Question

If tan $B=\frac{nsinAcosA}{1-nsi{n}^{2}A},$ then tan(A-B) is

• A. $ntanA$
• B. $(1-n)tanA$
• C. $(n-1)tanA$
• D. $n^2$tanA

Answer 1

The correct option is B $(1-n)tanA$

We know that $\tan (A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$

$=\frac{\frac{\sin A}{\cos A}-\frac{n\sin A\cos A}{1-n{\sin }^{2}A}}{1+\frac{\sin A}{\cos A}\times \frac{n\sin A\cos A}{1-n{\sin }^{2}A}}$

$=\frac{\frac{\sin A-n{\sin }^{3}A-n\sin A{\cos }^{2}A}{\cos A(1-n{\sin }^{2}A)}}{\frac{\cos A-n{\sin }^{2}A\cos A+n{\sin }^{2}A\cos A}{\cos A(1-n{\sin }^{2}A)}}$

$=\frac{\sin A-n\sin A({\sin }^{2}A+{\cos }^{2}A)}{\cos A}$

$=\frac{\sin A-n\sin A}{\cos A}$

$=\frac{(1-n)\sin A}{\cos A}$

$=(1-n)\tan A$