Question

If $A,B,C\ne \frac{(2n+1)\pi }{2}$, then the numerical value of $\frac{\sin (B-C)}{\cos B\cos C}+\frac{\sin (C-A)}{\cos C\cos A}+\frac{\sin (A-B)}{\cos A\cos B}$ is

• A. $1$
• B. $2$
• C. $0$
• D. $-2$

Answer 1

The correct option is C $0$

Here,
$\frac{\sin (B-C)}{\cos B\cos C}=\frac{\sin B\cos C-\cos B\sin C}{\cos B\cos C}=\tan B-\tan C\cdots \cdots \left(1\right)$

Similarly,
$\frac{\sin (C-A)}{\cos C\cos A}=\tan C-\tan A\cdots \cdots \left(2\right)$

$\frac{\sin (A-B)}{\cos A\cos B}=\tan A-\tan B\cdots \cdots \left(3\right)$

Adding the equations $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get
$\frac{\sin (B-C)}{\cos B\cos C}+\frac{\sin (C-A)}{\cos C\cos A}+\frac{\sin (A-B)}{\cos A\cos B}=0$