Question

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

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

Answer 1

The correct option is C $n\pi \pm \frac{\pi }{8}$

${\sec }^2\theta =\sqrt{2}(1-{\tan }^2\theta )$
$1+{\tan }^2\theta =\sqrt{2}(1-{\tan }^2\theta )$
$1+{\tan }^2\theta =\sqrt{2}-\sqrt{2}{\tan }^2\theta$
${\tan }^2\theta =\frac{\sqrt{2}-1}{1+\sqrt{2}}$

We know,
$\tan \frac{\pi }{4}=1=\frac{2\tan \frac{\pi }{8}}{1-{\tan }^2\frac{\pi }{8}}$
$\tan \frac{\pi }{8}=-1\pm \sqrt{2}$
${\tan }^2\frac{\pi }{8}=(-1\pm \sqrt{2}{)}^2=\frac{\sqrt{2}-1}{1+\sqrt{2}}$

So, principle solution $=\frac{\pi }{8}$
General $\Rightarrow$
$\theta =n\pi \pm \frac{\pi }{8}$