Question

1/secA-tanA = secA + tanA​

Answer 1

Step-by-step explanation:

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

Answer 2

Step-by-step explanation:

Given that

$\frac{1}{sec \: a - tan \: a} = sec \: a + tan \: a$

To solve it

taking L.H.S

$\frac{1}{sec \: a - tan \: a} \\ = \frac{ {sec}^{2} a - {tan}^{2} a}{sec \: a - \: tan \: a}$

$= \frac{(sec \: a - tan \: a)(sec \: a + \: tan \: a)}{sec \: a - \: tan \: a}$

$= sec \: a + tan \: a$

hence proved.

hence proved.L.H.S.=R.H.S.