Question
Prove.$\frac{1+\cos A}{1-\cos A}=\frac{\tan ^2 A}{(\sec A-1)^2}$
✓
Answer
$\begin{aligned} & \text { R.H.S }=\frac{\tan ^2 A}{(\sec A-1)^2}=\frac{\sec ^2 A-1}{(\sec A-1)^2} \quad \cdots\left[\sec ^2 \theta-\tan ^2 \theta=1\right. \\ & \left.\sec ^2 \theta-1=\tan ^2 \theta\right] \\ & =\frac{(\sec A+1)(\sec A-1)}{\sec A-1} \\ & =\frac{\sec A+1}{\sec A-1}=\frac{\frac{1}{\cos A}+1}{\frac{1}{\cos A}-1}=\frac{\frac{1+\cos A}{\cos A}}{\frac{1-\cos A}{\cos A}} \\ & =\frac{1+\cos A}{1-\cos A} \\ & \text { R.H.S. }=\text { L.H.S }\end{aligned}$
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