Question

If $\sqrt{2}\sec x+\tan x=1$, then the value of $x$ is

• A. $2n\pi +\frac{\pi }{3}$
• B. $2n\pi -\frac{\pi }{4}$
• C. $2n\pi +\frac{\pi }{6}$
• D. $2n\pi +\frac{\pi }{12}$

Answer 1

Answer: (B)

$2n\pi -\frac{\pi }{4}$

Given

$\sqrt{2}\sec x+\tan x=1$

$⟹\frac{\sqrt{2}}{\cos x}+\frac{\sin x}{\cos x}=1$

$(\sec x=\frac{1}{\cos x},\tan x=\frac{\sin x}{\cos x})$

$⟹\sqrt{2}+\sin x=\cos x$

$⟹\cos x-\sin x=\sqrt{2}$

$⟹\frac{1}{\sqrt{2}}\cos x-\frac{1}{\sqrt{2}}\sin x=1$

$⟹\cos \frac{\pi }{4}\cos x-\sin \frac{\pi }{4}\sin x=\cos 0$

$⟹\cos (\frac{\pi }{4}+x)=\cos 0$

$(\cos (A+B)=\cos A\cos B-\sin A\sin B)$

$⟹\cos (\frac{\pi }{4}+x)=\cos 0$

$⟹\frac{\pi }{4}+x=2n\pi$

$(\cos k=0⟹k=2n\pi )$

$⟹x=2n\pi -\frac{\pi }{4}$