Question
Prove that$\sqrt{\sec ^2 A+\operatorname{cosec} c^2 A}=\tan A+\cot A$
✓
Answer
$\text { LHS }=\sqrt{\sec ^2 A+\operatorname{cosec}{ }^2 A}$
$=\sqrt{\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}}$
$=\sqrt{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A \cos ^2 A}}$
$=\sqrt{\frac{1}{\sin ^2 A \cos ^2 A}}$
$=\frac{1}{\sin A \cos A}$
$\text { RHS }=\tan A+\cot A$
$=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$
$=\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A}$
$=\frac{1}{\sin A \cos A}$
$\text { LHS }=\text { RHS }$
$=\sqrt{\frac{1}{\cos ^2 A}+\frac{1}{\sin ^2 A}}$
$=\sqrt{\frac{\sin ^2 A+\cos ^2 A}{\sin ^2 A \cos ^2 A}}$
$=\sqrt{\frac{1}{\sin ^2 A \cos ^2 A}}$
$=\frac{1}{\sin A \cos A}$
$\text { RHS }=\tan A+\cot A$
$=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$
$=\frac{\sin ^2 A+\cos ^2 A}{\sin A \cos A}$
$=\frac{1}{\sin A \cos A}$
$\text { LHS }=\text { RHS }$
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