1-sin theta/1+cos thetA=(sec theta-tan theta)whole square

The question given is wrong. There should be 1+sin theta instead of 1+cos theta.    

 

 

 (1-sin\theta )/(1+sin\theta )
=(1-sin\theta )/(1+sin\theta )*(1-sin\theta )/(1-sin\theta )
=((1-sin\theta )(1-sin\theta ))/((1+sin\theta )(1-sin\theta ))
=((1-sin\theta )^(2))/((1-sin^(2)\theta ))
=((1-sin\theta )^(2))/(cos^(2)\theta )
=((1-sin\theta )/(cos\theta ))^(2)
=((1)/(cos\theta )-(sin\theta )/(cos\theta ))^(2)
=(sec\theta -tan\theta )^(2)
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