Prove the following trigonometric identities:\(\frac{1+sec θ}{sec θ}

1.	Prove the following trigonometric identities:\(\frac{1+sec θ}{sec θ}=\frac{sin^2 θ}{1-cos θ}\)
Answer» \(\frac{1+sec θ}{sec θ}\) = \(1+\frac{1}{\frac{cos θ}{\frac{1}{cos θ}}}\) = \(\frac{1+cos θ}{1}\) = \( \frac{(1+cos θ)(1-cos θ)}{(1-cos θ)}\) = \( \frac{1-cos^2 θ}{(1-cos θ)}\) = \(\frac{sin^2 θ}{1-cos θ}\) Hence Proved.

Prove the following trigonometric identities:\(\frac{1+sec θ}{sec θ}=\frac{sin^2 θ}{1-cos θ}\)

Answer»

\(\frac{1+sec θ}{sec θ}\) = \(1+\frac{1}{\frac{cos θ}{\frac{1}{cos θ}}}\)

= \(\frac{1+cos θ}{1}\)

= \( \frac{(1+cos θ)(1-cos θ)}{(1-cos θ)}\)

= \( \frac{1-cos^2 θ}{(1-cos θ)}\)

= \(\frac{sin^2 θ}{1-cos θ}\)

Hence Proved.