Trigonometric Values MCQ Quiz in मल्याळम - Objective Question with Answer for Trigonometric Values - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

tan 240°

Concept used:

Calculation:

tan240°

⇒ tan(180 + 60)°

⇒ tan60°

⇒ √3

∴ The required answer is √3.

Formula used:

Cos 2A = 2Cos² A  - 1

Calculation:

According to the formula, 

⇒ Cos 2A = 2Cos² A - 1

⇒ Let A = 15

⇒ Cos 30 = 2Cos² 15 - 1

⇒ \(\frac{\sqrt3} {2} +1 = 2cos^2 15\)

⇒ \( 2cos^2 15 = \frac{\sqrt3+2} {2}\)

⇒ \( cos^2 15 = \frac{\sqrt3+2} {4}\)

⇒ Hence, Option 2 is correct 

Concept used :

tan(90° - θ) = cot θ 

cot (90° - θ) = tan θ 

sec (90° - θ) = cosec θ 

cosec²θ - cot²θ = 1 

sec²θ - tan²θ = 1 

Calculations :

 sec2 54° - cot2 36° + \(\frac{3}{2}\) sin2 37° × sec2 53° + \(\frac{2}{\sqrt3}\) tan 60°

⇒  sec2 54° - cot2 (90° - 54°) + \(\frac{3}{2}\) sin2 37° × sec2 (90° - 37°) + \(\frac{2}{\sqrt3}\) tan 60°

⇒  sec2 54° - tan2 54° + \(\frac{3}{2}\) sin2 37° × cosec2 37° + \(\frac{2}{\sqrt3}\) tan 60°

⇒ 1 + 3/2 + 2 

⇒ 9/2

The value is 9/2.

Formula used:

Tan (- θ) = - tan θ 

Tan (360 + θ) = tan θ 

Calculation:

Tan (−1125°) = - tan 1125°

⇒ - tan 1125° = - tan {(360 × 3) + 45)

⇒ - tan {1080 + 45} = - tan 45° = - 1

∴ The correct answer is -1.

Given:

sin 23° = a/b

Concepts used:

In a right-angled triangle, applying Pythagoras Theorem,

H² = P² + B²

⇒ P² = H² – B²

P → Perpendicular, B → base, H → Hypotenuse of right-angled triangle

sinθ = P/H

cosθ = B/H

tanθ = P/B

Calculation:

sin23° = P/H = a/b

On comparing,

⇒ P = a units and H = b units

Using Pythagoras' Theorem,

⇒ B² = H² – P²

⇒ B² = b² – a²

⇒ B = √b2 – a2

⇒ sec23° = H/B = b/√b2 – a2

⇒ Cos23° = sin67° = √b2 – a2 /b

sec 23° - sin 67° ⇒  b/√b2 – a² - √b2 – a2 /b

sec 23° - sin 67° ⇒ \( \frac{a^2}{b \sqrt{b^2-a^2}}\)

The value is \( \frac{a^2}{b \sqrt{b^2-a^2}}\)

Given:

sin θ + cos θ = \(\sqrt5\) sin(90 - θ)

Concept used:

1. cot θ = cos θ ÷ sin θ

2. a² - b² = (a + b)(a - b)

Calculation:

sin θ + cos θ = \(\sqrt5\) sin(90 - θ)

⇒ sin θ + cos θ = \(\sqrt5\) cosθ

⇒ cos θ(\(\sqrt5\) - 1) = sinθ

⇒ cot θ = \(\frac {1}{\sqrt5 - 1}\)

⇒ cot θ = \(\frac {\sqrt5 + 1}{(\sqrt5 - 1)(\sqrt5 + 1)}\)

⇒ cot θ = \(\frac {\sqrt5 + 1}{5 - 1}\)

⇒ cot θ = \(\frac {\sqrt5 + 1}{4}\)

∴ The value of cot θ is \(\frac {\sqrt5 + 1}{4}\).

Formula used:

Sin (360 + θ) = sin θ

Calculation:

Sin (- 405°) = - sin (405°)

⇒ - sin (360 + 45)

⇒ - sin 45° = - (1/√2)

 ∴ The correct answer is - (1/√2).

Given:

PR = 17

QR = 8

PQ = 15

Concept used:

Sinθ = P/H

Cosθ = B/H

Here,

P = Perpendicular

B = Base

H = Hypotenuse

Calculation:

sin α + cos α

⇒ 8/17 + 15/17

⇒ (8 + 15)/17

⇒ 23/17

∴ The required answer is \(\frac{23}{17}\).

Given: 

tan A = \({{5} \over 12}\)

Calculation:

As we know tan = P/B and sin θ = P/H

So, H = √(P² + B²)

⇒ H = √(52 + 122)

⇒ H = √(25 + 144)

⇒ H = √169

⇒ H = 13

Now,

sin A + sin B + sin C

= sin A + sin 90° + sin C

= \({{5} \over 13}\) + 1 + \({{12} \over 13}\)

⇒ \({{5\,+\,13\,+12} \over 13}\)

⇒ \({{30} \over 13}\)

⇒ \(2{{4} \over 13}\)

∴ The required answer is \(2{{4} \over 13}\).

Concept Used:

Calculation:

cos60 = 1/2

cos240 = cos(180 + 60) = -cos60 = - 1/2

cos360 = cos(360 - 0) = cos0 = 1

Now according to the question, 

⇒ (1/2)³ - (-1/2)³ -  (1)³

⇒ 1/8 + 1/8 - 1 = 2/8 - 1 = -3/4

∴ The value of the given expression is -3/4.