Prove the following trigonometric identities: 1-sinθ/1-sinθ=(secθ-tanθ)^2

Question

Prove the following trigonometric identities:

\(\frac{1-sinθ}{1+sinθ}\) (secθ-tanθ)²

Answer 1

\(\frac{1-sinθ}{1+sinθ}\) = \(\frac{1-sinθ}{1+sinθ}\times\frac{1-sinθ}{1-sinθ}\)

=   \(\frac{(1-sinθ)^2}{1-sin^2θ}\)  

=    \(\frac{(1-sinθ)^2}{cos^2θ}\)​​

= ​​​​​ \(\Big(\frac{1-sinθ}{cosθ}\Big)^2\)  

= \(\Big(\frac{1}{cosθ}-\frac{sinθ}{cosθ}\Big)^2\)

= (secθ - tanθ)²

Hence Proved.