Question

Prove that $\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}=\sec \theta -\tan \theta$, (where $\theta$ is acute).

Answer 1

Given LHS ,$\sqrt{\frac{1-\sin \theta }{1+\sin \theta }}$

Multiplying numerator and denominator under root with $1-\sin \theta$

We get $\frac{1-\sin \theta }{\cos \theta }$ which will be equal to $\sec \theta -\tan \theta$