If tan B = frac{n sin A cos A}{1 - n cos^2 A} | vedclass

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(A) We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substituting $	an B = \frac{n \sin A \cos A}{1 - n \cos^2 A}$:$	an(A + B)

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

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(A) We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$.Substituting $	an B = \frac{n \sin A \cos A}{1 - n \cos^2 A}$:$	an(A + B) = \frac{	an A + \frac{n \sin A \cos A}{1 - n \cos^2 A}}{1 - 	an A \cdot \frac{n \sin A \cos A}{1 - n \cos^2 A}}$$= \frac{\frac{\sin A}{\cos A} + \frac{n \sin A \cos A}{1 - n \cos^2 A}}{1 - \frac{\sin A}{\cos A} \cdot \frac{n \sin A \cos A}{1 - n \cos^2 A}}$$= \frac{\sin A(1 - n \cos^2 A) + n \sin A \cos^2 A}{\cos A(1 - n \cos^2 A) - n \sin^2 A \cos A}$$= \frac{\sin A - n \sin A \cos^2 A + n \sin A \cos^2 A}{\cos A(1 - n(\cos^2 A + \sin^2 A))}$$= \frac{\sin A}{\cos A(1 - n(1))}$$= \frac{\sin A}{(1 - n) \cos A}$.

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

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