Pythagorean Identities MCQ Quiz in मल्याळम - Objective Question with Answer for Pythagorean Identities - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Given:

Sec θ + 1/cos θ = 2

Concept used:

Calculation:

According to the question:

Sec θ + 1/cos θ = 2

Sec θ + sec θ  = 2

So, θ = 0° 

Now putting the value of θ in equation:

⇒ Sec⁵⁵ θ + 1/sec⁵⁵ θ 

⇒ Sec55 0° + 1/sec55 0°

⇒ 1 + 1 = 2

∴ The correct answer is 2.

Given:

Cos² θ = 3/4

Formula used:

Cos 30° = Sin 60° = √3/2

Calculation:

Cos2 θ = 3/4

⇒ cos θ = √(3/4) = √3/2

⇒ cos θ  = cos 30° 

⇒ θ = 30° 

sin (θ + 30°) =  sin (30° + 30°)

⇒ sin 60° = √3/2

∴ The correct answer is √3/2.

Given:

\sin A \sin B

Formula Used:

2 \(\sin A \sin B = \left[ \cos (A - B) - \cos (A + B) \right]\)

Calculation:

We know that:

2 \(\sin A \sin B = \left[ \cos (A - B) - \cos (A + B) \right]\)

Therefore,

sin A sin B  =\(\frac{1}{2} \{ \cos(A - B) - \cos(A + B)\) }

Thus, the correct answer is option 4.

Formula used:

1 - sin² θ = cos² θ

Sec θ × cos θ = 1

Calculation:

Sec θ × √{1 - sin2 θ}

⇒ Sec θ × √{cos² θ} 

⇒ Sec θ × cos θ = 1

∴ The correct answer is 1. 

Given:

sec2A + tan2A = 3

Concept used:

sec2 α - tan2 α = 1

Calculation:

sec2A + tan2A = 3      ....(1)

sec2A - tan2A = 1      ....(2)

Solving (1) and (2) we get,

sec2A + tan2A - sec2A + tan2A = 3 - 1 = 2

2tan² A = 2

tan2 A = 1

So, tan A = √1 = 1

Now, cot A = 1/1 = 1

∴ The value of cot A is 1.

Given: 

If sec θ + tan θ = x, find sinθ.

Formulae Used:

(secθ + tanθ) (secθ - tanθ) = 1

Solution:

sec θ + tan θ = x   ---(1)

So, sec θ - tan θ = 1/x  ---(2)

Subtracting equation (2) from equation (1):

sec θ + tan θ - (sec θ - tan θ) = x - 1/x

⇒ sec θ + tan θ - sec θ + tan θ = (x² - 1)/x

⇒ 2 tan θ = (x² - 1)/x

⇒ tan θ = (x2 - 1)/2x

We know that tan θ = p/b

So, p = (x² - 1), b = 2x

h² = p² + b² = (x² - 1)² + (2x)²

⇒ h² = (x²)² + 1 - 2x² + 4x²

⇒ h2 = (x2)2 + 1 + 2x2

⇒ h2 = (x² + 1)²

⇒ h = (x² + 1)

So, sin θ = p/h = (x2 - 1) / (x2 + 1)

∴ The correct answer is option (3).

Formula Used : 

Sin²θ + Cos²θ = 1

Calculation : 

⇒ (sin θ + cos θ)2 + (sin θ - cos θ)2

⇒ sin²θ + cos²θ + 2sinθcosθ + sin2θ + cos2θ - 2sinθcosθ 

⇒ 2(sin²θ + cos²θ)

⇒ 2

∴ The correct answer is 2.

Given:

\(\tan(t) = \frac{1}{3}\)

Formula Used:

\(\sec(t) = \sqrt{1 + \tan^2(t)}\)

Calculation:

\(\tan(t) = \frac{1}{3}\)

\(\tan^2(t) = \left(\frac{1}{3}\right)^2 = \frac{1}{9}\)

\(\sec(t) = \sqrt{1 + \tan^2(t)}\)

⇒ \(\sec(t) = \sqrt{1 + \frac{1}{9}}\)

⇒ \(\sec(t) = \sqrt{\frac{9}{9} + \frac{1}{9}}\)

⇒ \(\sec(t) = \sqrt{\frac{10}{9}}\)

⇒ \(\sec(t) = \frac{\sqrt{10}}{3}\)

The value of sec(t) is \(\frac{\sqrt{10}}{3}\) .

Calculations:

\(\frac{1}{cosec θ - \cot θ} - \frac{1}{\sin θ}\)

We have,

cosec θ - cot θ = (1/sinθ) - (cosθ/sinθ) = (1 - cos θ)/sin θ

So,

\(\frac{1}{cosec θ - \cot θ} - \frac{1}{\sin θ}\)

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

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

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

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

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

⇒ \(\frac{\cos θ}{\sin θ}\)

⇒  cot θ

∴ The correct answer is option (3).

Given:

Cot A = 12/5

Formula used:

Pythagorean theorem:

H² = P² + B²

Cot A = B/P ; sin A = P/H ; cos A = B/H ; cosec A = H/P

Where, H = hypotenuse ; P = perpendicular ; B = Base

Calculation:

Cot A = 12/5 = B/P

Using Pythagorean theorem:

H2 = P2 + B2

⇒ H² = (5)² + (12)²

⇒ H = √(25 + 144) = √169

⇒ H = 13 cm

Now,

(sin A + cos A) × cosec A

⇒ {(5/13) + (12/13)} × 13/5

⇒ (17/13) × (13/5) = 17/5

∴ The correct answer is 17/5.