Question
Prove the following identities:
$1+\frac{\cot^2\theta}{(1+\text{cosec }\theta)}=\text{sec}\theta$
$1+\frac{\cot^2\theta}{(1+\text{cosec }\theta)}=\text{sec}\theta$
✓
Answer
$\text{L.H.S.}=1+\frac{\cot^2\theta}{(1+\text{cosec }\theta)}$
$=1+\frac{\text{cosec}^2\theta-1}{(1+\text{cosec}\theta)}$
$=\frac{1+\text{cosec}+\text{cosec}^2\theta-1^2}{1+\text{cosec}\theta}$
$=\frac{(1+\text{cosec}\theta)+(\text{cosec}\theta+1)(\text{cosec}\theta-1)}{(1+\text{cosec}\theta)}$
$=\frac{(1+\text{cosec}\theta)\big[1+\text{cosec}\theta-1)\big]}{(1+\text{cosec}\theta)}$
$=1+\text{cosec}\theta-1$
$=\text{cosec}\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
$=1+\frac{\text{cosec}^2\theta-1}{(1+\text{cosec}\theta)}$
$=\frac{1+\text{cosec}+\text{cosec}^2\theta-1^2}{1+\text{cosec}\theta}$
$=\frac{(1+\text{cosec}\theta)+(\text{cosec}\theta+1)(\text{cosec}\theta-1)}{(1+\text{cosec}\theta)}$
$=\frac{(1+\text{cosec}\theta)\big[1+\text{cosec}\theta-1)\big]}{(1+\text{cosec}\theta)}$
$=1+\text{cosec}\theta-1$
$=\text{cosec}\theta$
$=\text{R.H.S.}$
Hence, LHS = RHS.
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