Question

If $\frac{x}{y}=\frac{\cos A}{\cos B}$ then $\frac{x\tan A+y\tan B}{x+y}=$

• A. $\cot \frac{A+B}{2}$
• B. $\cot \frac{A-B}{2}$
• C. $\tan \frac{A-B}{2}$
• D. $\tan \frac{A+B}{2}$

Answer 1

The correct option is D $\tan \frac{A+B}{2}$

We have,

$\frac{x}{y}=\frac{\cos A}{\cos B}$

$x=\frac{y\cos A}{\cos B}$

Since,

$=\frac{x\tan A+y\tan B}{x+y}$…….. (1)

On putting the value of $x$ in equation (1), we get

$=\frac{\frac{y\cos A}{\cos B}\times \frac{\sin A}{\cos A}+\frac{y\sin B}{\cos B}}{\frac{y\cos A}{\cos B}+y}$

$=\frac{\frac{ysinA}{\cos B}+\frac{y\sin B}{\cos B}}{\frac{y\cos A+y\cos B}{\cos B}}$

$=\frac{\frac{y(sinA+\sin B)}{\cos B}}{\frac{y(\cos A+\cos B)}{\cos B}}$

$=\frac{(sinA+\sin B)}{(\cos A+\cos B)}$

We know that

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

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

Therefore,

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

$=\tan \left(\frac{A+B}{2}\right)$

Hence, the value is $\tan \left(\frac{A+B}{2}\right)$.