Question #9e76f

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Answer 1 · 2017-03-17T16:33:15

$LHS=(1 - sintheta)/(1 + sin theta)$

$=(1 - sintheta)/(1 + sin theta) xx (1 - sintheta)/(1 - sin theta)$

$=(1 - sintheta)^2/(1 - sin^2 theta)$

$=(1 - sintheta)^2/cos^2 theta$

$=((1 - sintheta)/cos theta)^2$

$=(1/costheta - sintheta/cos theta)^2$

$= (sec theta - tan theta)^2=RHS$