Question

If cosƟ + sinƟ = √2cosƟ, prove that cosƟ – sinƟ =√2 sinƟ

Answer 1

Answer:

hopefully this will help you

Answer 2

Given :-

• cosθ + sinθ = √2 cosθ ...eqn(1)

To prove :-

• cos θ - sinθ = √2 sinθ

Proof :-

Given that

⇨ cosθ + sinθ = √2 cosθ

squaring both sides

⇨ (cosθ + sinθ)² = (√2 cosθ)²

Using identity

(a+b)² = a² + b² + 2 ab in LHS

⇨ cos²θ + sin²θ + 2sinθcosθ = 2cos²θ

⇨ 2sinθcosθ = 2cos²θ - cos²θ - sin²θ

⇨ 2sinθcosθ = cos²θ - sin²θ

Using identity

a² - b² = (a+b) (a-b) in RHS

⇨ 2sinθcosθ = (cosθ + sinθ)(cosθ - sinθ)

Using eqn(1) cosθ + sinθ = √2 cosθ

⇨ 2sinθcosθ = (√2 cosθ) (cosθ - sinθ)

⇨ cosθ - sinθ = (2sinθcosθ)/(√2 cosθ)

⇨ cosθ - sinθ = √2 sinθ

Proved .

▶ Know some trigonometric identites & Ratios

→ sin²A + cos²A = 1

→ 1 + tan²A = sec²A

→ 1 + cot²A = cosec²A

→ sin(90 - A) = cos A

→ cos (90 - A) = sin A

→ cosec(90 - A) = sec A

→ sec(90-A) = cosec A

→ tan(90 - A) = cot A

→ cot (90 - A) = tan A

→ cosec A = 1 \ sin A

→ sec A = 1 \ cos A

→ tan A = sin A / cos A

→ cot A = cos A / sin A