Trigonometrical Equations 1 Question 23
23. General value of $\theta$ satisfying the equation $\tan ^{2} \theta+\sec 2 \theta=1$ is……. $\quad$ (1996, 1M)
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Answer:
Correct Answer: 23. $\theta=m \pi, n \pi \pm \frac{\pi}{3}$
Solution:
- Given, $\tan ^{2} \theta+\sec 2 \theta=1$
$ \begin{array}{lr} \Rightarrow & \tan ^{2} \theta+\frac{1}{\cos 2 \theta}=1 \\ \Rightarrow & \tan ^{2} \theta+\frac{1+\tan ^{2} \theta}{1-\tan ^{2} \theta}=1 \\ \Rightarrow & \tan ^{2} \theta\left(1-\tan ^{2} \theta\right)+\left(1+\tan ^{2} \theta\right)=1-\tan ^{2} \theta \\ \Rightarrow & 3 \tan ^{2} \theta-\tan ^{4} \theta=0 \\ \Rightarrow & \tan ^{2} \theta\left(3-\tan ^{2} \theta\right)=0 \\ \Rightarrow & \tan \theta=0 \\ \text { or } & \tan \theta= \pm \sqrt{3} \end{array} $
Now, $\tan \theta=0 \Rightarrow \theta=m \pi$, where $m$ is an integer.
$ \text { and } \begin{aligned} & & \tan \theta & = \pm \sqrt{3}=\tan \pm \frac{\pi}{3} \\ \Rightarrow & & \theta & =n \pi \pm \frac{\pi}{3} \end{aligned} $
$\therefore \theta=m \pi, n \pi \pm \frac{\pi}{3}$, where $m$ and $n$ are integers.
