Question

The general solution of the equation \(\tan^2x=1\) is

• A. \(n\pi+\frac{\pi}{4}\) only
• B. \(n\pi\pm\frac{\pi}{4}\)
• C. \(2n\pi\pm\frac{\pi}{4}\)
• D. \(n\pi-\frac{\pi}{4}\) only

Hint:

If \(\tan^2x=1\), then \(\tan x=\pm1\). Therefore, the general solution is \(x=n\pi\pm\frac{\pi}{4}\).

Answer 1

The Correct Option is B

We are given \[ \tan^2x=1. \] Taking square root on both sides: \[ \tan x=\pm 1. \] Now, \[ \tan x=1 \] when \[ x=\frac{\pi}{4}. \] The general solution of \[ \tan x=1 \] is \[ x=n\pi+\frac{\pi}{4}. \] Also, \[ \tan x=-1 \] when \[ x=-\frac{\pi}{4}. \] The general solution of \[ \tan x=-1 \] is \[ x=n\pi-\frac{\pi}{4}. \] Combining both cases: \[ x=n\pi\pm\frac{\pi}{4}. \] Therefore, the general solution is \[ n\pi\pm\frac{\pi}{4}. \]