Question

Principle value of $\theta$ satisfying $ta{n}^2\theta +sec2\theta =1$ is

• A. $\frac{\pi }{4}$
• B. $\frac{3\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{7}$

Answer 1

Answer: (C)

$\frac{\pi }{3}$

Given

${\tan }^2\theta +\sec 2\theta =1$

$⟹{\tan }^2\theta =1-\sec 2\theta$

$⟹{\tan }^2\theta =1-\frac{1}{\cos 2\theta }$

$⟹{\tan }^2\theta =\frac{\cos 2\theta -1}{\cos 2\theta }$

$⟹{\tan }^2\theta =\frac{-2{\sin }^2\theta }{\cos 2\theta }$

$(\cos 2\theta =1-2{\sin }^2\theta )$

$⟹\frac{{\sin }^2\theta }{{\cos }^2\theta }=\frac{-2{\sin }^2\theta }{\cos 2\theta }$

$⟹\frac{1}{{\cos }^2\theta }=\frac{-2}{\cos 2\theta }$

$⟹\cos 2\theta =-2{\cos }^2\theta$

$⟹2{\cos }^2\theta -1=-2co{s}^2\theta$

$(\cos 2\theta =2{\cos }^2\theta -1)$

$⟹4{\cos }^2\theta =1$

$⟹{\cos }^2\theta =\frac{1}{4}$

$⟹\cos \theta =\frac{1}{2}$

$⟹\theta =\frac{\pi }{3}$