Question

tanA+secA-1/tanA-secA+1=1+sinA/cosA​

Answer 1

To prove:-

$\large\rm { \frac{ \tan \ A + \sec \ A - 1}{ \tan \ A - \sec \ A +1} = \frac{1 + \sin \ A}{ \cos \ A}}$

Identity (s) used:-

$\large\rm { 1 + \tan^{2} \ A = \sec^{2} \ A}$

Proof:-

L.H.S.

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

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

$\large\rm { = \tan \ A + \sec \ A}$

$\large\rm { =( \sin \ A/ \cos \ A )+ \frac{1}{ \cos \ A}}$

$\large\rm { = \frac{1 + \sin \ A}{ \cos \ A}}$

Hence proven.✔️