Multiple Angle MCQ Quiz - Objective Question with Answer for Multiple Angle - Download Free PDF

Given:

A = 15°

Expression = (11√2 cos(3A) + 10 sin(2A)) / (7√3 sin(4A))

Formula used:

Use trigonometric values for cos(3A), sin(2A), and sin(4A) at A = 15°

Calculations:

Substitute A = 15° into the expression:

cos(3A) = cos(45°) = 1 / √2

sin(2A) = sin(30°) = 1/2

sin(4A) = sin(60°) = √3 / 2

Substitute these values into the expression:

(11√2 × (1 / √2) + 10 × (1/2)) / (7√3 × (√3 / 2))

⇒ (11 × 1 + 5) / (7 × 3 / 2)

⇒ (16) / (21 / 2)

⇒ (16 × 2) / 21

⇒ 32 / 21

The value of the expression is 32/21.

Given:

If 4tanθ - 3 = 0, then the value of (1 - cos2θ)/(1 + cos2θ) is:

Formula used:

4tanθ - 3 = 0

tanθ = 3/4

We know, tan²θ = (1 - cos2θ) / (1 + cos2θ)

Calculation:

4tanθ - 3 = 0

⇒ 4tanθ = 3

⇒ tanθ = 3/4

We know, tan²θ = (1 - cos2θ) / (1 + cos2θ)

⇒ tan²θ = (3/4)²

⇒ (1 - cos2θ) / (1 + cos2θ) = 9/16

∴ The correct answer is option (4).

Given:

$\frac{1-sin(2t)}{1+sin(2t)}\times \frac{cos(t)+sin(t)}{cos(t)-sin(t)}$

Formula used:

sin 2t = 2 sint cost, 

cos 2t = 1 - 2 sin²t 

Calculation:

$\frac{1-sin(2t)}{1+sin(2t)}\times \frac{cos(t)+sin(t)}{cos(t)-sin(t)}$

Multiplying by the second term:

$\frac{cos(t)+sin(t)-sin(2t)[cos(t)+sin(t)]}{cos(t)-sin(t)+sin(2t)[cos(t)-sin(t)]}$

Applying formula: sin 2t = 2 sint cost

$\frac{cos(t)+sin(t)-2sin(t)co{s}^2(t)-2si{n}^2(t)cos(t)]}{cos(t)-sin(t)+2sin(t)co{s}^2(t)-2si{n}^2(t)cos(t)]}$

$\frac{cos(t)[1-2si{n}^2(t)]-sin(t)[2co{s}^2(t)-1]}{cos(t)[1-2si{n}^2(t)]+sin(t)[2co{s}^2(t)-1]]}$

$\frac{cos(t).cos(2t)-sin(t).cos(2t)}{cos(t).cos(2t)+sin(t).cos(2t)}$

Taking cos(2t) common and cancel out in numerator and denominator:

$\frac{cos(t)-sin(t)}{cos(t)+sin(t)}$

Dividing by cos(t) :

$\frac{1-\tan (t)}{1+\tan (t)}$

Option 2 is the correct answer.

Given:

If tan 4θ × tan 6θ = 1, where 6θ is an acute angle, we need to find the value of cot 5θ.

Concepts Used:

If tan(A) × tan(B) = 1, then A + B = 90° (angles are complementary).

cot(θ) = 1/tan(θ).

Solution:

Use the given condition:

tan 4θ × tan 6θ = 1

This implies:

4θ + 6θ = 90°

10θ = 90°

θ = 9°

Find cot 5θ:

cot 5θ = cot(5 × 9°)

cot 5θ = cot 45°

We know that cot 45° = 1.

The value of cot 5θ = 1.

Calculation:

$\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }}$

⇒ By putting θ = 15° 

$\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }}$

⇒ $\sqrt{\frac{1-\sin 45}{1+\sin 45}}$

⇒ $\sqrt{\frac{1-(1/\sqrt{2})}{1+(1/\sqrt{2})}}$ = $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$

⇒ $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}$

⇒ $\sqrt{\frac{(\sqrt{2}-1{)}^2}{2-1}}$

⇒ $\sqrt{(\sqrt{2}-1{)}^2}$

⇒ √2 - 1

Again, In option 1  By putting θ = 15° we get,

sec 3θ - tan 3θ = sec 45° - tan 45°

⇒ √2 - 1

∴ The correct option is 1

Alternate Method

$\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }}$

⇒ $\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }\times \frac{1-\mathrm{s}\mathrm{i}\mathrm{n}3\theta }{1-\mathrm{s}\mathrm{i}\mathrm{n}3\theta }}$

⇒ $\sqrt{\frac{(1-\sin 3\theta {)}^2}{1^2-{\sin }^{2}3\theta }}$

⇒ $\sqrt{\frac{(1-\sin 3\theta {)}^2}{\mathrm{c}\mathrm{o}{\mathrm{s}}^{2}3\theta }}$

⇒ $\sqrt{(\frac{1-\sin 3\theta }{\mathrm{c}\mathrm{o}\mathrm{s}3\theta }}{)}^2$

⇒ $\frac{1-sin3\theta }{cos3\theta }$

⇒ $\frac{1}{cos3\theta }-\frac{sin3\theta }{cos3\theta }$

⇒ sec 3θ - tan 3θ

∴ The correct option is 1

Given:

sin 2x + 2 sin 4x + sin 6x 

Formula used:

sin C + sin D = 2 × sin (C + D)/2 × cos (C - D)/2

cos 2θ = (2 × cos 2 θ  - 1)

Calculation:

sin 6x + sin 2x + 2 sin 4x

⇒ 2 × sin (6x + 2x)/2 × cos (6x - 2x)/2 + 2 sin 4x

⇒ 2 × sin 4x × cos 2x + 2 sin 4x

⇒ 2 × sin 4x (cos 2x + 1)

⇒ 2 × sin 4x {(2 × cos²x - 1) + 1) }

⇒ (2 × sin 4x) × (2 × cos2 x) 

⇒ 4 cos2 x  sin 4x

∴ The correct answer is 4 cos2 x  sin 4x.

Calculation:

$\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }}$

⇒ By putting θ = 15° 

$\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }}$

⇒ $\sqrt{\frac{1-\sin 45}{1+\sin 45}}$

⇒ $\sqrt{\frac{1-(1/\sqrt{2})}{1+(1/\sqrt{2})}}$ = $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}$

⇒ $\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}\times \frac{\sqrt{2}-1}{\sqrt{2}-1}}$

⇒ $\sqrt{\frac{(\sqrt{2}-1{)}^2}{2-1}}$

⇒ $\sqrt{(\sqrt{2}-1{)}^2}$

⇒ √2 - 1

Again, In option 1  By putting θ = 15° we get,

sec 3θ - tan 3θ = sec 45° - tan 45°

⇒ √2 - 1

∴ The correct option is 1

Alternate Method

$\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }}$

⇒ $\sqrt{\frac{1-\sin 3\theta }{1+\sin 3\theta }\times \frac{1-\mathrm{s}\mathrm{i}\mathrm{n}3\theta }{1-\mathrm{s}\mathrm{i}\mathrm{n}3\theta }}$

⇒ $\sqrt{\frac{(1-\sin 3\theta {)}^2}{1^2-{\sin }^{2}3\theta }}$

⇒ $\sqrt{\frac{(1-\sin 3\theta {)}^2}{\mathrm{c}\mathrm{o}{\mathrm{s}}^{2}3\theta }}$

⇒ $\sqrt{(\frac{1-\sin 3\theta }{\mathrm{c}\mathrm{o}\mathrm{s}3\theta }}{)}^2$

⇒ $\frac{1-sin3\theta }{cos3\theta }$

⇒ $\frac{1}{cos3\theta }-\frac{sin3\theta }{cos3\theta }$

⇒ sec 3θ - tan 3θ

∴ The correct option is 1

Given:

tan (A/2) = x

Formula used:

tan (A/2) = ± $\frac{\sqrt{1-\cos \mathrm{A}}}{\sqrt{1+\cos \mathrm{A}}}$

Calculation:

tan (A/2) = x

⇒ $\frac{\sqrt{1-\cos \mathrm{A}}}{\sqrt{1+\cos \mathrm{A}}}$ = x

⇒ x = $\frac{\sqrt{1-\cos \mathrm{A}}}{\sqrt{1+\cos \mathrm{A}}}$

∴ The correct option is 3.
Alternate Method 

Given:

tan3θ⋅tan7θ = 1

Concept used:

Calculation:

tan3θ⋅tan7θ = 1

⇒  tan3θ = 1/tan7θ

⇒  tan3θ = cot7θ

⇒  tan3θ = tan(90° - 7θ)

⇒ 3θ = 90° - 7θ

⇒ 10θ = 90°

⇒ θ = 9°

So, cot15θ = cot135° = - tan45°

⇒ - 1

∴ The required answer is - 1.

Given:

sec 2θ = cosec (θ – 36°)

Concept used:

Calculation:

sec 2θ = cosec (θ – 36°)

⇒ sec 2θ = sec (90° – θ + 36°)

⇒ sec 2θ = sec (126° – θ)

⇒ 2θ = 126° – θ

⇒ 3θ = 126°

⇒ θ = 42°

∴ The required answer is 42°.

Given:

sin Y = x

Formula used:

cos 2Y = 1 - 2sin² Y

Calculations:

According to the question,

⇒ cos 2Y = 1 - 2sin2 Y

⇒ cos 2Y = 1 - 2(sin Y)²

⇒ cos 2Y = 1 - 2(x)²

∴ The value of cos 2Y is 1 - 2(x)2.

Given:

cot A = 15/8

Concept used:

1. tan 2α = $\frac{2\times tan\alpha }{(1-ta{n}^2\alpha )}$

2. tan α = 1/cot α 

Calculation:

cot A = 15/8

⇒ tan A = 8/15

Now, tan 2 A

⇒ $\frac{2\times tanA}{(1-ta{n}^{2}A)}$

⇒ $\frac{2\times \frac{8}{15}}{(1-(\frac{8}{15}{)}^2)}$

⇒ $\frac{\frac{16}{15}}{\frac{161}{225}}$

⇒ $\frac{16\times 225}{15\times 161}$

⇒ 240/161

∴ The value of tan 2A is 240/161.

Given:

sec 3θ = cosec (4θ - 15°)

Calculations:

According to question,

sec 3θ = cosec (4θ - 15)

⇒ cosec (90° - 3θ) = cosec (4θ – 15°)

⇒ 90° - 3θ = 4θ - 15°

⇒ 7θ = 90° + 15°

⇒ θ = 105°/7 = 15°

so, tan3θ

= tan(3 × 15°)

= tan45°

= 1

Hence, The Required value is 1.

Given:

The expression = $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta }}}$

Formula used:

1 + cosθ = 2cos²(θ/2)

Calculation:

According to the question

⇒ $\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta }}}$

⇒ $\sqrt{2+\sqrt{2+\sqrt{2(1+cos8\theta )}}}$

⇒ $\sqrt{2+\sqrt{2+\sqrt{2(2co{s}^{2}4\theta )}}}$

⇒ $\sqrt{2+\sqrt{2+2cos4\theta }}$

⇒ $\sqrt{2+\sqrt{2(1+cos4\theta )}}$

⇒ $\sqrt{2+\sqrt{2(2co{s}^{2}2\theta )}}$

⇒ $\sqrt{2+2\cos 2\theta }$

⇒ $\sqrt{2(1+\cos 2\theta )}$

⇒ $\sqrt{2(2{\cos }^2\theta )}$

⇒ 2cosθ 

∴ The required result will be 2cosθ. 

Given:

sin 2x + 2 sin 4x + sin 6x 

Formula used:

sin C + sin D = 2 × sin (C + D)/2 × cos (C - D)/2

cos 2θ = (2 × cos 2 θ  - 1)

Calculation:

sin 6x + sin 2x + 2 sin 4x

⇒ 2 × sin (6x + 2x)/2 × cos (6x - 2x)/2 + 2 sin 4x

⇒ 2 × sin 4x × cos 2x + 2 sin 4x

⇒ 2 × sin 4x (cos 2x + 1)

⇒ 2 × sin 4x {(2 × cos²x - 1) + 1) }

⇒ (2 × sin 4x) × (2 × cos2 x) 

⇒ 4 cos2 x  sin 4x

∴ The correct answer is 4 cos2 x  sin 4x.