Question

1/secA-tanA = secA + tanA​

Answer 1

Answer:

HENCE PROVED

Step-by-step explanation:

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