Question

State true or false:

$1+i\tan \alpha (-\pi <\alpha <\pi ,\alpha \ne \pm \frac{\pi }{2})=\frac{1}{\cos \alpha }(\cos \alpha +i\sin \alpha )$.

Answer 1

$1+itan\alpha$ $=sec\alpha (cos\alpha +isin\alpha )$ $=sec\alpha .e^{i\alpha }$
Now for $0\le \alpha \le \frac{\pi }{2}$ we get
$z=sec\alpha (cos\alpha +isin\alpha )$
for $\pi \le \alpha <\frac{\pi }{2}$
we get, $z=-sec\alpha (-cos\alpha +isin\alpha )$ $=sec\alpha (cos\alpha -isin\alpha )$
Hence answer is False.