Question #14663

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Answer 1 · 2016-08-04T20:15:09

The Proof is given below in 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,

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

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