The expression frac{tan A}{1 - cot A} + frac{ | vedclass

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(B) Given expression: $\frac{	an A}{1 - \cot A} + \frac{\cot A}{1 - 	an A}$Convert $	an A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:$= \frac{

Vedclass · Accepted Answer

$\sec A \csc A + 1$

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(B) Given expression: $\frac{	an A}{1 - \cot A} + \frac{\cot A}{1 - 	an A}$Convert $	an A$ and $\cot A$ in terms of $\sin A$ and $\cos A$:$= \frac{\frac{\sin A}{\cos A}}{1 - \frac{\cos A}{\sin A}} + \frac{\frac{\cos A}{\sin A}}{1 - \frac{\sin A}{\cos A}}$$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} + \frac{\cos^2 A}{\sin A(\cos A - \sin A)}$$= \frac{\sin^2 A}{\cos A(\sin A - \cos A)} - \frac{\cos^2 A}{\sin A(\sin A - \cos A)}$$= \frac{1}{\sin A - \cos A} \left[ \frac{\sin^3 A - \cos^3 A}{\sin A \cos A} ight]$Using the identity $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$:$= \frac{1}{\sin A - \cos A} \left[ \frac{(\sin A - \cos A)(\sin^2 A + \sin A \cos A + \cos^2 A)}{\sin A \cos A} ight]$$= \frac{\sin^2 A + \cos^2 A + \sin A \cos A}{\sin A \cos A}$$= \frac{1 + \sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1 = \sec A \csc A + 1$

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as:

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