Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively. The value of $(a^2-b^2-c^2)\tan A+(a^2-b^2+c^2)$ $\tan B$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

We can write given equation as

$\Rightarrow -\frac{(b^2+c^2-a^2)\left(2bc\right)\tan A}{2bc}+\frac{(a^2-b^2+c^2)\left(2ac\right)\tan B}{2ac}$

$=-2bc\cos A\tan A+2ac\cos B\tan B$

$=-2bc\sin A+2ac\sin B=4(-\frac{1}{2}bc\sin A+\frac{1}{2}ac\sin B)$

since, $\frac{1}{2}bc\sin A=\frac{1}{2}ac\sin B=\Delta$

$\therefore 4(-\frac{1}{2}bc\sin A+\frac{1}{2}ac\sin B)=4(\Delta -\Delta )=0$

Therefore, Answer is $B$