Question

A solution of the equation $(1-tan\theta )(1+tan\theta )se{c}^2\theta +2^{ta{n}^2\theta }=0$ where $\theta$ lies in the interval $(-\pi /2,\pi /2)$ is given by

• A. $\theta =0$
• B. $\theta =\pi /3$
• C. $\theta =-\pi /3$
• D. $\theta =\pi /6$

Answer 1

Answer: (B), (C)

$\theta =\pi /3$
$\theta =-\pi /3$

$(1-\tan \theta )(1+\tan \theta ).{\sec }^2\theta +2^{{\tan }^2\theta }=0$
$(1-{\tan }^2\theta ).(1+{\tan }^2\theta )+2^{{\tan }^2\theta }=0$

$(1-{\tan }^4\theta )+2^{{\tan }^2\theta }=0$

Let ${\tan }^2\theta =t$.

Hence
$1-t^2+2^t=0$

$t^2-1=2^t$.

From inspection of graph we get
$t^2=9$ is one solution.

$t=3$ since ${\tan }^2\theta =-3$ is not possible.

${\tan }^2\theta =3$
$\tan \theta =\pm \sqrt{3}$.

$\theta =\pm \frac{\pi }{3}$.