Question

plzzz ans......PROVE THE FOLLOWING ...

Answer 1

first one in question is wrong , there would be 2 in numerator not 1


hope this helps

Answer 2

HELLO DEAR,

let A=ALPHA


$\frac{sin \alpha + cos \alpha }{sin \alpha - cos \alpha } + \frac{sin \alpha - cos \alpha }{sin \alpha + cos \alpha} \\ \\ = \frac{ {(sin \alpha + cos \alpha })^{2} + {(sin \alpha - cos \alpha })^{2} }{( {sin}^{2} \alpha - {cos}^{2} \alpha ) } \\ \\ = \frac{ ({sin}^{2} \alpha + {cos}^{2} \alpha) + 2sin \alpha cos \alpha + ({sin}^{2} \alpha + {cos}^{2} \alpha) - 2sin \alpha cos \alpha }{ {sin}^{2} \alpha - {cos}^{2} \alpha } \\ \\ = \frac{1 + 1}{ {sin}^{2} \alpha - {cos}^{2} \alpha } \\ \\------------------------------------- \\ \\ = \frac{2}{(1 - {cos \alpha }^{2} - {cos}^{2} \alpha) } \\ \\ = \frac{2}{(1 - 2 {cos}^{2} \alpha) } \\ \\ ------------------------------------- \\ \\ = \frac{2}{1 - \frac{2}{ {sec}^{2} \alpha } } \\ \\ = \frac{2}{ \frac{( {sec}^{2} \alpha - 2) }{ {sec}^{2} \alpha } } \\ \\ = \frac{2 {sec}^{2} \alpha }{( {sec}^{2} \alpha - 1) - 1 } \\ \\ = \frac{2 {sec}^{2} \alpha }{ {tan}^{2} \alpha - 1}$
I HOPE ITS HELP YOU DEAR,
THANKS