Question
Prove that $\sin \theta ( 1 – \tan \theta ) − \cos \theta ( 1 − \cot \theta ) = cosec \theta − \sec \theta$
✓
Answer
$\text { L.H.S }=\sin \theta(1-\tan \theta)-\cos \theta(1-\cot \theta)$
$=\sin \theta\left(1-\frac{\sin \theta}{\cos \theta}\right)-\cos \theta\left(1-\frac{\cos \theta}{\sin \theta}\right)$
$=\sin \theta-\frac{\sin ^2 \theta}{\cos \theta}-\cos \theta+\frac{\cos ^2 \theta}{\sin \theta}$
$=\sin \theta+\frac{\cos ^2 \theta}{\sin \theta}-\frac{\sin ^2 \theta}{\cos \theta}-\cos \theta$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta}-\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta}\right)$
$=\frac{1}{\sin \theta}-\frac{1}{\cos \theta} \quad \ldots \ldots\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\operatorname{cosec} \theta-\sec \theta$
$=\text { R.H.S }$
$\therefore \sin \theta(1-\tan \theta)-\cos \theta(1-\cot \theta)=\operatorname{cosec} \theta-\sec \theta$
$=\sin \theta\left(1-\frac{\sin \theta}{\cos \theta}\right)-\cos \theta\left(1-\frac{\cos \theta}{\sin \theta}\right)$
$=\sin \theta-\frac{\sin ^2 \theta}{\cos \theta}-\cos \theta+\frac{\cos ^2 \theta}{\sin \theta}$
$=\sin \theta+\frac{\cos ^2 \theta}{\sin \theta}-\frac{\sin ^2 \theta}{\cos \theta}-\cos \theta$
$=\frac{\sin ^2 \theta+\cos ^2 \theta}{\sin \theta}-\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta}\right)$
$=\frac{1}{\sin \theta}-\frac{1}{\cos \theta} \quad \ldots \ldots\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$=\operatorname{cosec} \theta-\sec \theta$
$=\text { R.H.S }$
$\therefore \sin \theta(1-\tan \theta)-\cos \theta(1-\cot \theta)=\operatorname{cosec} \theta-\sec \theta$
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