Trigonometric Ratios and Identities MCQ Quiz - Objective Question with Answer for Trigonometric Ratios and Identities - Download Free PDF

Given:

Expression = sinθ/cosθ

Calculations:

The ratio of the sine of an angle to the cosine of the same angle is defined as the tangent of that angle.

⇒ \(\dfrac{\sin\theta}{\cos\theta}\) = tanθ

∴ The correct  answer is option 1.

Given:

In a right-angled triangle, perpendicular = base

Formula used:

sinθ = opposite / hypotenuse

cosθ = adjacent / hypotenuse

tanθ = opposite / adjacent

Calculations:

Since perpendicular = base, the triangle is isosceles right-angled.

Let each side = 1 unit.

Hypotenuse = √(1² + 1²) = √2

sinθ = 1 / √2 = 1 / 1.414 = 0.707 ≈ 1/√2

cosθ = 1 / √2 = 1/√2 (not 1/2)

tanθ = 1 / 1 = 1 (not √3)

sinθ ≠ 1/2

∴ The correct statement is: sinθ = 1/√2.

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).

Given:

tan2A - 6tanA + 9 = 0

Concept used:

tanθ = P/B

cotθ = 1/tanθ

cosθ = B/H

Here,

P = Perpendicular

B = Base

H = Hypotenuse 

Pythagoras theorem = H2 = P2 + B2

Calculation:

tan2A - 6tanA + 9 = 0

⇒ tan2A - 2 × 3 × tanA + 9 = 0

⇒ (tanA - 3)² = 0

⇒ (tanA - 3) = 0

⇒ tanA  = 3

So, P/B = 3/1

Now,

H² = 3² + 1²

⇒ H2 = 9 + 1

⇒ H2 = 10

⇒ H = √10

So, cosA = 1/√10

Now,

6 cot A + 8\(√{10} \) cos A = 6 × (1/3) + 8\(√{10} \) × (1/√10)

⇒ 2 + 8

⇒ 10

∴ Required answer is 10.

Shortcut Trick

tan2A - 6tanA + 9 = 0

⇒ (tanA - 3)2 = 0

⇒ (tanA - 3) = 0

⇒ tanA  = 3

⇒ cotA = 1/3

cosA = 1 ÷  \(\sqrt{sec^2A}\)

⇒ 1 ÷\(\sqrt{1 + tan^2A}\)

⇒ 1 ÷\(\sqrt{1 + 9}\) = \(\frac {1}{\sqrt{10}}\)

Now,

6 cot A + \(8\sqrt{10}\) cos A

⇒ 6 × (1/3) + \(8\sqrt{10}\) × \(\frac {1}{\sqrt{10}}\)

⇒ 2 + 8 = 10

∴ The value of 6 cot A + 8\(\sqrt{10} \) cos A is 10.

Given:

cos (x - y) = √3/2

sin (x + y) = 1/2

Formula:

cos30° = √3/2

sin30° = 1/2

Calculation:

cos(x - y) = √3/2

⇒ cos(x - y) = cos30°

(x - y) = 30°     ---- (1)

sin(x + y) = 1/2

⇒ sin(x + y) = sin30°

(x + y) = 30    ---- (2)

Add equation (1) and equation (2), we get

2x = 60°

∴ x = 30°

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.

Concept used: 

sec²(x) = 1 + tan²(x)

Calculation:

⇒ sec²θ + tan²θ = 5/3

⇒ 1 + tan²θ + tan²θ = 5/3

⇒ 2tan²θ = 2/3

⇒ tanθ = 1/√3

⇒ θ = 30

Given:

tanθ + cotθ = √3

Formula used:

(a + b)³ = a³ + b³ + 3ab(a + b)

a² + b² = (a + b)² - 2(a × b)

tanθ × cotθ = 1

Calculation:

tanθ + cotθ = √3

Taking cube on both sides, we get

(tanθ + cotθ)³ = (√3)³

⇒ tan³θ + cot³θ + 3 × tanθ × cotθ × (tanθ + cotθ) = 3√3

⇒ tan3θ + cot3θ + 3√3  = 3√3

⇒ tan3θ + cot3θ = 0  

Taking square on the both sides

(tan3θ + cot3θ)² = 0

⇒ tan⁶θ + cot⁶θ + 2 × tan3θ × cot3θ = 0

⇒ tan6θ + cot6θ + 2 = 0    

⇒ tan6θ + cot6θ = - 2

∴ The value of tan6θ + cot6θ is - 2.

As,

⇒ sec²θ = 1 + tan²θ

We have,

⇒ (sec²θ)² – sec²θ = 3

⇒ (1 + tan²θ)² – (1 + tan²θ) = 3

⇒ (1 + tan⁴θ + 2tan²θ) – (1 + tan²θ) = 3

⇒ 1 + tan⁴θ + 2tan²θ – 1 – tan²θ = 3

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.

Formula Used:

1/Cosec Ø = Sin Ø 

Sin²Ø + Cos²Ø = 1

Calculation:

Cos2Ø + 1/Cosec2Ø + 17 = x

⇒ Cos2Ø + Sin2Ø + 17 = x

⇒ 1 + 17 = x

⇒ x = 18

⇒ x² = 324

∴ The value of x² is 324.

Given:

secθ - cosθ = 14 and 14 secθ = x

Concept used:

\(Sec\theta =\frac{1}{Cos\theta}\)

Calculations:

According to the question,

⇒ \(sec\theta - cos\theta= 14\)

⇒ \(\sec\theta-\frac{1}{sec\theta}=14\)

⇒ \( sec²\theta-1=14sec\theta\)

⇒ \(\tan^2\theta=14sec\theta\)      ----(\(sec²\theta-1=tan^2\theta\))

\(\ tan²\theta=x\)

∴ The value of x is \(tan²\theta\).

Calculation:

cot⁴θ + cot²θ = 3

⇒ cos⁴θ/sin⁴θ + cos ²θ/sin ²θ 

⇒ cos 2θ(cos 2θ+ sin 2θ )/sin4θ = 3 (Taking LCM)

⇒ cos 2θ/sin⁴θ = 3

⇒ cot²θ . cosec²θ = 3

Now, cosec⁴θ – cosec²θ

⇒ cosec²θ(cosec²θ – 1)

⇒ cosec²θcot²θ = 3

∴ cosec4θ – cosec2θ = 3

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).