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

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

AskIITians Faculty