Question

If $\sin A\sin B\sin C=p$ and $\cos A\cos B\cos C=q$, where $A,B,C$ are the angles of a $\Delta ABC$, then $\tan A\tan B+\tan B\tan C+\tan C\tan A$ is equal to

• A. $\frac{p}{q}$
• B. $\frac{q+1}{q}$
• C. $\frac{p+1}{p}$
• D. $\frac{q+1}{p}$

Answer 1

The correct option is B $\frac{q+1}{q}$

For triangle $ABC$, $A+B+C=\pi$

$\Rightarrow -1=\cos (A+B+C)$

$\Rightarrow -1=\cos \left(\right(A+B)+C)$

$\Rightarrow -1=\cos (A+B)\cos C-\sin (A+B)\sin C$

$\Rightarrow -1=(\cos A\cos B-\sin A\sin B)\cos C-\left(\right(\sin A\cos B+\cos A\sin B\left)\sin C\right)$

$\Rightarrow -1=\cos A\cos B\cos C-\sin A\sin B\cos C-\sin A\cos B\sin C-\cos A\sin B\sin C$

$\Rightarrow -1=\cos A\cos B\cos C(1-\frac{\sin A\sin B}{\cos A\cos B}-\frac{\sin A\sin C}{\cos A\cos C}-\frac{\sin C\sin B}{\cos C\cos B})$

$\Rightarrow -1=q(1-\tan A\tan B-\tan A\tan C-\tan B\tan C)$

$\Rightarrow \frac{-1}{q}=1-(\tan A\tan B+\tan A\tan C+\tan B\tan C)$

$\Rightarrow \tan A\tan B+\tan A\tan C+\tan B\tan C=1+\frac{1}{q}$

$\Rightarrow \tan A\tan B+\tan A\tan C+\tan B\tan C=\frac{q+1}{q}$