Question

The most general solutions of the equation $se{c}^{2}x=\sqrt{2}(1-{\tan }^{2}x)$ are given by

• A. $n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{4}$
• B. $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$
• C. $n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{8}$
• D. None of these

Answer 1

The correct option is C

$n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{8}$

Explanation for the correct option:

Step 1: Simplify the given equation

In the question, an equation $se{c}^{2}x=\sqrt{2}(1-{\tan }^{2}x)$ is given.

Rewrite the given equation as follows:

$$\begin{gathered} 1+{\tan }^{2}x=\sqrt{2}(1-{\tan }^{2}x)\mathbf{[}\mathbf{\because }\mathbf{s}\mathbf{e}{\mathbf{c}}^{\mathbf{2}}\mathbf{x}\mathbf{=}\mathbf{1}\mathbf{+}\mathbf{t}\mathbf{a}{\mathbf{n}}^{\mathbf{2}}\mathbf{x}\mathbf{]} \\ \Rightarrow 1+{\tan }^{2}x=\sqrt{2}-\sqrt{2}{\tan }^{2}x \\ \Rightarrow {\tan }^{2}x+\sqrt{2}{\tan }^{2}x=\sqrt{2}-1 \\ \Rightarrow {\tan }^{2}x\left(1+\sqrt{2}\right)=\sqrt{2}-1 \\ \Rightarrow {\tan }^{2}x=\frac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)} \\ \Rightarrow {\tan }^{2}x=\frac{\sqrt{2}-1}{\left(1+\sqrt{2}\right)}·\frac{\left(1-\sqrt{2}\right)}{\left(1-\sqrt{2}\right)} \\ \Rightarrow {\tan }^{2}x=\frac{-{\left(\sqrt{2}-1\right)}^2}{\left(1-2\right)} \\ \Rightarrow {\tan }^{2}x={\left(\sqrt{2}-1\right)}^2...\left(1\right) \end{gathered}$$

So, the given equation can be written as ${\tan }^{2}x={\left(\sqrt{2}-1\right)}^2$.

Step 2: Find the general solution of the given equation

We know that, $\tan \left(2x\right)=\frac{2\tan \left(x\right)}{1-{\tan }^2\left(x\right)}$ and $\tan \left(\frac{\mathrm{\pi }}{4}\right)=1$.

So, substitute $x=\frac{\mathrm{\pi }}{8}$ in equation $1$.

Therefore,

$$\begin{gathered} \tan \left(\frac{\mathrm{\pi }}{4}\right)=\frac{2\tan \left({\displaystyle \frac{\mathrm{\pi }}{8}}\right)}{1-{\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)} \\ \Rightarrow 1=\frac{2\tan \left({\displaystyle \frac{\mathrm{\pi }}{8}}\right)}{1-{\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)} \\ \Rightarrow 1-{\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)=2\tan \left(\frac{\mathrm{\pi }}{8}\right) \\ \Rightarrow {\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)+2\tan \left(\frac{\mathrm{\pi }}{8}\right)-1=0 \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{8}\right)=\frac{-2\pm \sqrt{2^2+4}}{2} \\ \Rightarrow \tan \left(\frac{\mathrm{\pi }}{8}\right)=-1\pm \sqrt{2} \end{gathered}$$ {Since the solution of the equation $a{x}^2+bx+c=0$ can be given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.}

Since, $\tan \left(\frac{\mathrm{\pi }}{8}\right)$ is in the first quadrant. Therefore, $\tan \left(\frac{\mathrm{\pi }}{8}\right)=-1+\sqrt{2}$.

Also, ${\tan }^2\left(\frac{\mathrm{\pi }}{8}\right)={\left(\sqrt{2}-1\right)}^2$

We know that, ${\tan }^{2}x={\tan }^2(n\mathrm{\pi }\pm \mathrm{x})$.

Therefore, ${\tan }^2\left(n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{8}\right)={\left(\sqrt{2}-1\right)}^2...2$

From equation $\left(1\right)$ and equation $\left(2\right)$.

${\tan }^{2}x={\tan }^2\left(n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{8}\right)$

So, $x=n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{8}$.

Therefore, The most general solutions of the equation $se{c}^{2}x=\sqrt{2}(1-{\tan }^{2}x)$ are given by $x=n\mathrm{\pi }\pm \frac{\mathrm{\pi }}{8}$.

Hence, option $C$ is the correct answer.