Question

General solution of $\frac{1-{tan}^{2}x}{{sec}^{2}x}=\frac{1}{2}$ is

• A. $n\pi +\frac{\pi }{6},n\in Z$
• B. $n\pi -\frac{\pi }{6},n\in Z$
• C. $n\pi \pm \frac{\pi }{6},n\in Z$
• D. $2n\pi \pm \frac{\pi }{6},n\in Z$

Answer 1

Answer: (C)

$n\pi \pm \frac{\pi }{6},n\in Z$

Given $\frac{1-{\tan }^{2}x}{{\text{sec}}^{2}x}=\frac{1}{2}$

$⟹\frac{1-{\tan }^{2}x}{1+{\tan }^{2}x}=\frac{1}{2}$

$⟹2-2{\tan }^{2}x=1+{\tan }^{2}x$

$⟹{\tan }^{2}x=\frac{1}{3}={\tan }^2\frac{\pi }{6}$

$⟹x=n\pi \pm \frac{\pi }{6}$