# if y=log sqrt(1+tan x/1-tan x)) prove that dy/dx=sec 2x

Rinkoo Gupta
askIITians Faculty 80 Points
9 years ago

y=log sqrt(1+tan x/1-tan x))

dy/dx=logv[(tanp/4 +tanx)/(1-tanp/4.tanx)]

=logvtan(p/4+x)

=logtan(p/4+x)1/2

=(1/2). log tan(p/4+x)

Now diff.w.r. to x , we get

Dy/dx=(1/2). 1/tan(p/4+x) . sec2(p/4+x) .1

=(1/2).cos(p/4+x)/sin(p/4+x).1/cos2(p/4+x)

=1/[2sin(p/4+x)cos(p4+x)]

=1/sin(2p/4+2x)

=1/sin(p/2+2x)

=1/cos2x

=sec2x

Thanks & Regards

Rinkoo Gupta

Arun Kumar IIT Delhi
askIITians Faculty 256 Points
9 years ago
Hi

$\\y=log(\sqrt{\frac{1+tanx}{1-tanx}}) \\\frac{dy}{dx}=\frac{1}{2*\sqrt{\frac{1+tanx}{1-tanx}}}*(\frac{sec^2x}{1-tanx}+\frac{sec^2x*(1+tanx)}{(1-tanx)^2})*\frac{1}{\sqrt{\frac{1+tanx}{1-tanx}}} \\\frac{dy}{dx}=\frac{1}{2*{\frac{1+tanx}{1-tanx}}}*(\frac{sec^2x}{1-tanx}+\frac{sec^2x*(1+tanx)}{(1-tanx)^2}) \\\frac{dy}{dx}=\frac{sec^2x}{2*(1-tan^2x)}*((1+tanx)+(1-tanx))=\frac{sec^2x}{1-tan^x}=\frac{1}{cos^2x-sin^2x} \\=sec2x$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty