JEE MAIN - Mathematics (2013 (Offline) - No. 20)
The expression $${{\tan {\rm A}} \over {1 - \cot {\rm A}}} + {{\cot {\rm A}} \over {1 - \tan {\rm A}}}$$ can be written as:
$$\sin {\rm A}\,\cos {\rm A} + 1$$
$$\,\sec {\rm A}\,\cos ec{\rm A} + 1$$
$$\tan {\rm A} + \cot {\rm A}$$
$$\sec {\rm A} + \cos ec{\rm A}$$
Explanation
Given expression can be written as
$${{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}$$
(As $$\tan A = {{\sin A} \over {\cos A}}$$ and $$\cot A = {{\cos A} \over {\sin A}}$$ )
$$ = {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}$$
$$ = {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}$$
$$ = 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$$
$${{\sin A} \over {\cos A}} \times {{sin\,A} \over {\sin A - \cos A}} + {{\cos A} \over {\sin A}} \times {{\cos A} \over {\cos A - sin\,A}}$$
(As $$\tan A = {{\sin A} \over {\cos A}}$$ and $$\cot A = {{\cos A} \over {\sin A}}$$ )
$$ = {1 \over {\sin A - \cos A}}\left\{ {{{{{\sin }^3}A - {{\cos }^3}A} \over {\cos A\sin A}}} \right\}$$
$$ = {{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}\,A} \over {\sin A\cos A}}$$
$$ = 1 + \sec\, A{\mathop{\rm cosec}\nolimits} \,A$$
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