Q70 / 180 · COMEDK 2014 · Mathematics
If $1+i=(x+i y)(u+i v)$, then $\tan ^{-1}\left(\frac{y}{x}\right)+\cot ^{-1}\left(\frac{u}{v}\right)$ has the value
- A$n \pi+\frac{\pi}{6}, n \in I$
- B$2 n \pi+\frac{\pi}{3}, n \in I$
- Correct answerC$n \pi+\frac{\pi}{4}, n \in I$
- D$n \pi-\frac{\pi}{3}, n \in I$
Solution
Given, equation is
$$
\begin{gathered}
1+i=(x+i y)(u+i v) \\
\Rightarrow 1+i=(x u-y v)+i(x v+y u)
\end{gathered}
$$
Comparing real and imaginary parts, we get
$$
\begin{aligned}
&x u-y v=1 \quad \text{...(i)}\\
&x v+y u=1 \quad \text{...(ii)}
\end{aligned}
$$
Multiply Eq. (i) by $u$ and Eq. (ii) by $v$ and then adding, we get
$$
x\left(u^{2}+v^{2}\right)=u+v \Rightarrow x=\frac{u+v}{u^{2}+v^{2}}
$$
From Eq. (i)
$$
y=\frac{x u-1}{v}=\frac{u-v}{u^{2}+v^{2}}
$$
(Substituting the value of $x$ )
Now, $\tan ^{-1}(y / x)+\cot ^{-1}(u / v)$
$$
\begin{aligned}
\tan ^{-1}\left(\frac{u-v}{u+v}\right)+\cot ^{-1}(u / v) \\
&=\tan ^{-1}\left(\frac{1-\frac{v}{u}}{1+\frac{v}{u}}\right)+\cot ^{-1}(u / v) \\
&=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{v}{u}\right)+\tan ^{-1}\left(\frac{v}{u}\right) \\
&=n \pi+\pi / 4, n \in I
\end{aligned}
$$
Hence, the correct answer is option: 'C'.