Question

If ${\tan }^{-1}a+{\tan }^{-1}b={\sin }^{-1}1-{\tan }^{-1}c$ , then

• A. $a+b+c=abc$
• B. $ab+bc+ca=abc$
• C. $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$
• D. $ab+bc+ca=a+b+c$

Answer 1

The correct option is C

$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0$

Explanation for the correct option:

$\begin{array}{c}{\tan }^{-1}a+{\tan }^{-1}b={\sin }^{-1}1-{\tan }^{-1}c\\ \Rightarrow {\tan }^{-1}a+{\tan }^{-1}b+{\tan }^{-1}c={\sin }^{-1}1\\ \Rightarrow {\tan }^{-1}\frac{\left(a+b+c-abc\right)}{\left(1-ab-bc-ca\right)}=\frac{\pi }{2}\\ \Rightarrow 1-ab-bc-ca=0\\ \Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{1}{abc}=0\end{array}$

Hence option (C) is correct