Question

Express $1+itan\theta (-\pi <\theta <\pi ,\theta \ne \pm \pi /2)$ in the modulus-amplitude form

• A. $\{\begin{array}{c}sec\alpha (cos\alpha +isin\alpha )for\alpha in(-\pi /2,\pi /2)\\ -sec\alpha \left(cos\right(\pi -\alpha )+isin(\pi -\alpha \left)\right)for\alpha \in (-\pi .-\frac{\pi }{2})and\alpha \in (\frac{\pi }{2},\pi )\end{array}$
• B. $\{\begin{array}{c}sec\alpha (cos\alpha +isin\alpha )for\alpha in(-\pi /2,\pi /2)\\ -sec\alpha \left(cos\right(\pi +\alpha )+isin(\pi +\alpha \left)\right)for\alpha \in (-\pi .-\frac{\pi }{2})and\alpha \in (\frac{\pi }{2},\pi )\end{array}$
• C. $\{\begin{array}{c}sec\alpha (cos\alpha +isin\alpha )for\alpha in(-\pi /2,\pi /2)\\ sec\alpha \left(cos\right(\pi +\alpha )+isin(\pi +\alpha \left)\right)for\alpha \in (-\pi .-\frac{\pi }{2})and\alpha \in (\frac{\pi }{2},\pi )\end{array}$
• D. $\{\begin{array}{c}sec\alpha (cos\alpha -isin\alpha )for\alpha in(-\pi /2,\pi /2)\\ -sec\alpha \left(cos\right(\pi +\alpha )+isin(\pi +\alpha \left)\right)for\alpha \in (-\pi .-\frac{\pi }{2})and\alpha \in (\frac{\pi }{2},\pi )\end{array}$

Answer 1

The correct option is B $\{\begin{array}{c}sec\alpha (cos\alpha +isin\alpha )for\alpha in(-\pi /2,\pi /2)\\ -sec\alpha \left(cos\right(\pi +\alpha )+isin(\pi +\alpha \left)\right)for\alpha \in (-\pi .-\frac{\pi }{2})and\alpha \in (\frac{\pi }{2},\pi )\end{array}$

$1+itan\theta$
$=\frac{1}{cos\theta }(cos\theta +isin\theta )$
$=sec\theta .e^{i\theta }$
Hence
$|z|=sec\theta$
we get $arg\left(z\right)=\theta$
And
For $\theta \in (-\pi ,\frac{-\pi }{2})$ and $\theta \in (\frac{\pi }{2},\pi )$
we get
$z=-sec\theta \left(cos\right(\pi +\theta )+isin(\pi +\theta \left)\right)$