Question

The value of $ta{n}^{-1}\left(\frac{1}{2}tan2A\right)+ta{n}^{-1}\left(cotA\right)+ta{n}^{-1}\left(co{t}^{3}A\right)$ for $0<A<\left(\frac{\pi }{4}\right)$ is-

• A. $4ta{n}^{-1}\left(1\right)$
• B. $2ta{n}^{-1}\left(2\right)$
• C. None

Answer 1

By identity, $tan\left(2A\right)=\frac{2\times tan\left(A\right)}{1-ta{n}^{2}A}$
$\Rightarrow \left(\frac{1}{2}\right)\times tan\left(2A\right)=\frac{tan\left(A\right)}{1-ta{n}^{2}A}$
$\Rightarrow$ Left side is: $ta{n}^{-1}\left\{\frac{tan\left(A\right)}{(1-ta{n}^{2}A)}\right\}+ta{n}^{-1}\left\{\frac{1}{tan\left(A\right)}\right\}+ta{n}^{-1}\left\{\frac{1}{ta{n}^{3}A}\right\}$
This is of the form $ta{n}^{-1}x+ta{n}^{-1}y+ta{n}^{-1}z,wherex=\left\{\frac{tan\left(A\right)}{(1-ta{n}^{2}A)}\right\}$
$Y=\frac{1}{tan\left(A\right)}$ and $z=\frac{1}{ta{n}^{3}A}$
Also, $ta{n}^{-1}x+ta{n}^{-1}y+ta{n}^{-1}z=ta{n}^{-1}\left\{\frac{(x+y+z-xyz)}{(1-xy-yz-zx)}\right\}$
So, $x+y+z-xyz=\left\{\frac{tan\left(A\right)}{(1-ta{n}^{2}A)}\right\}+\frac{1}{tan\left(A\right)}+\frac{1}{\frac{ta{n}^{3}A-1}{\{tanA\times (1-ta{n}^{2}A\left)\right\}}}$
Taking LCM and simplifying, $\frac{\{ta{n}^{4}A+ta{n}^{2}A-ta{n}^{4}A+1-ta{n}^{2}A-1\}}{\{tanA\times (1-ta{n}^{2}A\left)\right\}}$
$=\frac{0}{tanA\times (1-ta{n}^{2}A)}=0$
Hence, $ta{n}^{-1}x+ta{n}^{-1}y+ta{n}^{-1}z=0$
$\Rightarrow ta{n}^{-1}\left\{\frac{tan\left(A\right)}{(1-ta{n}^{2}A)}\right\}+ta{n}^{-1}\left\{\frac{1}{tan\left(A\right)}\right\}+ta{n}^{-1}\left\{\frac{1}{ta{n}^{3}A}\right\}=0$ [Proved]