Question

Prove that (sinθ + cosecθ)² + (cosθ + secθ)² = 7 + tan²θ + cot²θ.

Answer 1

LHS = (sinθ + cosecθ)² + (cosθ + secθ)²

= sin²θ + cosec²θ + 2sinθ . cosecθ + cos²θ + sec²θ + 2cosθ . secθ  ...[∵ (a + b)² = a² + 2ab + b²]

= `(sin^2θ + cos^2θ) + (1 + cot^2θ) + 2sinθ 1/sinθ + (1 + tan^2θ) + 2cosθ1/(cosθ)`  `[∵ cosec^2θ = 1 + cot^2θ, sec^2θ = 1 + tan^2θ,
cosecθ = 1/(sinθ) and secθ = 1/cosθ ]`

= 1 + 1 + cot²θ + 2 + 1 + tan²θ + 2  ...[∵ sin²θ + cos²θ = 1]

= 7 + tan²θ + cot²θ

= RHS, Hence proved.

Answer 2

LHS = (sinθ + cosecθ)² + (cosθ + secθ)²

= sin²θ + cosec²θ + 2sinθ . cosecθ + cos²θ + sec²θ + 2cosθ . secθ  

= (sin²θ + cos²θ) + cosec²θ + sec²θ + 2 + 2      ...(∵ sinθ · cosecθ = 1 and secθ · cosθ = 1]

= 1 + cosec²θ + sec²θ + 4    ...[∵ sin²θ+ cos²θ = 1]

= 5 + (1 + cot²θ) + (1 + tan²θ)

= 7 + cot²θ + tan²θ = R.H.S.