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