Question

The expression $\frac{1+i\tan n\alpha }{1-i\tan n\alpha }-\frac{1-i\tan n\alpha }{1+\tan n\alpha }$ when simplified reduces to:

• A. Zero
• B. $2\sin n\alpha$
• C. $2\cos n\alpha$
• D. $2i\sin 2n\alpha$

Answer 1

The correct option is D $2i\sin 2n\alpha$

$1+i\tan n\alpha =\frac{\cos n\alpha +i\sin n\alpha }{\cos n\alpha }=\frac{e^{in\alpha }}{\cos n\alpha }$

$1-i\tan n\alpha =\frac{\cos n\alpha -i\sin n\alpha }{\cos n\alpha }=\frac{\cos (-n\alpha )+i\sin (-n\alpha )}{\cos n\alpha }$

$1+i\tan n\alpha =\frac{e^{-in\alpha }}{\cos n\alpha }$

$\to \frac{1+i\tan \alpha }{1-i\tan \alpha }-\frac{1-i\tan n\alpha }{1+i\tan n\alpha }$

$\to \frac{e^{in\alpha }}{e^{-in\alpha }}-\frac{e^{-in\alpha }}{e^{in\alpha }}$

$\to e^{2in\alpha }-e^{-2in\alpha }$

$\to 2i\sin \left(2n\alpha \right)$

$\to 2i\sin \left(2n\alpha \right)$