Question:
$\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}=\cot \theta$
Solution:
$\mathrm{LHS}=\frac{1+\cos \theta-\sin ^{2} \theta}{\sin \theta(1+\cos \theta)}$
$=\frac{(1+\cos \theta)-\left(1-\cos ^{2} \theta\right)}{\sin \theta(1+\cos \theta)}$
$=\frac{\cos \theta+\cos ^{2} \theta}{\sin \theta(1+\cos \theta)}$
$=\frac{\cos \theta(1+\cos \theta)}{\sin \theta(1+\cos \theta)}$
$=\frac{\cos \theta}{\sin \theta}$
$=\cot \theta$
$=\mathrm{RHS}$
Hence, L.H.S. = R.H.S.
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