Question

If $A+B+C=\pi$, $(A,B,C)>0$ and the angle $C$ is obtuse then

• A. $\tan A\tan B>1$
• B. $\tan A\tan B<1$
• C. $\tan A\tan B=1$
• D. None of these

Answer 1

The correct option is B

$\tan A\tan B<1$

Step 1: Determine the sign of $\tan C$

Given that $A+B+C=\pi$

$\begin{array}{llrl}A+B+C& =\pi & & \\ ⟹A+B& =\pi –C& & \\ ⟹\tan A+B& =tan(\pi –C)& & \\ ⟹\frac{\tan A+\tan B}{1–\tan A\tan B}& =-\tan C& ....& \left(1\right)\end{array}$

Also, given that $C$ is obtuse. Thus, $\tan C$ will be negative,

$\tan C<0\dots .\left(2\right)$

Step 2: Determine the sign of $1-\tan A\tan B$

$A$ and $B$ are acute that means less than $\frac{\pi }{2}$.

$\Rightarrow \tan A+\tan B>0\dots .\left(3\right)$

From $\left(1\right),\left(2\right)and\left(3\right)$, $1–\tan A\tan B>0$

Hence, $\tan A\tan B<1$. Hence option $\left(B\right)$ is the correct option.