Question

Prove the following identity : 

`[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`

Answer 1

LHS = `[1/((sec^2θ - cos^2θ)) + 1/((cosec^2θ - sin^2θ))](sin^2θcos^2θ) `

= `[1/((1/cos^2θ - cos^2θ)) + 1/((1/sin^2θ - sin^2θ))](sin^2θcos^2θ)`

= `[1/(((1 - cos^4θ)/cos^2θ)) + 1/(((1 - sin^4θ)/sin^2θ)]](sin^2θcos^2θ)`

= `[cos^2θ/(1 - cos^4θ) + sin^2θ/(1 - sin^4θ))](sin^2θcos^2θ)`

= `[(cos^2θ - cos^2θsin^2θ + sin^2θ  - sin^2θcos^4θ)/((1 - cos^4θ)(1 - sin^4θ))] (sin^2θcos^2θ)]`

= `[(cos^2θ + sin^2θ - cos^2θsin^2θ(cos^2θ + sin^2θ))/((1 - cos^2θ)(1 + cos^2θ)(1 - sin^2θ)(1 + sin^2θ))](sin^2θcos^2θ)`

= `[(1 - cos^2θsin^2θ)/(sin^2θ(1 + cos^2θ)cos^2θ(1 + sin^2θ))](sin^2θcos^2θ)`

(∵ `cos^2θ + sin^2θ = 1` , (`1 - cos^2θ`) = `sin^2θ` , (`1 - sin^2θ) = cos^2θ`)

= `(1 - cos^2θsin^2θ)/((1 + cos^2θ)(1 + sin^2θ)) = (1 - cos^2θsin^2θ)/(1 + sin^2θ + cos^2θ + sin^2θcos^2θ)`

= `(1 - cos^2θsin^2θ)/(1 + 1 + sin^2θcos^2θ) = (1 - sin^2θcos^2θ)/(2 + sin^2θcos^2θ)`