Question

can anyone solve this
2 sums​

Answer 1

Solution:

$= \frac{ \cot \gamma - 1}{2 - { \sec}^{2} \gamma } \\ \\ = \frac{ \cot \gamma \: - 1}{2 - ( { \tan}^{2} \gamma + 1)} [\because \: \sec {}^{2} \gamma = 1 + { \tan }^{2} \gamma </p><p>]\\ \\ = \frac{ \cot \gamma - 1}{2 - { \tan }^{2} \gamma - 1} \\ \\ = \frac{ \cot \gamma - 1}{1 - { \tan}^{2} \gamma } \\ \\ = \frac{ \cot \gamma - 1 }{(1 + \tan \gamma )(1 - \tan \gamma )} \\ \\ = \frac{ \cot \gamma - 1}{(1 + \tan \gamma )(1 - \frac{1}{ \cot \gamma } )} \\ \\ = \frac{ \cancel{\cot \gamma - 1}}{(1 + \tan \gamma ) \frac{ \cancel{(cot \gamma - 1)}}{cot \gamma }} \\ \\ = \frac{ \cot \gamma }{1 + \tan \gamma }$

Hence, [L.H.S] = R.H.S [proved]