Question

prove that (tanA+secA-1)/tanA-secA+1 =1+sinA/cosA

Answer 1

I wish to you meaning part can be solved by you

Answer 2

Chapter :

Trigonometrical Identities

To Prove :

$= \dfrac{tanA + secA - 1}{tanA - secA + 1} = \dfrac{1 + sinA}{cosA}$

Solution :

We have :

L.H.S = $= \dfrac{tanA + secA - 1}{tanA - secA + 1}$

$= \dfrac{(tanA + secA - 1) - ({sec}^{2}A - {tan}^{2} A)}{(tanA - secA + 1)}$

$= \dfrac{(secA + tanA) - ({sec}A + {tan} A)({sec}A - {tan} A)}{(tanA - secA + 1)}$

$= \dfrac{(secA + tanA)[1 - ({sec}A - {tan} A)]}{(tanA - secA + 1)}$

$= \dfrac{(secA + tanA) ({tanA - secA + 1})}{(tanA - secA + 1)}$

$= (secA + tanA)$

$= \dfrac{1}{cosA} + \dfrac{sinA}{cosA}$

$= \dfrac{(1 + sinA)}{cosA}$= R.H.S

$\therefore$L.HS = R.H.S

Hence, Proved.