Question

differentiate tan inverse of [root of 1+sinx/1-sinx] w.r.t tan inverse (of root of 1+x2 +x)

Answer 1

y=√((1-sinx)/(1+sinx))

1-sinx/1-sinx=

       =   1-cos(90-x)/1+cos(90-x)

       =  1-1+2Sin2(45-(x/2))/1+2cos2(45-(x/2))-1

       =    [tan(45-x/2)]2

 

hence y= [tan2(45-x/2)]1/2

            y= tan (45-x/2)

   dy/dx= sec2(45-x/2)  . {0-1/2}

   dy/dx=  -0.5sec2(45-x/2) 

0r        dy/dx+0.5sec2(45-x/2) = 0