Question
The value of $${\tan ^{ - 1}}\left( {\frac{1}{2}\left( {\tan 2A} \right) + {{\tan }^{ - 1}}\left( {\cot A} \right) + {{\tan }^{ - 1}}\left( {{{\cot }^3}A} \right)} \right)$$ is
A.
$$0{\text{ if}}\,\,\frac{\pi }{4} < A < \frac{\pi }{2}$$
B.
$$\pi {\text{ if}}\,\,0 < A < \frac{\pi }{4}$$
C.
both $$\left( A \right){\text{ and }}\left( B \right)$$
D.
None of these
Answer :
both $$\left( A \right){\text{ and }}\left( B \right)$$
Solution :
We know that $$\cot A > 1{\text{ if }}0 < A < \frac{\pi }{4}$$
$$\eqalign{
& {\text{and }}\cot A < 1{\text{ if }}\frac{\pi }{4} < A < \frac{\pi }{2} \cr
& {\tan ^{ - 1}}\left( {\cot A} \right) + {\tan ^{ - 1}}\left( {{{\cot }^3}A} \right) = \pi + {\tan ^{ - 1}}\frac{{\cot A + {{\cot }^3}A}}{{1 - {{\cot }^4}A}} \cr
& {\text{If }}0 < A < \frac{\pi }{4}{\text{ and }} = {\tan ^{ - 1}}\frac{{\cot A + {{\cot }^3}A}}{{1 - {{\cot }^4}A}}{\text{ if }}\frac{\pi }{4} < A < \frac{\pi }{2} \cr
& {\text{Also}},{\text{ }}\frac{{\cot A + {{\cot }^3}A}}{{1 - {{\cot }^4}A}} = \frac{{\cot A\,{\text{cose}}{{\text{c}}^2}A \cdot {{\sin }^4}A}}{{{{\sin }^4}A - {{\cot }^4}A}} \cr
& = \frac{{\sin A\cos A}}{{\left( {{{\sin }^2}A + {{\cos }^2}A} \right)\left( {{{\sin }^2}A - {{\cos }^2}A} \right)}} = - \frac{{\sin 2A}}{{2\cos 2A}} = - \frac{1}{2}\tan 2A \cr} $$
Hence, $${\tan ^{ - 1}}\left( {\frac{1}{2}\tan 2A} \right) + {\tan ^{ - 1}}\left( {\cot A} \right) + {\tan ^{ - 1}}\left( {{{\cot }^3}A} \right) = \pi ,$$
\[ = \left\{ {\begin{array}{*{20}{c}}
{\pi \,\,\,\,\,\,\,{\rm{ if }}\, 0 < A < \frac{\pi }{4}}\\
{0\,\,\,\,\,\,\,{\rm{ if }}\,\frac{\pi }{4} < A < \frac{\pi }{2}}
\end{array}\,\,\,\,\,\left[ {{\rm{Since,}}\,{{\tan }^{ - 1}}\left( { - x} \right) = - {{\tan }^{ - 1}}x} \right]} \right.\]