Question

please Solve the above question

Answer 1

sec theta+tan theta(1-sin theta)

1/cos theta + sin theta/cos theta(1-sin theta)

1+sin theta/cos theta(1-sin theta)

1-sin²theta/cos theta

cos²theta/cos theta=cos theta

Answer 2

Hey here is your answer. But here I am using x in place of theta for
convenience.
LHS:
1/cosx ( 1 - sinx ) [ 1/cosx + sinx/cosx]
= (1 - sinx)/ cosx [1 + sinx / cosx]
= (1 - sinx )(1 + sinx )/ cos^2x
= (1 - sin^2x) / cos^2x ...(1)
And we know that,
${ \sin( \alpha ) }^{2} + { \cos( \alpha ) }^{2} = 1$
So, cos^2x = 1 - sin^2x
Therefore, putting value of cos^2x in (1), we get,
cos^2x / cos^2x
=1
LHS = RHS
HENCE, PROVED