Question

Question #14663

Answer 1

The Proof is given below in Explanation.

Explanation:

We will use the Identity `: sec^2A-tan^2A=1`.

Let us divide the `Nr.` and `Dr.` of L.H.S by `cosA` to get,

The L.H.S. `={(sinA+1-cosA)/cosA}/{(sinA-1+cosA)/cosA`

`={sinA/cosA+1/cosA-cosA/cosA}/{sinA/cosA-1/cosA+cosA/cosA}`

`=(tanA+secA-1)/(tanA-secA+1}`

`={tanA+secA-(sec^2A-tan^2A)}/(tanA-secA+1)`

`={(tanA+secA)-(secA+tanA)(secA-tanA)}/(tanA-secA+1)`

`={(secA+tanA)cancel((1-secA+tanA))}/cancel((tanA-secA+1)`

`=secA+tanA`

`=` The R.H.S.

Hence, the Proof. Enjoy Maths.!

`II^(nd)` Method :-

We have, `1-sin^2A=cos^2A`

`rArr (1+sinA)(1-sinA)=cos^2A`

`rArr (1+sinA)/cosA=cosA/(1-sinA)`

`:.` Each Rratio `={(1+sinA)-(cosA)}/{(cosA)-(1-sinA)}`, i.e.,

Each Ratio`=(1+sinA-cosA)/(cosA-1+sinA)`.

In particular, `(1+sinA)/cosA=(1+sinA-cosA)/(cosA-1+sinA)`, or,

`1/cosA+sinA/cosA=(1+sinA-cosA)/(cosA-1+sinA)`.

`rArr secA+tanA=(1+sinA-cosA)/(cosA-1+sinA)`.