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

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:

sin θ + cos θ = √3 cos θ

We need to find the value of cot θ.

Solution:

Step 1: Rearrange the given equation:

sin θ + cos θ = √3 cos θ

Rearranging gives:

sin θ = √3 cos θ - cos θ

sin θ = cos θ (√3 - 1).

Step 2: Use the definition of cot θ:

cot θ = cos θ / sin θ.

Substitute sin θ = cos θ (√3 - 1):

cot θ = cos θ / [cos θ (√3 - 1)].

Cancel cos θ from numerator and denominator:

cot θ = 1 / (√3 - 1).

Step 3: Rationalize the denominator:

cot θ = 1 / (√3 - 1) × (√3 + 1) / (√3 + 1).

cot θ = (√3 + 1) / (3 - 1).

cot θ = (√3 + 1) / 2.

The value of cot θ is (√3 + 1) / 2.

Given:

3 sin A + 4 cos A = 5

Formula Used:

Using the identity sin² A + cos² A = 1

Calculation:

Let sin A = x and cos A = y

3x + 4y = 5

And,

x² + y² = 1

⇒ x = 3/5 and y = 4/5

tan A = sin A / cos A

⇒ tan A = (3/5) / (4/5)

⇒ tan A = 3/4

The value of tan A is 3/4.

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:

\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.

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).

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:

cos(40° + x) = sin 30°

Formula used:

Cos 60° = sin 30° = 1/2

Calculation:

Cos (40° + x) = sin 30°

⇒ Cos (40° + x) = 1/2

⇒ Cos (40° + x) = cos 60

⇒ (40° + x) = 60°

⇒ x = 20° 

∴ The correct answer is 20°.