Question

plzzz ans.......Prove the following

Answer 1

HELLO DEAR,

$\frac{1}{sec \alpha + tan \alpha } - \frac{1}{cos \alpha } \\ \\ = \frac{1}{sec \alpha + tan \alpha } \times \frac{sec \alpha - tan \alpha }{sec \alpha - tan \alpha } - \frac{1}{cos \alpha } \\ \\ = \frac{sec \alpha - tan \alpha }{ {(sec}^{2} \alpha - {tan}^{2} \alpha) } - \frac{1}{cos \alpha } \\ \\ = sec \alpha - tan \alpha - sec \alpha \\ \\ = - tan \alpha ................(1)$


-----------------------L.H.S.--------------------------
$\frac{1}{cos \alpha } - \frac{1}{sec \alpha - tan \alpha } \\ \\ = \frac{1}{cos \alpha } - \frac{1}{sec \alpha - tan \alpha } \times \frac{sec \alpha + tan \alpha }{sec \alpha + tan \alpha } \\ \\ = \frac{1}{cos \alpha } - \frac{sec \alpha + tan \alpha }{ ({sec}^{2} \alpha - {tan}^{2} \alpha) } \\ \\ = sec \alpha - sec \alpha - tan \alpha \\ \\ = - tan \alpha .......(2)$
-----------------------R.H.S.--------------------------

from--(1) and---(2)

L.H.S. =R.H.S



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THANKS