Question

Let $k$ be a real number such that $\tan ({\tan }^{-1}2+{\tan }^{-1}20k)=k.$ Then the sum of all possible values of $k$ is

• A. $-\frac{19}{40}$
• B. $-\frac{21}{40}$
• C. $0$
• D. $\frac{1}{5}$

Answer 1

The correct option is A $-\frac{19}{40}$

We have, $\tan ({\tan }^{-1}2+{\tan }^{-1}20k)=k$
Let ${\tan }^{-1}2=A$ and ${\tan }^{-1}20k=B$
Then, $\tan A=2$ and $\tan B=20k$
$\Rightarrow \tan (A+B)=k$
$\Rightarrow \frac{\tan A+\tan B}{1-\tan A\cdot \tan B}=k$
$\Rightarrow \frac{2+20k}{1-2\cdot 20k}=k$
or, $40k^2+19k+2=0$
Sum of solutions, $k_1+k_2=-\frac{19}{40}$