Question

If $\tan B=\frac{n\sin A\cos A}{1-n{\cos }^{2}A}$ then $\tan (A+B)$ equals

• A. $\frac{\sin A}{(1-n)\cos A}$
• B. $\frac{(n-1)\cos A}{\sin A}$
• C. $\frac{\sin A}{(n-1)\cos A}$
• D. $\frac{\sin A}{(n+1)\cos A}$

Answer 1

The correct option is A $\frac{\sin A}{(1-n)\cos A}$

$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$
$=\frac{\tan A+\left(\frac{n\sin A\cos A}{1-n{\cos }^{2}A}\right)}{1-\tan A\left(\frac{n\sin A\cos A}{1-n{\cos }^{2}A}\right)}$
$=\frac{\frac{\sin A}{\cos A}+\frac{n\sin A\cos A}{1-n{\cos }^{2}A}}{1-\frac{\sin A}{\cos A}\left(\frac{n\sin A\cos A}{1-n{\cos }^{2}A}\right)}$
$\frac{\frac{\sin A-n{\cos }^{2}A\sin A+n\sin A{\cos }^{2}A}{\cos A(1-n{\cos }^{2}A)}}{\frac{\cos A-n{\cos }^{2}A-n{\sin }^{2}A\cos A}{\cos A(1-n{\cos }^{2}A)}}=\frac{\sin A}{\cos A(1-n{\cos }^{2}A-n{\sin }^{2}A)}$
$\Rightarrow \frac{\sin A}{\cos A(1-n({\sin }^{2}A+{\cos }^{2}A))}$
$\Rightarrow \frac{\sin A}{\cos A(1-n)}$