Question

A solution of the equation $(1-tanx)(1+tanx)se{c}^{2}x+2^{ta{n}^2}=0$, where x lies in the interval $[\frac{-\pi }{2},\frac{\pi }{2}]$ is given by

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

Answer 1

The correct option is B $x=\frac{\pi }{3}$

$(1-tanx)(1+tanx)\left(se{c}^{2}x\right)+2ta{n}^{2}x=0$

(a) let $x=0$

$(1-0)(1+0)\left(1\right)+2^{\circ }=0$

$1+1\ne 0$

(b) $x=\frac{\pi }{3}$

$(1-tan\frac{\pi }{3})(1+tan\frac{\pi }{3})\left(se{c}^2\frac{\pi }{3}\right)+2ta{n}^2\frac{\pi }{3}=0$

$(1-\sqrt{3})(1+\sqrt{3})\left(4\right)+2^3=0$

$(1-3)\left(4\right)+8=0$

$-8+8=0$

it is satisfying the given Eq${}^n$ so

it will be the answer of this question.