - sin 2θ 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 + sin² θ = 1
⇒ sin² θ = 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


∴  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
∴

[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 cos² 48° – 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 2θ +  a sin 2θ




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
= (sin²x + 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 – a<sup>2
∴    |sin x – cos x| =    

Hence, the value of the filler is</sup>

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 sin² x cos² x (sin² x + cos² x)]
= 3[1 + 4 sin² x 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).