Question

Prove that (sin θ + cosec θ)² + (cos θ + sec θ)² = 7 + tan² θ + cot² θ. 

Answer 1

L.H.S = (sin θ + cosec θ)² + (cos θ + sec θ)² 

= (sin²θ + cosec²θ + 2 sin θ cosec θ + cos²θ + sec²θ + 2cos θ sec θ)

= (sin²θ + cos²θ) + (cosec²θ + sec²θ) + 2 sin θ `(1/("sin"theta)) + 2 cos theta (1/("cos" theta))`

= (1) + (1 + cot²θ + 1 + tan²θ) + (2) + (2)

= 7 + tan²θ + cot²θ 

= R.H.S

Answer 2

L.H.S = (sin θ + cosec θ)² + (cos θ + sec θ)² 

= (sin²θ + cosec²θ + 2 sin θ cosec θ + cos²θ + sec²θ + 2cos θ sec θ)

= (sin²θ + cos²θ ) + 1 + cot²θ + 2 sin θ x `1/sin θ` + 1 + tan² θ + 2cos θ. `1/cos θ`

= 1 + 1 + 1 + 2 + 2 + tan² θ + cot²θ

= 7 + tan² θ + cot²θ

= RHS

Hence proved.