IIT-JEE 1996Question 1
General value of $\theta$ satisfying the equation $\tan^{2}\theta+\sec2\theta=1$ is ______.
Trigonometric Ratio & Identities
Answer$\theta=n\pi$ or $\theta=n\pi\pm\dfrac{\pi}{3}$, $n\in\mathbb Z$
Solution
Concept. With $t=\tan^2\theta$, $\sec2\theta=\frac{1+t}{1-t}$, and the equation becomes a quadratic in $t$. Formulas used. (1) $\cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$Step 1 - Substitute. By (1), with $t=\tan^2\theta$ ($t\ne1$): $$t+\frac{1+t}{1-t}=1\ \Longrightarrow\ t(1-t)+1+t=1-t\ \Longrightarrow\ 3t-t^2=0.$$ Step 2 - Solve. $t=0$ or $t=3$ (both allowed: $\cos2\theta=1$ or $-\frac12$). Step 3 - General values. $\tan\theta=0$ gives $\theta=n\pi$; $\tan\theta=\pm\sqrt3$ gives $\theta=n\pi\pm\frac\pi3$. Answer: $\theta=n\pi$ or $\theta=n\pi\pm\dfrac{\pi}{3}$, $n\in\mathbb Z$