Question

How to Prove ??? `(sinA+cosA)/(cosA-SinA) = tan2A+sec2A`

Answer 1

`LHS=(sinA+cosA)/(cosA-SinA)`

`=(sinA+cosA)^2/((cosA-SinA)(cosA+sinA)`
`=(sin^2A+cos^2A+2sinAcosA)/(cos^2A-Sin^2A)`

`=(2sinAcosA+1)/(cos2A)`

`=(sin2A)/(cos2A)+1/(cos2A)`

`= tan2A+sec2A=RHS`

Answer 2

It seemed easier to start from the right; please see proof below.

Explanation:

We need the crudest cosine double angle formula:

`cos(2A) = cos(A+A) = cos^2 A - sin ^2 A`

`sin(2A) = sin(A+A) = 2 sin A cos A`

`tan 2A + sec 2A`

`= {sin 2A}/{cos 2A} + 1/{cos 2A }`

`= {1 + sin 2A}/{cos 2A }`

`= { 1 + 2 sin A cos A}/{cos^2 A - sin ^2A }`

`={ sin ^2 A + cos ^2 A + 2 sin A cos A }/{cos ^2 A- sin ^2 A}`

`= {(sin A + cos A)^2 }/{(cos A - sin A)(cos A+sin A)}`

`= {sin A + cos A}/{cos A - sin A} quad sqrt`