Trigonometry MCQ Quiz - Objective Question with Answer for Trigonometry - Download Free PDF

Formula Used:

(a + b)² = a² + 2ab + b²,

sin2 θ+cos² θ = 1 and 1 - sin2 θ = cos² θ

cot θ = cos θ/sin θ and cosec θ = 1/sin θ

and for equations a/b = c/d

Using the Componendo and Dividendo rule, we get 

(a + b)/(a - b) = (c + d)/(c - d)

or (a-b)/(a+b) = (c-d)/(c+d)

Calculation:

1/sin θ + cos θ/sin θ = p

⇒ (1+cos θ)/sin θ = p/1

Squaring both sides of the given equation:
(1 + cos θ)²/sin² θ = p²/1

⇒ (1 + cos² θ + 2cos θ)/sin2 θ = p²/1

Applying Componendo and Dividendo, we get

(1 + cos² θ + 2cos θ - sin2 θ)/(1 + cos² θ + 2cos θ + sin2 θ) = (p²-1)/(p²+1)

⇒ (1 - sin2 θ + cos² θ + 2cos θ )/(1 + cos² θ+ sin2 θ + 2cos θ ) = (p²-1)/(p2+1)

⇒ (cos² θ + cos² θ + 2cos θ )/(1 + 1 + 2cos θ ) = (p²-1)/(p2+1)

⇒ (2cos² θ + 2cos θ )/(2 + 2cos θ ) = (p²-1)/(p2+1)

⇒ 2cos θ(cos θ + 1 )/2(1 + cos θ ) = (p²-1)/(p2+1)

⇒ cos θ = (p²-1)/(p2+1)

Thus, value of  (p²-1)/(p2+1) is cos θ.

Given:

If p sin A - cos A = 1

Formula Used:

sin 90^∘ = 1

cos 90∘ = 0

Calculation:

p sin A - cos A = 1

To get the value of p put A = 90^∘

\( (p \sin 90^\circ - \cos 90^\circ) = 1 \)

p × 1 - 0 = 1

p = 1

Putting A = 90° and p = 1 in p2 - (1 + p2) cos A:

p2 - (1 + p2) cos A = 1² - (1 - 1²)cos 90° 

⇒ 1 - (1 + 1) × 0 

⇒ 1 - 2 × 0 = 1 - 0 = 1

∴ The value of p2 - (1 + p2) cos A = 1.

Alternate Method

Concept used:

if cosecθ + cotθ = x, then cosecθ - cotθ = 1/x

Calculation:

p sin A - cos A = 1

⇒ p sin A = 1 + cos A

⇒ p = (1 + cos A)/ sinA

⇒ p = cosec A + cot A

So, cosec A - cot A = 1/p

Now,

p + 1/p = 2 cosec A

p - 1/p = 2 cot A

So, p2 - (1 + p2) cos A

⇒ p²- p(1/p + p) cosA

⇒ p² - p(2cosecA) cosA

⇒ p²- p (2/sin A) cosA

⇒ p² - p (2 cotA)

⇒ p² - p (p - 1/p)

⇒ p² - p² + 1 = 1

∴ The correct answer is option (1).

Explanation:

tan θ = tan α

∴ θ = nπ + α 

Concept:

tan-1 x + cot-1 x = \(\rm \frac {\pi}{2}\)

Calculation:

As we know tan-1 x + cot-1 x = \(\rm \frac {\pi}{2}\)

∴ cot (tan-1 x + cot-1 x) = cot \(\rm \frac {\pi}{2}\)= 0

Given:

3sin θ + 5cos θ = 5

Concept:

sin² θ + cos² θ = 1

a² + b² + 2ab = (a + b)²

a2 + b2 - 2ab = (a - b)2

Calculation:

Let 5sin θ - 3cos θ = x    ....(1)

And,

3sin θ + 5cos θ = 5       ....(2)

Squaring in both the equation:

9sin² θ + 25cos² θ + 30sin θ. cos θ= 25                     ....(3)

25sin2 θ + 9cos2 θ - 30sin θ. cos θ  = x²               .....(4)

From equation (3) and equation (4)

⇒ 34 (sin2 θ + cos2 θ) = 25 + x2

⇒ 9 = x2

x = 3 and -3

∴ 5sin θ - 3cos θ = 3

Formula used:

sec (180° - θ) = - sec θ 

cosec (180° - θ) = cosec θ

cos θ × sec θ = 1 ; sin θ × cosec θ = 1

Calculation:

cos 47° sec 133° + sin 44° cosec 136°

⇒ cos 47° × sec (180° - 47) + sin 44° cosec (180° - 44°)

⇒ cos 47° × (- sec 47°) + sin 44° × (cosec 44°)

⇒ -1 + 1 = 0

∴ The correct answer is 0.

Given:

\(\frac{\cos 45^{\circ }}{\sec 30^{\circ}+ cosec30^{\circ}}\)

Concept used:

Calculation:

\(\frac{\cos 45^{\circ }}{\sec 30^{\circ}+ cosec30^{\circ}}\)

⇒ \(\frac {\frac {1}{\sqrt2}} {\frac {2}{\sqrt3}+ \frac {2}{1}}\)

⇒ \(\frac {\frac {1}{\sqrt2}} {2(\frac {\sqrt3 + 1}{\sqrt3})}\)

⇒ \(\frac {\sqrt3} {2{\sqrt2}({\sqrt3 + 1})}\)

⇒ \(\frac {\sqrt3({\sqrt3 - 1})} {2{\sqrt2}({\sqrt3 + 1})({\sqrt3 - 1})}\)

⇒ \(\frac {\sqrt3({\sqrt3 - 1})} {2{\sqrt2}({3 - 1)}}\)

⇒ \(\frac {({3 - \sqrt3})} {4{\sqrt2}}\)

⇒ \(\frac {({3\sqrt2 - \sqrt6})} {8}\)

∴ The required answer is \(\frac {({3\sqrt2 - \sqrt6})} {8}\).

Given:

tan2θ + cot2θ  - sec2θ cosec2θ
Concept used:

1. tanα = sinα/cosα

2. cotα = 1/tanα

3. secα = 1/cosα

4. cosecα = 1/sinα

5. (a + b)² - 2ab = a² + b²

6. sin²α + cos²α = 1

Calculation:

tan2θ + cot2θ  - sec2θ cosec2θ

⇒ \(\frac {sin^2θ}{cos^2θ} + \frac {cos^2θ}{sin^2θ} - \frac {1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {sin^4θ + cos^4θ - 1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {(sin^2θ + cos^2θ)^2 - 2sin^2θ cos^2θ - 1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {(1)^2 - 2sin^2θ cos^2θ - 1}{sin^2θ \times cos^2θ}\)

⇒ \(\frac {-2sin^2θ cos^2θ}{sin^2θ \times cos^2θ}\)

⇒ -2

∴ The required answer is -2.

Shortcut Trick 

Use value putting method to solve this question, 

Use θ = 45° 

tan2θ + cot2θ  - sec2θ cosec2θ
⇒ 12 + 12  - (√2)2(√2)2

⇒ 1 + 1 - 4

⇒ 2 - 4 = - 2

∴ The correct answer to this question is -2.

Given:

cos 2A cos 2B + sin2(A - B) - sin2(A + B)

Concept used:

cos (a + b) = cos a cos b - sin a sin b

sin²a - sin²b = sin(a + b) sin(a - b)

Calculation:

cos 2A cos 2B + sin2(A - B) - sin2(A + B)

⇒ cos 2A cos 2B - [sin2(A + B) - sin2(A - B)] 

{sin2a - sin2b = sin(a + b) sin(a - b)}

⇒ cos 2A cos 2B - [sin(A + B + A - B) sin(A + B - A + B)]

⇒ cos 2A cos 2B - [sin(A + A) sin(B + B)]

⇒ cos 2A cos 2B - sin 2A sin 2B

⇒ cos (2A + 2B)

∴ The required answer is cos (2A + 2B).

Given:

sec θ + tan θ = 5

Concept used:

If sec θ + tan θ = y

then sec θ - tan θ = 1/y

Calculation:

sec θ + tan θ = 5  ----- (1)

then,

sec θ - tan θ = 1/5 ------- (2)

Subtracting the eq. (1) and (2)

⇒ (sec θ + tan θ) - (sec θ - tan θ) = (5 - 1/5)

⇒ sec θ + tan θ - sec θ + tan θ = 24/5

⇒ 2 × tan θ = 24/5

⇒ tan θ = 12/5

∴ The correct answer is 12/5.

Given:

cos (36° - A) cos (36° + A) + cos (54° - A) cos (54° + A)

Formula used:

cos (a - b) = cos a cos b + sin a sin b.

sin (90 - a) = cos a

Calculation:

⇒ sin[90 – (36 – A)]sin[90 – (36 + A)] + cos (54° – A) cos (54° + A)

⇒ sin(54^º + A)sin(54^º – A) + cos (54° – A)cos (54° + A)

⇒ Using the identity cos(A – B),

⇒ cos(54 + A – 54 + A) = cos(2A)

Therefore, the value of cos (36° - A) cos (36° + A) + cos (54° - A) cos (54° + A) is cos(2A).

Concept:

sin (2nπ ± θ) = ±  sin θ

sin (90 + θ) = cos θ

Calculation:

Given: sin (1920°)

⇒ sin (1920°) = sin(360° × 5° + 120°) = sin (120°)

Given:

{(3Sinθ – Cosθ)/(Cosθ + Sinθ)} = 1

Calculation:

We have a trigonometric equation 

{(3Sinθ – Cosθ)/(Cosθ + Sinθ)} = 1

On dividing numerator and denominator by sin θ, we get

⇒ [{(3sinθ – cosθ)/Sinθ}/{(cosθ + sinθ)/sinθ}] = 1

⇒ {(3 – cotθ)/(cotθ + 1)} = 1

⇒ 3 – cotθ = 1 + cotθ

⇒ 2cotθ = 2

cotθ = 1

The value 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.

Given:

a cot θ + b cosec θ = p

b cot θ + a cosec θ = q

Formula used:

Cosec² θ - cot² θ = 1

Calculation:

a cot θ + b cosec θ = p

Squaring both sides

(a cot θ + b cosec θ)² = (p)²

a² cot² θ  + b2 cosec2 θ  + 2 × ab cot θ × cosec θ = p² ----- (1)

b cot θ + a cosec θ = q

Squaring both sides

(b cot θ + a cosec θ)2 = (q)2

b2 cot2 θ  + a2 cosec2 θ  + 2 × ab cot θ × cosec θ = q2 ----- (2)

Substracting the eq.(1) and (2)

⇒ (p² - q²) = a2 cot2 θ  + b2 cosec2 θ  + 2 × ab × cot θ × cosec θ - (b2 cot2 θ + a2 cosec2 θ  + 2 × ab × cot θ × cosec θ)

⇒ a2 cot2 θ - a2 cosec2 θ + b2 cosec2 θ - b2 cot2 θ