Question

A value of $\theta$ in the interval $(-\frac{\pi }{2},\frac{\pi }{2})$ satisfying the equation $(1-tan\theta )(1+tan\theta )se{c}^2\theta +2ta{n}^2\theta +2=0$ is

• A. $-\frac{\pi }{6}$
• B. $-\frac{\pi }{3}$
• C. $\frac{\pi }{6}$
• D. $\frac{\pi }{3}$

Answer 1

Answer: (B), (D)

$-\frac{\pi }{3}$
$\frac{\pi }{3}$

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

$\Rightarrow (1-ta{n}^2\theta )se{c}^2\theta +2(1+ta{n}^2\theta )=0$

$\Rightarrow (1-ta{n}^2\theta )se{c}^2\theta +2se{c}^2\theta =0$

$\Rightarrow (3-ta{n}^2\theta )se{c}^2\theta =0$

since $\mid sec\theta \mid \ge 1$ $\Rightarrow 3-ta{n}^2\theta =0$

$\Rightarrow tan\theta =\pm \sqrt{3}$

Hence solution in given interval is,

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