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

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Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

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

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

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