Question
Prove that:
$\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}$
$\sec \theta+\tan \theta=\frac{\cos \theta}{1-\sin \theta}$
✓
Answer
Taking RHS
$=\frac{\cos \theta}{1-\sin \theta}$
$=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}+\frac{\left.\sin \sin ^2 \theta+\cos ^2 \theta=1\right]}{\cos \theta}$
= secθ + tanθ
= LHS
Proved !
$=\frac{\cos \theta}{1-\sin \theta}$
$=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta}$
$=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}$
$=\frac{1+\sin \theta}{\cos \theta}$
$=\frac{1}{\cos \theta}+\frac{\left.\sin \sin ^2 \theta+\cos ^2 \theta=1\right]}{\cos \theta}$
= secθ + tanθ
= LHS
Proved !
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