Prove the following trigonometric identities:\(\frac{1-sinθ}{1+sinθ}

1.	Prove the following trigonometric identities:\(\frac{1-sinθ}{1+sinθ}\) (secθ-tanθ)2
Answer» \(\frac{1-sinθ}{1+sinθ}\) = \(\frac{1-sinθ}{1+sinθ}\times\frac{1-sinθ}{1-sinθ}\) = \(\frac{(1-sinθ)^2}{1-sin^2θ}\) = \(\frac{(1-sinθ)^2}{cos^2θ}\) = \(\Big(\frac{1-sinθ}{cosθ}\Big)^2\) = \(\Big(\frac{1}{cosθ}-\frac{sinθ}{cosθ}\Big)^2\) = (secθ - tanθ)² Hence Proved.

Prove the following trigonometric identities:\(\frac{1-sinθ}{1+sinθ}\) (secθ-tanθ)2

Answer»

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

= \(\frac{(1-sinθ)^2}{1-sin^2θ}\)

= \(\frac{(1-sinθ)^2}{cos^2θ}\)

= \(\Big(\frac{1-sinθ}{cosθ}\Big)^2\)

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

= (secθ - tanθ)²

Hence Proved.