Question
Prove the following identities:
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
$\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}=\frac{1+\sin\theta}{\cos\theta}$
✓
Answer
$\text{LHS}=\frac{1+\cos\theta+\sin\theta}{1+\cos\theta-\sin\theta}$
On dividing numerator and denominator of LHS $\cos^\theta,$
We, get
$\text{LHS}=\frac{\sec\theta+1+\tan\theta}{\sec\theta+1-\tan\theta}$
$=\frac{(\sec\theta+\tan\theta)+\big(\sec^2\theta-\tan^2\theta\big)}{1+\sec\theta-\tan\theta}$
Writing 1 $=\big(\sec^2\theta-\tan^2\theta\big)$
$=\frac{(\sec\theta+\tan\theta)+(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)}$
$=\frac{(\sec\theta+\tan\theta)(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)}$
$=(\sec\theta+\tan\theta)=\Big(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\Big)$
$=\Big(\frac{1+\sin\theta}{\cos\theta}\Big)$
$=\text{R.H.S.}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
On dividing numerator and denominator of LHS $\cos^\theta,$
We, get
$\text{LHS}=\frac{\sec\theta+1+\tan\theta}{\sec\theta+1-\tan\theta}$
$=\frac{(\sec\theta+\tan\theta)+\big(\sec^2\theta-\tan^2\theta\big)}{1+\sec\theta-\tan\theta}$
Writing 1 $=\big(\sec^2\theta-\tan^2\theta\big)$
$=\frac{(\sec\theta+\tan\theta)+(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)}$
$=\frac{(\sec\theta+\tan\theta)(\sec\theta+\tan\theta)(\sec\theta-\tan\theta)}{(1+\sec\theta-\tan\theta)}$
$=(\sec\theta+\tan\theta)=\Big(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta}\Big)$
$=\Big(\frac{1+\sin\theta}{\cos\theta}\Big)$
$=\text{R.H.S.}$
$\therefore\text{L.H.S.}=\text{R.H.S.}$
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