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