A solution of the equation $(1 - t a n x) (1 + t a n x) s e c^{2} x + 2^{t a n^{2}} = 0$ , where x lies in the interval $[\frac{- π}{2}, \frac{π}{2}]$ is given by

x = 0

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x = \frac{π}{3}

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x = \frac{- π}{3}

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x = \frac{π}{6}

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Solution

The correct option is B $x = \frac{π}{3}$
$(1 - t a n x) (1 + t a n x) (s e c^{2} x) + 2 t a n^{2} x = 0$
(a) let $x = 0$
$(1 - 0) (1 + 0) (1) + 2^{\circ} = 0$
$1 + 1 \neq 0$
(b) $x = \frac{π}{3}$
$(1 - t a n \frac{π}{3}) (1 + t a n \frac{π}{3}) (s e c^{2} \frac{π}{3}) + 2 t a n^{2} \frac{π}{3} = 0$
$(1 - \sqrt{3}) (1 + \sqrt{3}) (4) + 2^{3} = 0$
$(1 - 3) (4) + 8 = 0$
$- 8 + 8 = 0$
it is satisfying the given Eq $^{n}$ so
it will be the answer of this question.

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The correct option is B x=π3(1−tanx)(1+tanx)(sec2x)+2tan2x=0(a) let x=0 (1−0)(1+0)(1)+2∘=01+1≠0(b) x=π3(1−tanπ3)(1+tanπ3)(sec2π3)+2tan2π3=0 (1−√3)(1+√3)(4)+23=0(1−3)(4)+8=0−8+8=0it is satisfying the given Eqn soit will be the answer of this question.

A solution of the equation $(1 - t a n x) (1 + t a n x) s e c^{2} x + 2^{t a n^{2}} = 0$ , where x lies in the interval $[\frac{- π}{2}, \frac{π}{2}]$ is given by