Question

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

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

Answer 1

The correct option is B

$2n\mathrm{\pi },2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}$

Explanation for the correct option:

Find the general solution of the given equation

Given equation, $secx-1=(\sqrt{2}-1)\tan x...\left(1\right)$

We know that, $se{c}^{2}x-1={\tan }^{2}x$.

So,

$$\begin{gathered} \left(secx-1\right)(secx+1)={\tan }^{2}x \\ \Rightarrow (secx+1)=\frac{{\tan }^{2}x}{secx-1} \end{gathered}$$

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

$$\begin{gathered} (sex+1)=\frac{{\tan }^{2}x}{(\sqrt{2}-1)\tan x} \\ \Rightarrow (secx+1)=\frac{\tan x}{(\sqrt{2}-1)} \\ \Rightarrow (secx+1)=\frac{\tan x}{(\sqrt{2}-1)}·\frac{(\sqrt{2}+1)}{(\sqrt{2}+1)} \\ \Rightarrow (secx+1)=(\sqrt{2}+1)\tan x...\left(2\right) \end{gathered}$$

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

$$\begin{gathered} 2secx=2\sqrt{2}\tan x \\ \Rightarrow \frac{1}{\cos x}=\sqrt{2}\frac{\sin x}{\cos x} \end{gathered}$$

Which is possible when, $\sin x=\frac{1}{\sqrt{2}}$ and $\cos x=1$.

Therefore, The most general solutions of the equation $secx-1=(\sqrt{2}-1)\tan x$ are given by $2n\mathrm{\pi },2\mathrm{n\pi }+\frac{\mathrm{\pi }}{4}$.

Hence, the correct option is $B$.