Question

Let $A,B,C$ be the three angles of a triangle. If $\tan A.\tan B=2$, then the value of $\frac{\cos A.\cos B}{\cos C}$ is.

• A.
  $0$
• B.
  $1$
• C.
  $-1$
• D.
  $-2$

Answer 1

The correct option is B

$1$
Given $A,B,C$ are angles of a triangle.
$\Rightarrow A+B+C=\pi .\Rightarrow C=\pi -(A+B)$
Now,
$\frac{\cos A.\cos B}{\cos C}=\frac{\cos A.\cos B}{\cos (\pi -(A+B))}\Rightarrow \frac{\cos A.\cos B}{\cos C}=\frac{-\cos A.\cos B}{\cos \left(A+B\right)}$
Now, we know the expression
$\cos (A+B)=\cos A\cos B-\sin A\sin B$
So, we get our expression as
$\frac{-\cos A.\cos B}{\cos \left(A+B\right)}=\frac{-\cos A.\cos B}{\cos A\cos B-\sin A\sin B}$
Now, dividing both numerator and denominator by numerator, we get
$\frac{-\cos A.\cos B}{\cos A\cos B-\sin A\sin B}=\frac{-1}{1-\tan A\tan B}\Rightarrow \frac{-\cos A.\cos B}{\cos A\cos B-\sin A\sin B}=\frac{-1}{1-2}=1$
Thus, Option b. is correct.