Question : What is the value of $\frac{1+\tan A}{\operatorname{cosec} A}+\frac{1+\cot A}{\sec A}$?
Option 1: $2\sec^2A$
Option 2: $\sec \mathrm{A} - \mathrm{cosec A}$
Option 3: $\sec \mathrm{A} + \mathrm{cosec A}$
Option 4: $2 \;\mathrm{cosec^2 A}$
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Correct Answer: $\sec \mathrm{A} + \mathrm{cosec A}$
Solution :
$\frac{1+\tan A}{\operatorname{cosec} A}+\frac{1+\cot A}{\sec A}$
$=\frac{1+\frac{\sin A}{\cos A}}{\frac{1}{\sin A}}+\frac{1+\frac{\cos A}{\sin A}}{\frac{1}{\cos A}}$
$=\sin A(1+\frac{\sin A}{\cos A})+\cos A(1+\frac{\cos A}{\sin A})$
$=\sin A(\frac{\cos A+\sin A}{\cos A})+\cos A(\frac{\cos A+\sin A}{\sin A})$
$=(\sin A+\cos A)(\frac{\sin A}{\cos A}+\frac {\cos A}{\sin A}))$
$=(\sin A+\cos A)(\frac{\sin ^2A+\cos^2A}{\sin A \cos A}))$
$=(\frac{\sin A+\cos A}{\sin A \cos A})$ [$\because \left ( \sin ^2A+\cos^2A=1 \right )$]
$=(\frac{\sin A}{\sin A \cos A})+(\frac{\cos A}{\sin A \cos A})$
$=(\frac{1}{ \cos A})+(\frac{1}{\sin A})$
$=\sec \mathrm{A} + \mathrm{cosec A}$
Hence, the correct answer is $\sec \mathrm{A} + \mathrm{cosec A}$.
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