Question

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

Answer 1

This is the answer......

First simplify LHS and then do the RHS

In LHS convert sec and tan to sin and cos

In RHS use the trigonometric identity and then the quadratic Identity

Answer 2

$\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}}=\dfrac{\cos ^{2}\theta }{(1+\sin \theta)^{2}}$

Step-by-step explanation:

L.H.S. $=\frac{\sec \theta-\tan \theta}{\sec \theta+\tan \theta}}$

$= \dfrac{\dfrac{1}{\cos \theta} -\dfrac{\sin \theta}{\cos \theta}}{\dfrac{1}{\cos \theta}+ \dfrac{\sin \theta}{\cos \theta}}$

$= \dfrac{1-\sin \theta}{1+\sin \theta}$

R.H.S.$=\dfrac{\cos ^{2}\theta }{(1+\sin \theta)^{2}}$

$=\dfrac{1-\sin ^{2}\theta }{(1+\sin \theta)^{2}}$

[ ∵ $\sin ^{2}\theta+\cos ^{2}\theta=1$]

$=\dfrac{(1+\sin \theta)(1-\sin \theta)}{(1+\sin \theta)^{2}}$

$= \dfrac{1-\sin \theta}{1+\sin \theta}$

∴ L.H.S.=R.H.S. $=\dfrac{1-\sin \theta}{1+\sin \theta}$, proved.