Question

if sec A - tan A= x, prove that secA =1/2(x+1/x) and tan A =1/2(x-1/x)

Answer 1

HELLO DEAR,

i think something is mistake in your questions

right questions is like that:-


if sec A - tan A= x, prove that secA =1/2(x+1/x) and (-tan A )=1/2(x-1/x)

given that:-


x = secA-tanA


1/x = 1/secA-tanA

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


------------------------(1)------------------------

$\frac{1}{2} (x + \frac{1}{x} ) \\ \\ = > \frac{1}{2} (sec \alpha - tan \alpha + sec \alpha + tan \alpha ) \\ \\ = > \frac{1}{2} (2sec \alpha ) \\ \\ = > sec \alpha$


------------------------(2)------------------------


$\frac{1}{2} (x - \frac{1}{x} ) \\ \\ = > \frac{1}{2}[(sec \alpha - tan \alpha - (sec \alpha + tan \alpha )] \\ \\ = > \frac{1}{2} (sec \alpha - tan \alpha - sec \alpha - tan \alpha ) \\ \\ = > \frac{1}{2} ( - 2tan \alpha ) \\ \\ = > - tan \alpha$

I HOPE ITS HELP YOU DEAR,
THANKS