Question
Prove that $\sqrt{\frac{1+\cos A}{1}}=\operatorname{cosec} A+\cot A$
✓
Answer
$ \text { L.H.S }=\sqrt{\frac{1+\cos A}{1-\cos A}}$
$ =\sqrt{\frac{1+\cos A}{1-\cos A} \times \frac{1+\cos A}{1+\cos A}} \ldots \ldots [ $On rationalising the denominator$]$
$ =\sqrt{\frac{(1+\cos A)^2}{1-\cos ^2 A}}$
$=\sqrt{\frac{(1+\cos A)^2}{\sin ^2 A}} \quad \ldots \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =\frac{1+\cos A}{\sin A}$
$ =\frac{1}{\sin A}+\frac{\cos A}{\sin A}$
$ =\operatorname{cosec} A+\cot A$
$ =\text { R.H.S }$
$ \therefore \sqrt{\frac{1+\cos A}{1-\cos A}}=\operatorname{cosec} A+\cot A$
$ =\sqrt{\frac{1+\cos A}{1-\cos A} \times \frac{1+\cos A}{1+\cos A}} \ldots \ldots [ $On rationalising the denominator$]$
$ =\sqrt{\frac{(1+\cos A)^2}{1-\cos ^2 A}}$
$=\sqrt{\frac{(1+\cos A)^2}{\sin ^2 A}} \quad \ldots \ldots\left[\begin{array}{l}\because \sin ^2 A+\cos ^2 A=1 \\ \therefore 1-\cos ^2 A=\sin ^2 A\end{array}\right]$
$ =\frac{1+\cos A}{\sin A}$
$ =\frac{1}{\sin A}+\frac{\cos A}{\sin A}$
$ =\operatorname{cosec} A+\cot A$
$ =\text { R.H.S }$
$ \therefore \sqrt{\frac{1+\cos A}{1-\cos A}}=\operatorname{cosec} A+\cot A$
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