Question

In triangle $ABC$, if $\tan A\cdot \tan C=2$ and $\tan B\cdot \tan C=18$, then the value of $\cos C$ is

• A. $\frac{1}{\sqrt{17}}$
• B. $-\frac{1}{\sqrt{17}}$
• C. $\frac{4}{\sqrt{17}}$
• D. Cannot be determined as the angle $C$ can be acute or obtuse

Answer 1

The correct option is A $\frac{1}{\sqrt{17}}$

Let $\tan C=x\Rightarrow \tan A=\frac{2}{x},\tan B=\frac{18}{x}$
We know that, in $\Delta ABC$,
$\tan A+\tan B+\tan C=\tan A\cdot \tan B\cdot \tan C$

$\Rightarrow \frac{2}{x}+\frac{18}{x}+x=\frac{36}{x}$

$\Rightarrow x^2=16$
$\Rightarrow x=4(\because \tan A\cdot \tan C>0\Rightarrow A,C<\frac{\pi }{2})$
$\Rightarrow \tan C=4\Rightarrow \cos C=\frac{1}{\sqrt{17}}$