​ If y = √(1+tan x/1-tan x) then dy/dx = ? A. 1/2 sec^2x.tan(x+π/4) ​ B. sec^2(x+π/4)/2√tan(x+π/4) ​C. sec^2(x/4)/√tan(x+π/4)

Question

If y = \(\sqrt{\frac{1+tan\,x}{1-tan\,x}}\) then dy/dx = ?

√(1+tan x/1-tan x)

A. 1/2 sec²x.tan(x+π/4)

B. \(\frac{sec^2(x+\pi/4)}{2\sqrt{tan(x+\pi/4)}}\)

sec²(x+π/4)/2√tan(x+π/4)

C. \(\frac{sec^2(x/4)}{\sqrt{tan(x+\pi/4)}}\)

sec²(x/4)/√tan(x+π/4)

D. none of these

Answer 1

Answer is:  B. \(\frac{sec^2(x+\pi/4)}{2\sqrt{tan(x+\pi/4)}}\)

Differentiating with respect to x, we get

Hence, dy/dx = \(\frac{sec^2(x+\pi/4)}{2\sqrt{tan(x+\pi/4)}}\)