Question

Prove that 1/(sinθ+cosθ)+1/(sinθ–cosθ)=2sinθ/(1–2cos^2θ)

Answer 1

To prove :

1/(sinθ + cosθ) + 1/(sinθ - cosθ)

= 2 sinθ/(1 - 2 cos²θ)

Proof :

Now, 1/(sinθ + cosθ) + 1/(sinθ - cosθ)

= {(sinθ - cosθ) + (sinθ + cosθ)}/{(sinθ + cosθ)(sinθ - cosθ)}

= (sinθ + cosθ + sinθ - cosθ)/(sin²θ - cos²θ)

= 2 sinθ/(1 - cos²θ - cos²θ)

= 2 sinθ/(1 - 2 cos²θ)

⇒ 1/(sinθ + cosθ) + 1/(sinθ - cosθ)

= 2 sinθ/(1 - 2 cos²θ)

Hence, proved.

Trigonometric Rules :

• sin²θ + cos²θ = 1

• sec²θ - tan²θ = 1

• cosec²θ - cot²θ = 1