Question

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

• A. $\theta =0$
• B. $\theta =\frac{\mathrm{\pi }}{3}or\frac{-\mathrm{\pi }}{3}$
• C. $\theta =\frac{-\mathrm{\pi }}{6}$
• D. $\theta =\frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is B

$\theta =\frac{\mathrm{\pi }}{3}or\frac{-\mathrm{\pi }}{3}$

Explanation for the correct option:

Step 1. Using the identity $\mathit{s}\mathit{e}{\mathit{c}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{–}\mathbf{}\mathit{t}\mathit{a}{\mathit{n}}^{\mathbf{2}}\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$,

$(1–tan\theta )(1+tan\theta )se{c}^2\theta +2^{{\tan }^2\theta }=0$

$\Rightarrow$ $(1–ta{n}^2\theta )(1+ta{n}^2\theta )+2^{{\tan }^2\theta }=1$

Step 1. Put ${\tan }^2\theta =x$

$\Rightarrow$ $(1–x)(1+x)+2^x=0$

$\Rightarrow$ $1–x^2+2^x=0$

$\Rightarrow$ $x^2-1=2^x$

Step 3. Draw the curve $y=x^2-1andy=2^x$

It is clear from the graph that two curves are intersecting at one point $\left(3,8\right)$

$\Rightarrow {\tan }^2\theta =3$

$\Rightarrow$$\tan \theta =\pm \surd 3$

Thus, $\theta =\frac{\pi }{3}or\frac{-\pi }{3}$

Hence, Option ‘B’ is Correct.