Question

tan A +sec A -1 / tan A -Sec A +1 = 1+ sin A / cos A ​

Answer 1

EXPLANATION.

$\sf : \implies \: \dfrac{ \tan(a) + \sec(a) - 1}{ \tan(a) - \sec(a) + 1} = \dfrac{1 + \sin(a) }{ \cos(a) } \\ \\ \sf : \implies \: 1 = (\sec {}^{2} (a) - \tan {}^{2} (a) ) \\ \\ \sf : \implies \: \frac{ \tan(a) + \sec(a) - ( \sec {}^{2} (a) - \tan {}^{2} (a)) }{ \tan(a) - \sec(a) + 1} \\ \\ \sf : \implies \: \frac{ \tan(a) + \sec(a) - [( \sec(a) - \tan(a))( \sec(a) + \tan(a) )]}{ \tan(a) - \sec(a) + 1}$

$\sf : \implies \: \dfrac{( \sec(a) + \tan(a)(1 - \sec(a) + \tan(a)) }{ \tan(a) - \sec(a) + 1} \\ \\ \sf : \implies \: \sec(a) + \tan(a) \\ \\ \sf : \implies \: \frac{1}{ \cos(a )} + \frac{ \sin(a) }{ \cos(a) } \\ \\ \sf : \implies \: \frac{1 + \sin(a) }{ \cos(a) } = proved$

Answer 2

Given :

$\frac{tan A + sec A - 1}{tan A - sec A + 1} = ( \frac{1 + sin A}{cos A})$

L. H. S :

= $\frac{tan A + sec A - 1}{tan A - sec A + 1}$

= $\frac{( tan A + sec A ) - ( sec^2 A - tan^2 A )}{tan A - sec A + 1}$

= $\frac{( sec A + tan A ) - ( sec A + tan A )(sec A - tan A)}{tan A - sec A + 1}$

= $\frac{( sec A + tan A ) - [( 1 - sec A - tan A )]}{tan A - sec A + 1}$

= $\frac{( sec A + tan A ) - ( tan A - sec A + 1 )}{tan A - sec A + 1}$

= $( sec A + tan A ) = ( \frac{1}{cos A}) + ( \frac{sin A}{cos A}) = ( \frac{( 1 + sin A )}{cos A}) = R. H. S$

Therefore, L. H. S = R. H. S