Question
The expression $$\frac{{\tan A}}{{1 - \cot A}} + \frac{{\cot A}}{{1 - \tan A}}$$ can be written as:
A.
$$\sin A\cos A + 1$$
B.
$$\sec A\,{\text{cosec}}\,A + 1$$
C.
$$\tan A + \cot A$$
D.
$$\sec A + \,{\text{cosec}}\,A$$
Answer :
$$\sec A\,{\text{cosec}}\,A + 1$$
Solution :
Given expression can be written as
$$\eqalign{
& \frac{{\sin A}}{{\cos A}} \times \frac{{\sin A}}{{\sin A - \cos A}} + \frac{{\cos A}}{{\sin A}} \times \frac{{\cos A}}{{\cos A - \sin A}} \cr
& \left( {\because \,\,\tan A = \frac{{\sin A}}{{\cos A}}\,{\text{and }}\cot A = \frac{{\cos A}}{{\sin A}}} \right) \cr
& = \frac{1}{{\sin A - \cos A}}\left\{ {\frac{{{{\sin }^3}A - {{\cos }^3}A}}{{\cos A\sin A}}} \right\} \cr
& = \frac{{{{\sin }^2}A + \sin A\cos A + {{\cos }^2}A}}{{\sin A\cos A}} \cr
& = 1 + \sec A\,{\text{cosec}}\,A \cr} $$