Question

tan0/(sec0+1)= (sec0-1)/tan0​

Answer 1

$LHS = \frac{ tan0}{(sec 0 + 1 )} \\=\frac{ tan0(sec 0 - 1)}{(sec 0 + 1 )(sec 0 - 1 )} \\=\frac{ tan0(sec 0 - 1)}{sec^{2} 0 - 1^{2} } \\= \frac{ tan0(sec 0 - 1)}{tan^{2} 0 }$

$\boxed { \pink { Since, sec^{2} 0 - 1 = tan^{2} 0 }}$

$= \frac{ (sec 0 - 1)}{tan 0 } \\=RHS$

$Hence\: proved$