The number of all values of theta in the inte | vedclass

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(C) Given equation: $(1-	an 	heta)(1+	an 	heta) \sec ^2 	heta+2 	an ^2 	heta=0$Using $(1-	an 	heta)(1+	an 	heta) = 1-	an^2 	heta$ and $\s

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$2$

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(C) Given equation: $(1-	an 	heta)(1+	an 	heta) \sec ^2 	heta+2 	an ^2 	heta=0$Using $(1-	an 	heta)(1+	an 	heta) = 1-	an^2 	heta$ and $\sec^2 	heta = 1+	an^2 	heta$,we get:$(1-	an^2 	heta)(1+	an^2 	heta) + 2	an^2 	heta = 0$Let $x = 	an^2 	heta$. Since $	heta \in \left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$,$x \ge 0$.The equation becomes $(1-x)(1+x) + 2x = 0$$1 - x^2 + 2x = 0$$x^2 - 2x - 1 = 0$Solving for $x$ using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{2 \pm \sqrt{4 - 4(1)(-1)}}{2} = \frac{2 \pm \sqrt{8}}{2} = 1 \pm \sqrt{2}$Since $x = 	an^2 	heta \ge 0$,we must have $x = 1 + \sqrt{2}$ (as $1 - \sqrt{2} Thus,$	an^2 	heta = 1 + \sqrt{2}$.Since $1 + \sqrt{2} > 0$,there are two values of $	an 	heta$,namely $	an 	heta = \pm \sqrt{1 + \sqrt{2}}$.For each value of $	an 	heta$,there is exactly one value of $	heta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}ight)$.Therefore,there are $2$ values of $	heta$ that satisfy the equation.

The number of all values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ satisfying the equation $(1-\tan \theta)(1+\tan \theta) \sec ^2 \theta+2 \tan ^2 \theta=0$ is

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