Question

1 + cos 0 + sino 1 + cos 0 - sin sec 0 + tan 0

i need to prove the identity

Answer 1

we have the left hand side as

$\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}$

we multiply numerator and denominator by

1+ sin e

$=\frac{1+\cos \theta+\sin \theta}{1+\cos \theta-\sin \theta}*\frac{1+sin \theta}{1+\sin \theta}$

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

now using identity

$1-\sin^2\theta=\cos^2\theta$

$=\frac{(1+\cos\theta+\sin \theta)(1+\sin \theta)}{\cos^2\theta+\cos \theta+\sin \theta\cos \theta}$

$=\frac{(1+\cos\theta+\sin \theta)(1+\sin \theta)}{\cos\theta(1+\cos \theta+\sin \theta)}$

$=\frac{1+\sin \theta}{\cos \theta}$

$=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta}$

$=\boldsymbol{\sec \theta+\tan \theta}$

hence proved ,