Question

The number of values of $\theta$ in the interval (- $\frac{\pi }{2}$, $\frac{\pi }{2}$) satisfying the eqautions

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

• A. 4
• B. 3
• C. 5
• D. 1

Answer 1

The correct option is A

4

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

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

1 + $2^{ta{n}^2\theta }$ = $ta{n}^4\theta$ By observation $ta{n}^2\theta$ = 3

$\theta$ = n$\pi$ ± $\frac{\pi }{3}$

Moreover there will be values the given equation as if f(x) = $x^2$ - $2^x$ - 1 then $f\left(3^{+}\right)f\left(4^{-}\right)$ < 0

So numbers of values of $\theta$ = 4