Question
Prove the identity $(\sin \theta + \cos \theta) (\tan \theta + \cot \theta ) = \sec \theta + \cos ec θ.$
✓
Answer
$\text { L.H.S. }=(\sin \theta+\cos \theta)(\tan \theta+\cot \theta)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)$
$=(\sin \theta+\cos \theta) \times \frac{1}{\sin \theta \cos \theta}\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}$
$=\frac{\sin \theta}{\cos \theta \sin \theta}+\frac{\cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$
$=\sec \theta+\cos e c \theta$
$= R.H.S$
Hence proved.
$=(\sin \theta+\cos \theta)\left(\frac{\sin \theta}{\cos \theta}+\frac{\cos \theta}{\sin \theta}\right)$
$=(\sin \theta+\cos \theta)\left(\frac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \sin \theta}\right)$
$=(\sin \theta+\cos \theta) \times \frac{1}{\sin \theta \cos \theta}\left(\because \sin ^2 \theta+\cos ^2 \theta=1\right)$
$=\frac{\sin \theta+\cos \theta}{\cos \theta \sin \theta}$
$=\frac{\sin \theta}{\cos \theta \sin \theta}+\frac{\cos \theta}{\cos \theta \sin \theta}$
$=\frac{1}{\cos \theta}+\frac{1}{\sin \theta}$
$=\sec \theta+\cos e c \theta$
$= R.H.S$
Hence proved.
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