Question

If $(1-\tan \theta )(1+\tan \theta ){\sec }^2\theta +2^{{\tan }^2\theta }=0$, then the number of values of $\theta$ in the interval $(-\frac{\pi }{2},\frac{\pi }{2})$

• A. $0$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $0$

$(1-\tan \theta )(1+\tan \theta ){\sec }^2\theta +2^{{\tan }^2\theta }=0$
$\Rightarrow 1-ta{n}^4\theta +2^{{\tan }^2\theta }=0$
$\Rightarrow 2^x=x^2-1$ where $x=ta{n}^2\theta$ and thus positive.
Let there be a function $f\left(x\right)=2^x-x^2+1$
$f\left(0\right)=2,f\left(1\right)=2,f\left(2\right)=1$ and so on.
Thus, f is always non zero $\forall x\in (0,\infty )$