Question

solve this question immediately ​

Answer 1

Solution :-

We need to prove

cosθ + sinθ = √2 cosθ

Given

cosθ - sinθ = √2 sinθ

Squaring on both sides

→ ( cosθ - sinθ )² = ( √2 sinθ )²

Simplifying using

(a - b)² = a² + b² - 2ab

→ cos²θ + sin²θ - 2cosθsinθ = ( √2 )² sin²θ

→ cos²θ + sin²θ - 2cosθsinθ = 2sin²θ

Assuming as equation 1

Let cosθ + sinθ = x = √2 cosθ

Now, squaring on both sides

→ ( cosθ + sinθ )² = x²

Simplifying using

(a + b)² = a² + b² + 2ab

→ cos²θ + sin²θ + 2cosθsinθ = x²

Assuming as equation 2

Now , equation ( 1 + 2)

→ cos²θ + sin²θ + 2cosθsinθ + cos²θ + sin²θ - 2cosθsinθ = 2sin²θ + x²

→ 2cos²θ + 2sin²θ = 2sin²θ + x²

→ 2cos²θ = x²

→ x = √(2cos²θ)

→ x = √2 cosθ

Here , x = √2 cosθ

→ √2 cosθ = √2 cosθ

LHS = RHS

Hence, proved !!

Answer 2

Answer:

Solution :-

We need to prove

cosθ + sinθ = √2 cosθ

Given

cosθ - sinθ = √2 sinθ

Squaring on both sides

→ ( cosθ - sinθ )² = ( √2 sinθ )²

Simplifying using

(a - b)² = a² + b² - 2ab

→ cos²θ + sin²θ - 2cosθsinθ = ( √2 )² sin²θ

→ cos²θ + sin²θ - 2cosθsinθ = 2sin²θ

Assuming as equation 1

Let cosθ + sinθ = x = √2 cosθ

Now, squaring on both sides

→ ( cosθ + sinθ )² = x²

Simplifying using

(a + b)² = a² + b² + 2ab

→ cos²θ + sin²θ + 2cosθsinθ = x²

Assuming as equation 2

Now , equation ( 1 + 2)

→ cos²θ + sin²θ + 2cosθsinθ + cos²θ + sin²θ - 2cosθsinθ = 2sin²θ + x²

→ 2cos²θ + 2sin²θ = 2sin²θ + x²

→ 2cos²θ = x²

→ x = √(2cos²θ)

→ x = √2 cosθ

Here , x = √2 cosθ

→ √2 cosθ = √2 cosθ

LHS = RHS

Hence, proved !!