MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers
Trigonometric Functions Class 11 Maths MCQs Questions with Answers
Check the below NCERT MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions with Answers Pdf free download. MCQ Questions for Class 11 Maths with Answers were prepared based on the latest exam pattern. We have Provided Trigonometric Functions Class 11 Maths MCQs Questions with Answers to help students understand the concept very well.
Class 11 Maths Chapter 3 Quiz
Class 11 Maths Chapter 3 MCQ Online Test
You can refer to NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.
Trigonometric Functions Class 11 Maths MCQ online test
| Q1. | The value of sin 15 + cos 15 is |
A.1 |
|
B.1/2 |
|
C.√3/2 |
|
D.√3 |
Ans: √3/2
Given, sin 15 + cos 15
= sin 15 + cos(90 – 15)
= sin 15 + sin 15
= 2 × sin 45 × cos 30
= 2 × (1/√2) × (√3/2)
= √3/2
Given, sin 15 + cos 15
= sin 15 + cos(90 – 15)
= sin 15 + sin 15
= 2 × sin 45 × cos 30
= 2 × (1/√2) × (√3/2)
= √3/2
| Q2. | The value of tan A/2 – cot A/2 + 2cot A is |
A.0 |
|
B.1 |
|
C.-1 |
|
D.None of these |
Ans: 0
Given, tan A/2 – cot A/2 + 2cot A
= {sin(A/2)/cos(A/2)} – {cos(A/2)/sin(A/2)} + 2cotA
= {sin² (A/2) – cos² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cos² (A/2) – sin² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cosA}/{cos(A/2) × sin(A/2)} + 2cotA (since cos² A – sin² A = cos²A )
= -{cos(2A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{2 × cosA}/{2 × cos(A/2) × sin(A/2)} + 2cotA
= -{2cosA}/{sin(2A/2)} + 2cotA
= {-(2cosA)/(sinA)} + 2cotA (since sin2A = 2 × sinA × cosA)
= -2cotA + 2cotA
= 0
Given, tan A/2 – cot A/2 + 2cot A
= {sin(A/2)/cos(A/2)} – {cos(A/2)/sin(A/2)} + 2cotA
= {sin² (A/2) – cos² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cos² (A/2) – sin² (A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{cosA}/{cos(A/2) × sin(A/2)} + 2cotA (since cos² A – sin² A = cos²A )
= -{cos(2A/2)}/{cos(A/2) × sin(A/2)} + 2cotA
= -{2 × cosA}/{2 × cos(A/2) × sin(A/2)} + 2cotA
= -{2cosA}/{sin(2A/2)} + 2cotA
= {-(2cosA)/(sinA)} + 2cotA (since sin2A = 2 × sinA × cosA)
= -2cotA + 2cotA
= 0
| Q3. | The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is |
A.sin x |
|
B.sin 2x |
|
C.sin 3x |
|
D.sin 4x |
Ans: sin 3x
Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3)
= 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3}
= 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}
= 4 × sin x × {-(sin² x)/4 + (3 × cos² x)/4}
= sin x × {-sin² x + 3 × cos² x}
= sin x × {-sin² x + 3 × (1 – sin² x)}
= sin x × {-sin² x + 3 – 3 × sin² x}
= sin x × {3 – 4 × sin² x}
= 3 × sin x – 4sin³ x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x
Given, 4 × sin x × sin(x + π/3) × sin(x + 2π/3)
= 4 × sin x × {sin x × cos π/3 + cos x × sin π/3} × {sin x × cos 2π/3 + cos x × sin 2π/3}
= 4 × sin x × {(sin x)/2 + (√3 × cos x)/2} × {-(sin x)/2 + (√3 × cos x)/2}
= 4 × sin x × {-(sin² x)/4 + (3 × cos² x)/4}
= sin x × {-sin² x + 3 × cos² x}
= sin x × {-sin² x + 3 × (1 – sin² x)}
= sin x × {-sin² x + 3 – 3 × sin² x}
= sin x × {3 – 4 × sin² x}
= 3 × sin x – 4sin³ x
= sin 3x
So, 4 × sin x × sin(x + π/3) × sin(x + 2π/3) = sin 3x
| Q4. | If tan x = (cos 9 + sin 9)/(cos 9 – sin 9), then x = |
A.45 |
|
B.54 |
|
C.36 |
|
D.None of these |
Ans: 54
Given, tan x = (cos 9 + sin 9)/(cos 9 – sin 9)
⇒ tan x = {cos 9(1 + sin 9/cos 9)}/{cos 9(1 – sin 9/cos 9)}
⇒ tan x = (1 + tan 9)}/(1 – tan 9)
⇒ tan x = (tan 45 + tan 9)}/(1 – tan 45 × tan 9) {since tan 45 = 1}
⇒ tan x = tan(45 + 9) {Apply tan(A + B) formula}
⇒ tan x = tan(54)
⇒ x = 54
Given, tan x = (cos 9 + sin 9)/(cos 9 – sin 9)
⇒ tan x = {cos 9(1 + sin 9/cos 9)}/{cos 9(1 – sin 9/cos 9)}
⇒ tan x = (1 + tan 9)}/(1 – tan 9)
⇒ tan x = (tan 45 + tan 9)}/(1 – tan 45 × tan 9) {since tan 45 = 1}
⇒ tan x = tan(45 + 9) {Apply tan(A + B) formula}
⇒ tan x = tan(54)
⇒ x = 54
| Q5. | In a triangle ABC, sin A – cos B = cos C, then angle B is |
A.π/2 |
|
B.π/3 |
|
C.π/4 |
|
D.π/6 |
Ans: π/2
Given, sin A – cos B = sin C
⇒ sin A = cos B + sin C
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {(B + C)/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {π/2 – A/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × sin (A/2) × cos {(B – C)/2}
⇒ cos (A/2) = cos {(B – C)/2}
⇒ A/2 = (B – C)/2
⇒ A = B – C
⇒ B = A + C
⇒ B = π – B {Since A + B + C = π}
⇒ 2B = π
⇒ B = π/2
Given, sin A – cos B = sin C
⇒ sin A = cos B + sin C
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {(B + C)/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × cos {π/2 – A/2} × cos {(B – C)/2}
⇒ 2 × sin (A/2) × cos (A/2) = 2 × sin (A/2) × cos {(B – C)/2}
⇒ cos (A/2) = cos {(B – C)/2}
⇒ A/2 = (B – C)/2
⇒ A = B – C
⇒ B = A + C
⇒ B = π – B {Since A + B + C = π}
⇒ 2B = π
⇒ B = π/2
| Q6. | The value of cos 420° is |
A.0 |
|
B.1 |
|
C.1/2 |
|
D.√3/2 |
Ans: 1/2
cos 420° = cos(360° + 60° ) = cos 60° = 1/2
cos 420° = cos(360° + 60° ) = cos 60° = 1/2
| Q7. | If in a triangle ABC, tan A + tan B + tan C = 6 then the value of cot A × cot B × cot C is |
A.1/2 |
|
B.1/3 |
|
C.1/4 |
|
D.1/6 |
Ans: 1/6
Given tanA + tanB + tanC = 6
Now tan(A + B + C) = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
We know that,
A + B + C = π
⇒ tan(A + B + C) = tan π
⇒ tan(A + B + C) = 0
Now
0 = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
⇒ tanA + tanB + tanC – tanA × tanB × tanC = 0
⇒ tanA + tanB + tanC = tanA × tanB × tanC
⇒ tanA × tanB × tanC = 6
⇒ (1/cotA) × (1/cotB) × (1/cotC) = 6
⇒ 1/(cot A × cot B × cot C) = 6
⇒ cot A × cot B × cot C = 1/6
Given tanA + tanB + tanC = 6
Now tan(A + B + C) = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
We know that,
A + B + C = π
⇒ tan(A + B + C) = tan π
⇒ tan(A + B + C) = 0
Now
0 = {(tanA + tanB + tanC) – tanA × tanB × tanC}/{1 – (tanA × tanB + tanB × tanC + tanA × tanC)}
⇒ tanA + tanB + tanC – tanA × tanB × tanC = 0
⇒ tanA + tanB + tanC = tanA × tanB × tanC
⇒ tanA × tanB × tanC = 6
⇒ (1/cotA) × (1/cotB) × (1/cotC) = 6
⇒ 1/(cot A × cot B × cot C) = 6
⇒ cot A × cot B × cot C = 1/6
| Q8. | If a × cos x + b × cos x = c, then the value of (a × sin x – b × cos x)² is |
A.a² + b² + c² |
|
B.a² – b² – c² |
|
C.a² – b² + c² |
|
D.a² + b² – c² |
Ans: a² + b² – c²
We have
(a × cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²
We have
(a × cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b²
⇒ c² + (a × sin x – b × cos x)² = a² + b²
⇒ (a × sin x – b × cos x)² = a² + b² – c²
| Q9. | When the length of the shadow of a pole is equal to the height of the pole, then the elevation of source of light is |
A.30° |
|
B.60° |
|
C.75° |
|
D.45° |
Ans: 45°
Let AB is the length of the pole and BC is the shadow of the pole.
Given AB = BC
Now from triangle ABC,
tan θ = AB/BC
⇒ tan θ = 1
⇒ θ = 45°
So, the elevation of source of light is 45°
Let AB is the length of the pole and BC is the shadow of the pole.
Given AB = BC
Now from triangle ABC,
tan θ = AB/BC
⇒ tan θ = 1
⇒ θ = 45°
So, the elevation of source of light is 45°
| Q10. | In any triangle ABC, if cos A/a = cos B/b = cos C/c and the side a = 2, then the area of the triangle is |
A.√3 |
|
B.√3/4 |
|
C.√3/2 |
|
D.1/√3 |
Ans: √3
Given cos A/a = cos B/b = cos C/c
= cos A/k × sin A = cos B/k × sin B = cos C/k × sin C {since sin A/a = sin B/b = sin C/c = k}
= cot A = cot B = cot C
⇒ A = B = C = 60
So, triangle ABS is equilateral.
Now area of the triangle = (√3/4) × a² = (√3/4) × 2² = (√3/4) × 4 = √3
Given cos A/a = cos B/b = cos C/c
= cos A/k × sin A = cos B/k × sin B = cos C/k × sin C {since sin A/a = sin B/b = sin C/c = k}
= cot A = cot B = cot C
⇒ A = B = C = 60
So, triangle ABS is equilateral.
Now area of the triangle = (√3/4) × a² = (√3/4) × 2² = (√3/4) × 4 = √3
| Q11. | The least values of cos² θ + sec² θ is |
A.0 |
|
B.1 |
|
C.2 |
|
D.more than 2 |
Ans: 2
If a × cos² θ + b × sec² θ is given,
then the least value = 2√ab
Now given, cos² θ + sec² θ
Here, a = 1, b = 1
Now, least value = 2√(1 × 1) = 2 × 1 = 2
If a × cos² θ + b × sec² θ is given,
then the least value = 2√ab
Now given, cos² θ + sec² θ
Here, a = 1, b = 1
Now, least value = 2√(1 × 1) = 2 × 1 = 2
| Q12. | The equation (cos p – 1) x² + cos p × x + sin p = 0, where x is a variable, has real roots. Then the interval of p may be any one of the following: |
A.(0, π) |
|
B.(−π/2, π/2) |
|
C.(0, π) |
|
D.(−π, 0) |
Ans: (0, π)
The equation (cos p – 1)
x² + cos p × x + sin p = 0, where x is a variable, has real roots.
Now, for real roots,
Discriminant ≥ 0
⇒ cos² p – 4(cosp – 1)sinp ≥ 0
⇒ (cosp – 2sinp)² – 4sin² p + 4sinp ≥ 0
⇒ (cosp – 2sinp)² + 4sin p(1 – sinp) ≥ 0 ………..1
Now, 1 – sinp ≥ 0
⇒ For all real p such that 0 < p < π
So that 4sin p(1 – sinp) ≥ 0
So, p ∈ (0, π)
The equation (cos p – 1)
x² + cos p × x + sin p = 0, where x is a variable, has real roots.
Now, for real roots,
Discriminant ≥ 0
⇒ cos² p – 4(cosp – 1)sinp ≥ 0
⇒ (cosp – 2sinp)² – 4sin² p + 4sinp ≥ 0
⇒ (cosp – 2sinp)² + 4sin p(1 – sinp) ≥ 0 ………..1
Now, 1 – sinp ≥ 0
⇒ For all real p such that 0 < p < π
So that 4sin p(1 – sinp) ≥ 0
So, p ∈ (0, π)
| Q13. | The value of (sec 8A – 1)/(sec 4A – 1) is |
A.0 |
|
B.1 |
|
C.tan 8A/tan 2A |
|
D.tan 2A/tan 8A |
Ans: tan 8A/tan 2A
Given, (sec 8A – 1)/(sec 4A – 1)
= (1/cos 8A – 1)/(1/cos 4A – 1)
= {(1 – cos 8A)/cos 8A}/{(1 – cos 4A)/cos 4A}
= {(1 – cos 8A) × cos 4A}/{(1 – cos 4A) × cos 8A}
= (2sin² 4A × cos 4A}/{2sin² 2A × cos 8A} {since cos 2A = 1 – 2sin² A}
= (2sin 4A × sin 4A × cos 4A}/{2sin 2A × sin 2A × cos 8A}
= (sin 8A × sin 4A}/{2sin 2A × sin 2A × cos 8A} {since sin 2A = 2×sin A × cos A}
= (sin 8A × 2sin 2A × cos 2A}/{2sin 2A × sin 2A × cos 8A}
= (sin 8A × cos 2A}/{sin 2A × cos 8A}
= (sin 8A/cos 8A)/(sin 2A/cos 2A)
= tan 8A/tan 2A
So, (sec 8A – 1)/(sec 4A – 1) = tan 8A/tan 2A
Given, (sec 8A – 1)/(sec 4A – 1)
= (1/cos 8A – 1)/(1/cos 4A – 1)
= {(1 – cos 8A)/cos 8A}/{(1 – cos 4A)/cos 4A}
= {(1 – cos 8A) × cos 4A}/{(1 – cos 4A) × cos 8A}
= (2sin² 4A × cos 4A}/{2sin² 2A × cos 8A} {since cos 2A = 1 – 2sin² A}
= (2sin 4A × sin 4A × cos 4A}/{2sin 2A × sin 2A × cos 8A}
= (sin 8A × sin 4A}/{2sin 2A × sin 2A × cos 8A} {since sin 2A = 2×sin A × cos A}
= (sin 8A × 2sin 2A × cos 2A}/{2sin 2A × sin 2A × cos 8A}
= (sin 8A × cos 2A}/{sin 2A × cos 8A}
= (sin 8A/cos 8A)/(sin 2A/cos 2A)
= tan 8A/tan 2A
So, (sec 8A – 1)/(sec 4A – 1) = tan 8A/tan 2A
| Q14. | The value of (sin7x + sin5x) /(cos7x + cos5x) + (sin9x + sin3x) / (cos9x + cos3x) is |
A.tan6x |
|
B.2 tan6x |
|
C.3 tan6x |
|
D.4 tan6x |
Ans: 2 tan6x
Given, (sin7x + sin5x) /(cos7x + cos5x) + (sin9x + sin3x) / (cos9x + cos3x)
⇒ [{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 × sin6x × cosx}/{2 × cos6x × cosx}] + [{2 × sin6x × cosx}/{2 × cos6x × cosx}]
⇒ (sin6x/cos6x) + (sin6x/cos6x)
⇒ tan6x + tan6x
⇒ 2 tan6x
Given, (sin7x + sin5x) /(cos7x + cos5x) + (sin9x + sin3x) / (cos9x + cos3x)
⇒ [{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 × sin6x × cosx}/{2 × cos6x × cosx}] + [{2 × sin6x × cosx}/{2 × cos6x × cosx}]
⇒ (sin6x/cos6x) + (sin6x/cos6x)
⇒ tan6x + tan6x
⇒ 2 tan6x
| Q15. | If x > 0 then the value of f(x) = -3 × cos√(3 + x + x²) lie in the interval |
A.[-1, 1] |
|
B.[-2, 2] |
|
C.[-3, 3] |
|
D.None of these |
Ans: [-3, 3]
Given x > 0 then 3 + x + x² > 0
Now, -1 ≤ cos√(3 + x + x² ) ≤ 1 {Since -1 ≤ cosx ≤ 1}
⇒ 3 ≥ -3 × cos√(3 + x + x² ) ≥ -3 {Multiply by -3}
⇒ -3 ≤ f(x) ≤ 3
⇒ f(x) ∈ [-3, 3]
Given x > 0 then 3 + x + x² > 0
Now, -1 ≤ cos√(3 + x + x² ) ≤ 1 {Since -1 ≤ cosx ≤ 1}
⇒ 3 ≥ -3 × cos√(3 + x + x² ) ≥ -3 {Multiply by -3}
⇒ -3 ≤ f(x) ≤ 3
⇒ f(x) ∈ [-3, 3]
| Q16. | The value of cos 4A – cos 4B is |
A.(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B) |
|
B.2(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B) |
|
C.4(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B) |
|
D.8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B) |
Ans: 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
Given, cos 4A – cos 4B
= 2cos² 2A – 1 – (2cos2 2B – 1) {since 2cos² x – 1 = cos 2x}
= 2cos² 2A – 1 – 2cos² 2B + 1
= 2cos² 2A – 2cos² 2B
= 2(cos² 2A – cos² 2B)
= 2(cos 2A – cos 2B) × (cos 2A + cos 2B)
= 2{2cos² A – 1 – (2cos² B – 1)} × {2cos² A – 1 + 1 – 2sin² B} {since 1 – 2sin² x = cos 2x}
= 2{2cos² A – 1 – 2cos² B + 1} × {2cos² A – 1 + 1 – 2sin² B}
= 2{2cos² A – 2cos² B} × {2cos² A – 2sin² B}
= 2 × 2 × 2{cos² A – cos² B} × {cos² A – sin² B}
= 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
So, cos 4A – cos 4B = 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cosA + sin B)
Given, cos 4A – cos 4B
= 2cos² 2A – 1 – (2cos2 2B – 1) {since 2cos² x – 1 = cos 2x}
= 2cos² 2A – 1 – 2cos² 2B + 1
= 2cos² 2A – 2cos² 2B
= 2(cos² 2A – cos² 2B)
= 2(cos 2A – cos 2B) × (cos 2A + cos 2B)
= 2{2cos² A – 1 – (2cos² B – 1)} × {2cos² A – 1 + 1 – 2sin² B} {since 1 – 2sin² x = cos 2x}
= 2{2cos² A – 1 – 2cos² B + 1} × {2cos² A – 1 + 1 – 2sin² B}
= 2{2cos² A – 2cos² B} × {2cos² A – 2sin² B}
= 2 × 2 × 2{cos² A – cos² B} × {cos² A – sin² B}
= 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cos A + sin B)
So, cos 4A – cos 4B = 8(cos A – cos B) × (cos A + cos B) × (cos A – sin B) × (cosA + sin B)
| Q17. | The value of cos 420° is |
A.0 |
|
B.1 |
|
C.1/2 |
|
D.√3/2 |
Ans: 1/2
cos 420° = cos(360° + 60° ) = cos 60° = 1/2
cos 420° = cos(360° + 60° ) = cos 60° = 1/2
| Q18. | In a ΔABC, (b + c) cos A + (c + a) cos B + (a + b) cos C is equal to |
A.a + b + c |
|
B.0 |
|
C.none of these |
|
D.Rr |
Ans: a + b + c
Given (b + c) cos A + (c + a) cos B + (a + b) cos C
= b × cos A + c × cos A + c × cos B + a × cos B + a × cos C + b × cos C
= (b × cos C + c × cos B) + (c × cos A + a × cos C) + (b × cos A + a × cos B)
= a + b + c {since b × cos C + c × cos B = a, c × cos A + a × cos C = b, b × cos A + a × cos B = c}
Given (b + c) cos A + (c + a) cos B + (a + b) cos C
= b × cos A + c × cos A + c × cos B + a × cos B + a × cos C + b × cos C
= (b × cos C + c × cos B) + (c × cos A + a × cos C) + (b × cos A + a × cos B)
= a + b + c {since b × cos C + c × cos B = a, c × cos A + a × cos C = b, b × cos A + a × cos B = c}
| Q19. | tan² θ = 1 – a² then the value of sec θ + tan³ θ × cosec θ is |
A.(2 – a²) |
|
B.(2 – a²)1/2 |
|
C.(2 – a²)3/2 |
|
D.None of these |
Ans: (2 – a²)3/2
Given, tan² θ = 1 – a²
⇒ tan θ = √(1 – a²)
From the figure and apply Pythagorus theorem,
AC² = AB² + BC²
⇒ AC² = {√(1 – a²)}² + 12
⇒ AC² = 1 – a² + 1
⇒ AC² = 2 – a²
⇒ AC = √(2 – a²)
Now, sec θ = √(2 – a²)
cosec θ = √(2 – a²)/√(1 – a²)
and tan θ = √(1 – a²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – a²) + {(1 – a²)3/2 × √(2 – a²)/√(1 – a²)}
= √(2 – a²) + {(1 – a²) × (1 – a²) × √(2 – a²)/√(1 – a²)}
= √(2 – a²) + (1 – a²) × √(2 – a²)
= √(2 – a²) × (1 + 1 – a²)
= √(2 – a²) × (2 – a²)
= (2 – a²)3/2
Given, tan² θ = 1 – a²
⇒ tan θ = √(1 – a²)
From the figure and apply Pythagorus theorem,
AC² = AB² + BC²
⇒ AC² = {√(1 – a²)}² + 12
⇒ AC² = 1 – a² + 1
⇒ AC² = 2 – a²
⇒ AC = √(2 – a²)
Now, sec θ = √(2 – a²)
cosec θ = √(2 – a²)/√(1 – a²)
and tan θ = √(1 – a²)
Given, sec θ + tan³ θ × cosec θ
= √(2 – a²) + {(1 – a²)3/2 × √(2 – a²)/√(1 – a²)}
= √(2 – a²) + {(1 – a²) × (1 – a²) × √(2 – a²)/√(1 – a²)}
= √(2 – a²) + (1 – a²) × √(2 – a²)
= √(2 – a²) × (1 + 1 – a²)
= √(2 – a²) × (2 – a²)
= (2 – a²)3/2
| Q20. | The value of cos(π/7) × cos(2π/7) × cos(4π/7) is |
A.-1/2 |
|
B.-1/4 |
|
C.-1/6 |
|
D.-1/8 |
Ans: -1/8
We know that cos A × cos 2A × cos 2² A × ……………… × cos 2n-1 A = sin (2ⁿ A)/{2ⁿ × sin A} ……………1
Given, cos(π/7) × cos(2π/7) × cos(4π/7)
= cos(π/7) × cos(2π/7) × cos(2² π/7)
= [sin (2³ × π/7) ]/{2³ × sin (π/7)} ……………..from equation 1
= [sin (8π/7) ]/{8 × sin (π/7)}
= [sin (π + π/7) ]/{8 × sin (π/7)}
= -sin (π/7)/{8 × sin (π/7)}
= -1/8
So, cos(π/7) × cos(2π/7) × cos(4π/7) = -1/8
We know that cos A × cos 2A × cos 2² A × ……………… × cos 2n-1 A = sin (2ⁿ A)/{2ⁿ × sin A} ……………1
Given, cos(π/7) × cos(2π/7) × cos(4π/7)
= cos(π/7) × cos(2π/7) × cos(2² π/7)
= [sin (2³ × π/7) ]/{2³ × sin (π/7)} ……………..from equation 1
= [sin (8π/7) ]/{8 × sin (π/7)}
= [sin (π + π/7) ]/{8 × sin (π/7)}
= -sin (π/7)/{8 × sin (π/7)}
= -1/8
So, cos(π/7) × cos(2π/7) × cos(4π/7) = -1/8
MCQ Questions for Class 11 Maths
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MCQ Questions for Class 11 Maths Chapter 1 Sets
MCQ Questions for Class 11 Maths Chapter 2 Relations and Functions
MCQ Questions for Class 11 Maths Chapter 3 Trigonometric Functions
MCQ Questions for Class 11 Maths Chapter 4 Principle of Mathematical Induction
MCQ Questions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations
MCQ Questions for Class 11 Maths Chapter 6 Linear Inequalities
MCQ Questions for Class 11 Maths Chapter 7 Permutations and Combinations
MCQ Questions for Class 11 Maths Chapter 8 Binomial Theorem
MCQ Questions for Class 11 Maths Chapter 9 Sequences and Series
MCQ Questions for Class 11 Maths Chapter 10 Straight Lines
MCQ Questions for Class 11 Maths Chapter 11 Conic Sections
MCQ Questions for Class 11 Maths Chapter 12 Introduction to Three Dimensional Geometry
MCQ Questions for Class 11 Maths Chapter 13 Limits and Derivatives
MCQ Questions for Class 11 Maths Chapter 14 Mathematical Reasoning
MCQ Questions for Class 11 Maths Chapter 15 Statistics
MCQ Questions for Class 11 Maths Chapter 16 Probability
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