Question

If $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$ and ${\tan }^2\theta \in W$, then the number of solution(s) of the equation
$(1-\tan \theta )(1+\tan \theta ){\sec }^2\theta +2^{{\tan }^2\theta }=0$ is

• A. $2$
• B. $0$
• C. $4$
• D. Infinitely many

Answer 1

The correct option is A $2$

Given:
$(1-\tan \theta )(1+\tan \theta ){\sec }^2\theta +2^{{\tan }^2\theta }=0$
$\Rightarrow (1-{\tan }^2\theta )(1+{\tan }^2\theta )+2^{{\tan }^2\theta }=0$
Let ${\tan }^2\theta =t$
$\Rightarrow (1-t)(1+t)+2^t=0\Rightarrow 1-t^2+2^t=0\Rightarrow 2^t=t^2-1$

As ${\tan }^2\theta \in W$
So, by observation, $t=3$ is the only solution.
$\because 2^3=3^2-1$
Now,
${\tan }^2\theta =3$
$\Rightarrow \tan \theta =\pm \sqrt{3}$
As $\theta \in (-\frac{\pi }{2},\frac{\pi }{2})$, so
$\theta =\pm \frac{\pi }{3}$

Hence, the number of solution is $2$.