Question

In $△ABC$ the value of $\frac{si{n}^{2}A+si{n}^{2}B-si{n}^{2}C}{sinAsinBsinC}=?$

• A. $2\cot B$
• B. $\frac{1}{2}\cos A$
• C. $\sin B$
• D. $2\sin B$

Answer 1

The correct option is A $2\cot B$

$\frac{{\sin }^{2}A+{\sin }^{2}B-{\sin }^{2}C}{\sin A\sin B\sin C}$

$\cos 2A=1-2{\sin }^{2}A$

$=\frac{1-\cos 2A+1-\cos 2B+1+\cos 2C}{2\sin A\sin B\sin C}$

$=\frac{1-(\cos 2A+\cos 2B-\cos 2C)}{2\sin A\sin B\sin C}$

$=\frac{1-(\cos 2A-2\sin (2B+2C)/2\cdot \sin (2B-2C\left)/2\right)}{2\sin A\sin B\sin C}$.

$=\frac{1-(\cos 2A-2\sin (B+C\left)\sin \right(B-C\left)\right)}{2\sin A\sin B\sin C}$

$A+B+C={100}^o$

$=\frac{1-(\cos 2A-2\sin (180-A\left)\sin \right(B-C\left)\right)}{2\sin A\cdot \sin B\cdot \sin C}$

$=\frac{1-(\cos 2A+2\sin A\cdot \sin (B-C\left)\right)}{2\sin A\sin B\sin C}$

$=\frac{1-(1-2{\sin }^{2}A+2\sin A\sin (B-C\left)\right)}{2\sin A\sin B\sin C}$

$=\frac{1-(1+2{\sin }^{2}A-2\sin A\sin (B-C)}{2\sin A\sin B\sin C}$

$=\frac{2\sin A(\sin A-\sin (B-C\left)\right)}{2\sin A\sin B\sin C}$

$=\frac{2\cos \frac{(A+B-C)}{2}\sin \frac{(A-B+C)}{2}}{\sin B\cdot \sin C}$

$=\frac{2\cos \left(\frac{180-2C}{2}\right)\sin \left(\frac{180-2B}{2}\right)}{\sin B\cdot \sin C}$

$=\frac{2\sin C\cdot \cos B}{\sin B\cdot \sin C}$

$=\frac{2\cos B}{\sin B}=2\cot B$.