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

• A

  1
• B

  0
• Correct answer
  C

  2
• D

  infinitely many.

Solution

$\begin{aligned} & (1-\tan \theta)(1+\tan \theta) \sec ^2 \theta+2 \tan ^2 \theta=0 \\ & \Rightarrow\left(1-\tan ^2 \theta\right)\left(1+\tan ^2 \theta\right)+2 \tan ^2 \theta=0 \end{aligned}$ Put $\tan ^2 \theta=x$ $\begin{aligned} & \Rightarrow(1-x)(1+x)+2 x=0 \\ & \Rightarrow 1-x^2+2 x=0 \\ & \Rightarrow x^2-1=2 x \end{aligned}$ Let us draw graph of $y=x^2-1$ and $y=2 x$
From the graph the two curves are intersecting at 2 points. $\therefore \quad$ There are 2 values of $x$. Only one value of $x$ exists, for $\theta \in\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ $\therefore \quad x=\tan ^2 \theta$ $\therefore \quad$ Two values of $\theta$ satisfies above equation

Hence, the correct answer is option: 'C'.