Question

Prove that:

`(cotA + tanA − 1) (sinA + cosA)/sin^3A + cos^3A = secA × cosec A`

Answer 1

`​(cot A + tan A − 1)/(sin^3 A + cos^3 A)`

Rewriting cot⁡A and tan⁡A in terms of sine and cosine:

`((cos A/sin A + sin A/cos A − 1) (sin A + cos A))/((sin A + cos A) [(sin A + cos A)^2] − 3 sin A cos A)`

Simplifying the numerator:

= `(((cos^2A + sin^2A − sin A cos A)/(sin A cos A)) (sin A + cos A))/((sin A cos A) (sin^2A + cos^2 A + 2 sin A cos A − 3 sin A cos 4))`

Using sin² A + cos² A = 1

= `(1 − sin A cos A)/(sin A cos A (1 − sin A cos A))`    ...[Cancelling common terms]

= `1/(sin A cos A)`

= sec A . cosec A

Hence proved.