frac{tan A}{1-cot A} + frac{cot A}{1-tan A} = | vedclass

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Question

(A) 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{\f

Vedclass · Accepted Answer

$1 + \sec A \operatorname{cosec} A$

Vedclass · Answer

(A) 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 $a^3 - b^3 = (a - b)(a^2 + b^2 + ab)$: $= \frac{(\sin A - \cos A)(\sin^2 A + \cos^2 A + \sin A \cos A)}{(\sin A - \cos A)(\sin A \cos A)}$ $= \frac{1 + \sin A \cos A}{\sin A \cos A} = \frac{1}{\sin A \cos A} + 1 = \sec A \operatorname{cosec} A + 1$

$\frac{\tan A}{1-\cot A} + \frac{\cot A}{1-\tan A} = ?$

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