Prove that \[ \tan\left(\frac{\pi}{4}+x\right) + \tan\left(\frac{\pi}{4}-x\right) = 2\sec2x \]

Proof: Using the identity \[ \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} \] we get \[ \tan\left(\frac{\pi}{4}+x\right) = \frac{1+\tan x}{1-\tan x} \] Similarly, \[ \tan\left(\frac{\pi}{4}-x\right) = \frac{1-\tan x}{1+\tan x} \] Therefore, \[ LHS = \frac{1+\tan x}{1-\tan x} + \frac{1-\tan x}{1+\tan x} \] Taking LCM: \[ LHS = \frac{(1+\tan x)^2+(1-\tan x)^2}{1-\tan^2x} \] Expanding numerator: \[ = \frac{ 1+2\tan x+\tan^2x + 1-2\tan x+\tan^2x }{1-\tan^2x} \] \[ = \frac{2+2\tan^2x}{1-\tan^2x} \] \[ = \frac{2(1+\tan^2x)}{1-\tan^2x} \] Using the identity \[ \sec2x=\frac{1+\tan^2x}{1-\tan^2x} \] we get \[ LHS=2\sec2x \] Hence proved, \[ \boxed{ \tan\left(\frac{\pi}{4}+x\right) + \tan\left(\frac{\pi}{4}-x\right) = 2\sec2x } \]