Question

prove tan theta upon sec theta + 1 is equal to sec theta minus 1 upon 10 theta​

Answer 1

$\frac{ \tan( \alpha ) }{\sec( \alpha ) + 1 } \times \frac{ \sec( \alpha ) - 1 }{ \sec( \alpha ) - 1}$

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

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