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
$\begin{array}{l}=\frac{\cos \theta}{1-\sin \theta} \\
=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \text { (Multiplying both Numerator and Denominator by } 1+\operatorname{Sin} \theta \text { ) } \\
=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta} \\
=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\frac{1+\sin \theta}{\cos \theta} \\
=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \\
=\sec \theta+\tan \theta \\
=\text { LHS }
\end{array}$
Proved!
$\begin{array}{l}=\frac{\cos \theta}{1-\sin \theta} \\
=\frac{\cos \theta}{1-\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta} \text { (Multiplying both Numerator and Denominator by } 1+\operatorname{Sin} \theta \text { ) } \\
=\frac{\cos \theta(1+\sin \theta)}{1-\sin ^2 \theta} \\
=\frac{\cos \theta(1+\sin \theta)}{\cos ^2 \theta}\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\frac{1+\sin \theta}{\cos \theta} \\
=\frac{1}{\cos \theta}+\frac{\sin \theta}{\cos \theta} \\
=\sec \theta+\tan \theta \\
=\text { LHS }
\end{array}$
Proved!
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