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Class 11 Maths MCQ Questions of Trigonometric Functions with Answers are prepared for college students of class 11 to assist them score better marks within the examination. the entire topic of trigonometric functions is meant by the topic experts following the newest guidelines of CBSE. 

The class 11 maths trigonometric functions important MCQ questions feature the step-by-step solutions for straightforward to difficult questions consistent with the understanding level of the students. These MCQ Questions can also help students to fast revision, because the students solve these important questions, they’re going to easily develop a command over the subject of trigonometric functions.

Practice MCQ Questions for class 11 Maths Chapter-Wise

1. The value of cos²x + cos² y – 2cosx × cosy × cos (x + y) is

(a) sin (x + y)
(b) sin² (x + y)
(c) sin³ (x + y)
(d) sin⁴ (x + y)

2. If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is

(a) a² + b² + c²
(b) a² – b² – c²
(c) a² – b² + c²
(d) a² + b² – c²

3. The value of cos 5π is

(a) 0
(b) 1
(c) -1
(d) None of these

4. In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals

(a) none of these
(b) c/a
(c) 1
(d) a/c

5. The value of cos 180° is

(a) 0
(b) 1
(c) -1
(d) infinite  

6. The perimeter of a triangle ABC is 6 times the arithmetic mean of the sines of its angles. If the side b is 2, then the angle B is

(a) 30°
(b) 90°
(c) 60°
(d) 120°

7. If 3 × tan(x – 15) = tan(x + 15), then the value of x is

(a) 30
(b) 45
(c) 60
(d) 90

8. If the sides of a triangle are 13, 7, 8 the greatest angle of the triangle is

(a) π/3
(b) π/2
(c) 2π/3
(d) 3π/2

9. The value of tan 20 × tan 40 × tan 80 is

(a) tan 30
(b) tan 60
(c) 2 tan 30
(d) 2 tan 60

10.  The value of cos 20 + 2sin² 55 – \(\sqrt2\;sin65\)

(a) 0
(b) 1
(c) -1
(d) None of these

11. If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is

(a) 2π/3
(b) π/3
(c) π/2
(d) π/6

12. The value of 4 × sinx × sin(x + π/3) × sin(x + 2π/3) is

(a) sin x
(b) sin 2x
(c) sin 3x
(d) sin 4x

13. The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is

(a) tan 6x
(b) 2 tan 6x
(c) 3 tan 6x
(d) 4 tan 6x

14. What is the value of cot (– 870°)?     

(a) \(\sqrt3\)
(b) \(\frac{1}{\sqrt3}\)
(c) \(-\sqrt3\)
(d) \(-\frac{1}{\sqrt3}\)

15. The value of cos2 48° - sin2 12° is

(a) \(\frac{\sqrt{5}+1}{8}\)
(b) \(\frac{\sqrt 5-1}{8}\)
(c) \(\frac{\sqrt{5+1}}{5}\)
(d) \(\frac{\sqrt{5}+\frac{1}{2}}{\sqrt2}\)

16. The value of tan 20° + 2 tan 50° – tan 70° is :

(a) 1
(b) 0
(c) tan 50°
(d) none of these

17. The value of tan 20° + 2 tan 50° – tan 70° is :

(a) \(\frac{1}{\sqrt2}\)
(b) 0
(c) 1
(d) -1

18.  If A = sin2 x + cos4x, then for all real x :

(a) 13/16≤ A ≤ 1
(b) 1 ≤ A ≤ 2
(c) 3/4≤ A ≤  13/16
(d) 3/4≤ A ≤ 1

19. The value of cos12° + cos84° + cos156° + cos 132° is

(a) 1/2
(b) 1
(c) -1/2 
(d) 1/8

20. The value of cos 420° is

(a) 0
(b) 1
(c) 1/2
(d) \(\frac{\sqrt3}{2}\)

Answer:

1.  Answer: (b) sin² (x + y)

Explanation: cos²x + cos²y – 2co x × cosy × cos(x + y)

{since cos(x + y) = cosx x cos y – sin x × sin y }

= cos²x + cos²y – 2cosx × cosy × (cosx × cosy – sinx × siny)

= cos²x + cos²y – 2cos²x × cos²y + 2cosx × cosy × sinx × siny

= cos²x + cos²y – cos²x × cos²y – cos²x × cos²y + 2cosx × cosy × sinx × siny

= (cos²x – cos² x × cos² y) + (cos²y – cos²x × cos²y) + 2cosx × cosy × sinx × siny

= cos²x(1- cos²y) + cos2 y(1 – cos²x) + 2cosx × cosy × sinx × siny

= sin²y × cos²x + sin²x × cos²y + 2cosx × cosy × sinx × siny (since sin²x + cos²x = 1 )

= sin²x × cos²y + sin²y × cos²x + 2cosx × cosy × sinx × siny

= (sinx × cosy)² + (siny × cosx)² + 2cosx × cosy × sinx × siny

= (sinx × cosy + siny × cosx)²

= {sin (x + y)}²

= sin² (x + y)

2. Answer: (d) a² + b² – c²

Explanation: a × cosx + b × sinx)² + (a × sinx – b × cosx)² = a² + b²

⇒ c²+ (a × sinx – b × cosx)² = a² + b²

⇒ (a × sinx – b × cosx)² = a² + b² – c²

3. Answer: (c) -1

Explanation: cos 5π = cos (π + 4π) 

cos ( 5 × 180° )

= cos 900°

= cos ( 10 × 90° + 0° ​​​)

= - cos 0° ​​​​​​

= - 1

4. Answer: (c) 1

Explanation: Given cosec A (sin B cos C + cos B sin C)

= cosecA × sin(B+C)

= cosecA × sin(180 – A)

= cosecA × sin A

= cosecA × 1/cosec A

= 1

5. Answer: (c) -1

Explanation: cos 180 = cos(90° + 90°)

= -sin 90°

= -1 {sin90° = 1}

cos180° = -1

6. Answer: (b) 90°

Explanation: Let the lengths of the sides if ∆ABC be a, b and c

Perimeter of the triangle = 2s = a + b + c = 6(sinA + sinB + sinC)/3

⇒ (sinA + sinB + sinC) = ( a + b + c)/2

⇒ (sinA + sinB + sinC)/( a + b + c) = 1/2

From sin formula,Using

sinA/a = sinB/b = sinC/c = (sinA + sinB + sinC)/(a + b + c) = 1/2

Now, sinB/b = 1/2

Given b = 2

So, sinB/2 = 1/2

⇒ sinB = 1

⇒ B = π/2

7. Answer: (b) 45

Explanation: Given, 3×tan(x – 15) = tan(x + 15)

⇒ tan(x + 15)/tan(x – 15) = 3/1

⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = (3 + 1)/(3 – 1)

⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 4/2

⇒ {tan(x + 15) + tan(x – 15)}/{tan(x + 15) – tan(x – 15)} = 2

⇒ sin(x + 15 + x – 15)/sin(x + 15 – x + 15) = 2

⇒ sin 2x/sin 30 = 2

⇒ sin 2x/(1/2) = 2

⇒ 2 × sin 2x = 2

⇒ sin 2x = 1

⇒ sin 2x = sin 90

⇒ 2x = 90

⇒ x = 45

8. Answer: (c) 2π/3

Explanation: Given, the sides of a triangle are 13, 7, 8

Since greatest side has greatest angle,

Now Cos A = (b² + c² – a²)/2bc

⇒ Cos A = (7² + 8⁸ – 13²)/(2×7×8)

⇒ Cos A = (49 + 64 – 169)/(2×7×8)

⇒ Cos A = (113 – 169)/(2×7×8)

⇒ Cos A = -56/(2×56)

⇒ Cos A = -1/2

⇒ Cos A = Cos 2π/3

⇒ A = 2π/3

So, the greatest angle is = 2π/3

9. Answer: (b) tan 60

Explanation: tan20 × tan40 × tan80

= tan40 × tan80 × tan20

= [{sin40 × sin80}/{cos40 × cos80}] × (sin20/cos20)

= [{2 x sin40 × sin80}/{2 × cos40 × cos80}] × (sin20/cos20)

= [{cos40 – cos120}/{cos120 + cos40}] × (sin20/cos20)

= [{cos40 – cos (90 + 30)}/{cos(90 + 30) + cos40}] × (sin 20/cos 20)

= [{cos40 + sin30}/{-sin30 + cos40}] × (sin20/cos 20)

= [{(2 × cos40 + 1)/2}/{(-1 + cos40)/2}] × (sin20/cos20)

= [{2 × cos40 + 1}/{-1 + cos40}] × (sin 20/cos20)

= [{2 × cos40 × sin20 + sin20}/{-cos20 + cos40 × cos20}]

= (sin60 – sin20 + sin20)/(-cos20 + cos60 + cos20)

= sin60/cos60

= tan 60

10. Answer: (b) 1

Explanation: Given, cos 20 + 2sin² 55 – \(\sqrt2\;sin65\)

= cos20 + 1 – cos110 – \(\sqrt 2\;sin 65\) {since cos 2x = 1 – 2sin² x}

= 1 + cos20 – cos 110 – √\(\sqrt \;sin 65\)

= 1 – 2 × sin {(20 + 110)/2 × sin{(20 – 110)/2} – \(\sqrt 2\;sin 65\) {Apply cos C – cos D formula}

= 1 – 2 × sin65 × sin (-45) – \(\sqrt2\;sin 65\)

= 1 + 2 × sin65 × sin45 – \(\sqrt 2\;sin 65\)

= 1 + (2 × sin 65)/\(\sqrt2\) – \(\sqrt 2\;sin 65\)

= 1 + \(\sqrt2\) ( sin 65 – √2 sin 65

= 1

11. Answer: (a) 2π/3 

Explanation: In ΔPQR PQ = PR = R₁ and Q = R where R₁ is circumradius

 from sine rule: PQ/sin R = 2R₁

Sin R = 1/2

R = 1/2

R = π/6

 P = π−Q −R  

= π - 2R

= 2π/3

12. Answer: (c) sin 3x

Explanation: 4 × sinx × sin(x + π/3) × sin(x + 2π/3)

= 4 × sinx × {sinx × cos π/3 + cosx × sin π/3} × {sinx × cos 2π/3 + cos x × sin 2π/3}

= 4 × sin x × {(sin x)/2 + ((c) sin 3x cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}

= 4 × sin x × {-(sin 2x)/4 + (3 × cos 2x)/4}

= sinx × {-sin 2x + 3 × cos 2x}

= sinx × {-sin 2x + 3 × (1 – sin 2x)}

= sinx × {-sin 2x + 3 – 3 × sin 2x}

= sinx × {3 – 4 × sin 2x}

= 3 × sinx – 4 sin 3x

= sin 3x

 So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x    

13. Answer: (b) 2 tan 6x

Explanation: (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x)

⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x 3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}]

⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos6x × cosx}]

⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x)

⇒ tan 6x + tan 6x

⇒ 2 tan 6x

14. Answer: (a) 

Explanation: cot(−870°)=−cot(2×360°+150°)

=−cot150°

=−cot(180°−30°)

cot30° = \(\sqrt3\)

15. Answer: (a) 

Explanation: cos2 48°– sin2 12° 

= cos (48°+ 12°) cos (48° – 12°)

= cos60° cos36°

= \(\frac{1}{2}\times\frac{\sqrt5+1}{4}\)

\(=\frac{\sqrt5+1}{8}\)

16. Answer: (b) 0

Explanation: 50°= 70° – 20°

Taking “tan” on both sides,

tan 50° = tan(70° – 20°)

= (tan 70° – tan 20°)/(1 + tan 70° tan 20°)

= (tan 70° – tan 20°)/ [1 + tan(90° – 20°) tan 20°]

= (tan 70° – tan 20°)/ (1 + cot 20° tan 20°)

= (tan 70° – tan 20°)/(1 + 1)

2 tan 50° = tan 70° – tan 20°

2 tan 50° + tan 20° – tan 70° 

= 0

17. Answer: (b) 0

Explanation: cos1°, cos2°, cos3° .............cos179°

= cos1°, cos2°, cos3° .............cos90° .....cos179°

= cos1°, cos2°, cos3° .............cos90° .....cos179° = 0 As [cos90°= 0]

18. Answer: (d) 3/4≤ A ≤ 1

Explanation: A = sin2x + cos2x

We have cos4x ≤ cos2x

sin2x = sin2x

Adding sin2x + cos4x ≤ sin2x + cos2x

∴ A ≤ 1.

Again A = t + (1 – t)2 = t2 – t + 1, t ≥ 0, where minimum is 3/4. Thus 3/4 ≤ A ≤ 1 .

19. Answer: (c) -1/2  

Explanation: cos 12°+cos 84°+cos 156°+cos 132°

= (cos 12°+cos 132°)+(cos 84°+cos 156°)

= 2cos 72°cos 60°+2cos 120°cos 36°

=cos 72°—cos 36°

=sin18°—cos 36° 

=−1/2

20. Answer: (c) 1/2  

Explanation: cos(360°+A) = cos(A)

420° = 360°+ 60°

cos (420°) = cos(60°)

 = 1/2

Click here to practice MCQ Questions for Trigonometric Functions class 11

2 sin θ = 3 ∴ sin θ = \(\cfrac{3}{2}\)

This is not possible because -1 ≤ sin θ ≤ 1 for any θ.

∴ 2 sin θ = 3 does not have any solution.

The expression involving integer ‘n’ which gives all solutions of a trigonometric equation is called the general solution. 

Equations involving trigonometric functions of a variable are called trigonometric equations.

The general solution of sin θ = sin \(\propto\) is

θ = nπ + (-1 )ⁿ \(\propto\), n ∈ Z

Now, sinθ = \(\cfrac{1}{2}\) = sin \(\cfrac{\pi}{6}\).....[∵ sin\(\cfrac{\pi}{6}\) = \(\cfrac{1}{2}\)]

∴ the required general solution is

θ = nπ + (-1)ⁿ \(\cfrac{\pi}{6}\), n ∈ Z.

LHS = (secx sec y + tanx tan y)² – (secx tan y + tanx sec y)²

= [(secx sec y)² + (tanx tan y)² + 2 (secx sec y) (tanx tan y)] – [(secx tan y)² + (tanx sec y)² + 2 (secx tan y) (tanx sec y)]

= [sec²x sec² y + tan²x tan² y + 2 (secx sec y) (tanx tan y)] – [sec²x tan² y + tan²x sec² y + 2 (sec²x tan² y) (tanx sec y)]

= sec²x sec² y - sec²x tan²y + tan²x tan²y - tan²x sec² y

= sec²x (sec²y - tan²y) + tan²x (tan²y - sec²y)

= sec²x (sec²y - tan²y) - tan²x (sec²y - tan²y)

We know that sec²x – tan²x = 1.

= sec²x × 1 – tan²x × 1 

 sec²x – tan²x

= 1

= RHS

Hence proved.

Correct option is: (b) Isosceles triangle

The solutions of a trigonometric equation for which 0 ≤ x ≤ 2π are called principal solutions.

Correct option is: (a) 1/√2

LHS = (1 + tan α tan β)² + (tan α – tan β)²

= 1+ tan² α tan² β + 2 tan α tan β + tan² α + tan² β – 2 tan α tan β

= 1 + tan² α tan² β + tan² α + tan² β

= tan² α (tan² β + 1) + 1 (1 + tan² β)

= (1 + tan² β) (1 + tan² α)

We know that 1 + tan² θ = sec² θ

= sec² α sec² β

= RHS

Hence proved.

Given,

sec β = \(\frac{-5}{3}\)

sec²β = \(\frac{25}{9}\)

tan²β = \(\frac{25}{9}\)-1 = \(\frac{16}{9}\)

tan β = ±\(\frac{4}{3}\)

tan β = \(\frac{-4}{3}\)   (\(\frac{\pi}{2}\)< β < π)

And,  cot α = \(\frac{1}{2}\)

∴ tan α = 2,  (∵ \(\frac{\pi}{2}\)< α <\(\frac{3\pi}{2}\))

Now,

tan (α + β) = \(\frac{tanα+tanβ}{1-tanα\,tanβ}\)

\(= \frac{2+(\frac{-4}{3})}{1-2(\frac{-4}{3})}\)

= \(\frac{6-4}{3+8}\)

= \(\frac{2}{11}\)

Let us consider the LHS

sin 5x

Then,

sin 5x = sin(3x + 2x)

But as we know,

Sin(x + y) = sin x cos y + cos x sin y…..(i)

Therefore,
sin 5x = sin 3x cos 2x + cos 3x sin 2x

= sin(2x + x) cos 2x + cos (2x + x) sin 2x……..(ii)

And

cos(x + y) = cos x cos y – sin x sin y……(iii)

Then, substituting equation (i) and (iii) in equation (ii), we get

sin 5x = (sin 2x cos x + cos 2x sin x) cos 2x + (cos 2x cos x – sin 2x sin x) sin 2x

= sin 2x cos 2x cos x + cos² 2x sin x + (sin 2x cos 2x cos x – sin² 2x sin x)

= 2sin 2x cos 2x cos x + cos² 2x sin x – sin² 2x sin x …….(iv)

Then sin 2x = 2sin x cos x………(v)

And cos 2x = cos²x – sin²x………(vi)

By substituting equation (v) and (vi) in equation (iv), we get

sin 5x = 2(2sin x cos x) (cos²x – sin²x) cos x + (cos²x – sin²x)² sin x – (2sin x cos x)² sin x

= 4(sin x cos² x) ([1 – sin²x] – sin²x) + ([1 – sin²x] – sin²x)² sin x – (4sin² x cos² x)sin x

(as cos²x + sin²x = 1 ⇒ cos²x = 1 – sin²x)

sin 5x = 4(sin x [1 – sin²x]) (1 – 2sin²x) + (1 – 2sin²x)² sin x – 4sin³ x [1 – sin²x]

= 4sin x (1 – sin²x) (1 – 2sin² x) + (1 – 4sin²x + 4sin⁴x) sin x – 4sin³ x + 4sin⁵x

= (4sin x – 4sin³x) (1 – 2sin²x) + sin x – 4sin³x + 4sin⁵x – 4sin³ x + 4sin⁵x

= 4sin x – 8sin³x – 4sin³x + 8sin⁵x + sin x – 8sin³x + 8sin⁵x

= 5sin x – 20sin³x + 16sin⁵x

= RHS

Thus proved.

 Correct option is: b. \(\cfrac{7\pi}{6}\),\(\cfrac{11\pi}{6}\)

To find: Value of cosec (-750°)

We have,

cosec (-750°) = - cosec(750°)

[∵ cosec(-θ) = -cosec θ]

= - cosec [90° × 8 + 30°]

Clearly, 405° is in Ist Quadrant and the multiple of 90° is even

= - cosec 30°

= -2 [∵ cosec 30° = 2]

For maximum value of 3 – 7 cos 5x, cos 5x must be minimum so minimum value is – 1 and hence maximum value = 3 – 7 (– 1) 

= 3 + 7 

= 10

We know that, general solution of \(sinx=0\) is \(x=n\pi\) 

∴ \(sin(x+\frac{\pi}{12})=0\) 

⇒ \(x+\frac{\pi}{12}=n\pi\) 

⇒ \(x=n\pi-\frac{\pi}{12}\)

Given, \(sin12θ+sin4θ\) 

= \(2sin(\frac{12θ+4θ}{2})cos(\frac{12θ-4θ}{2})\) 

= \(sin8θcos4θ\) 

[Using the formulae \(sinC+sinD\) \(=2sin(\frac{C+D}{2})cos(\frac{C-D}{2})\)]

Using, 

2 cos A sin B = sin (A + B) – sin (A – B) 

We get 

2 cos 4x sin 2x = sin (4x + 2x) – sin (4x - 2x) 

= sin 6x – sin 2x

(i) 2sin 6x cos 4x = sin (6x+4x) + sin (6x-4x)

= sin 10x + sin 2x

Using,

2sinAcosB = sin (A+ B) + sin (A - B)

(ii) 2cos 5x sin3x = sin (5x + 3x) – sin (5x – 3x)

= sin8x – sin2x

Using,

2cosAsinB = sin(A + B) – sin (A - B)

(iii) 2cos7xcos3x = cos (7x+3x) + cos (7x – 3x)

= cos10x + cos 4x

Using,

2cosAcosB = cos (A+ B) + cos (A - B)

(iv) 2sin8 x sin2 x = cos (8x - 2x) – cos (8x + 2x)

= cos6x – cos10x

Using,

2sinAsinB = cos (A - B) – cos (A+ B)

Given: sinx = \(\frac{\sqrt{5}}{3}\)

To find: cos2x

We know that,

cos2x = 1 – 2sin²x

Putting the value, we get

cos2x = 1 - 2\((\frac{\sqrt{5}}{3})^{2}\)

cos2x = 1 - 2 x \(\frac{5}{9}\)

cos2x = 1 - \(\frac{10}{9}\)

cos2x = \(\frac{9\,-\,10}{9}\)

∴ cos2x = -\(\frac{1}{9}\)

(i) sin80°cos20° - cos80°sin20° = sin(80° - 20°)

(using sin(A - B) = sinAcosB - cosAsinB)

= sin60°

= \(\frac{\sqrt{3}}2{}\)

(ii) cos45°cos15° - sin45°sin15° = cos(45° + 15°)

(Using cos(A + B) = cosAcosB - sinAsinB)

= cos60°

= \(\frac{1}{2}\)

(iii) cos75°cos15° + sin75°sin15° = cos(75° - 15°)

(using cos(A - B) = cosAcosB + sinAsinB)

= cos60°

= \(\frac{1}{2}\)

(iv) sin40°cos20° + cos40°sin20° = sin(40° + 20°)

(using sin(A + B) = sinAcosB + cosAsinB)

= sin60°

(v) cos130°cos40° + sin130°sin40° = cos(130° - 40°)

(using cos(A - B) = cosAcosB + sinAsinB)

= cos90°

= 0

B. cos 2(θ + ϕ)

Given that, cos 2θ cos 2ϕ + sin²(θ–ϕ) – sin²(θ + ϕ)

= cos 2θ cos 2ϕ + sin(θ–ϕ+ θ + ϕ)sin(θ–ϕ- θ – ϕ)

[∵sin²A – sin²B = sin (A + B) sin (A – B)]

= cos 2θ cos 2ϕ + sin 2θ. sin(- 2ϕ)

= cos 2θ cos 2ϕ - sin 2θ. Sin 2ϕ [∵ sin(-θ) = -sin θ]

= cos 2(θ + ϕ) [∵ cos x cos y – sin x sin y = cos(x + y)]

Let s be the length of the arc subtending an angle θ^c at the center of a circle of radius r.

Then,

θ = \(\frac{s}{r}\)

Here, 

r = 5 and θ = 15°

= \((15\times\frac{\pi}{180})^c\)

= \((\frac{\pi}{12})^c\)

θ = \(\frac{s}{r}\)

⇒ \(\frac{\pi}{12}=\frac{s}{5}\)

⇒ s = \(\frac{5\pi}{12}\) cm

A + B = π - C

\(\frac {tanA+tanB}{1-tanAtanB}\) = - tan C

tan A + tan B + tan C = tanA tanB tanC

\(\frac{tanA +tanB+tanC}{tanA\,tanB\,tanC}\) = 1

cos p = \(\frac{1}7\),sin P = \(\frac{\sqrt{48}}7\)

cos Q = \(\frac{13}{14}\),sin Q = \(\frac{\sqrt{27}}{14}\)

Cos(p-q) = cos(p)cos(q) + sin(p)sin(q)

cos P -Q = \(\frac{1}2\)

P - Q = \(\frac{π}3\)

cot (α + β) = 0

(α + β) = \(\frac{π}2\)

sin (α + 2 β)= sin (α +β) cos(β) + sin(β) cos (α +β) 

put  (α + β) = \(\frac{π}2\)

sin (α + 2 β) = cos(β)