CBSE Class 12-science Answered

tan^-1 (cosx /1-sinx)

Apply : cos y = cos² (y/2) - sin ² (y/2) and 1-siny = [cos(y/2) - sin(y/2) ]²

tan^-1 ( { cos² (x/2) - sin ² (x/2) } / [cos(x/2) - sin(x/2) ]² )

Apply : a² - b ² = (a+b)(a-b)

tan^-1 ( {( cos(x/2) + sin(x/2) ) * (cos(x/2) - sin(x/2)) } / [cos(x/2) - sin(x/2) ]² )

tan^-1 ( {( cos(x/2) + sin(x/2) ) * ( 1 )} / [cos(x/2) - sin(x/2) ]  )

tan^-1 ( {( cos(x/2) + sin(x/2) ) } / [cos(x/2) - sin(x/2) ] )

Divide the Numerator and Denominator by "cos(x/2)"

tan^-1 ( { ( cos(x/2) + sin(x/2) ) / cos(x/2) } / [ {cos(x/2) - sin(x/2)} / cos(x/2) ] )

We get :

tan^-1( { 1 + tan(x/2) } / [ 1- tan(x/2) ] )

tan^-1( { tan?/4 + tan(x/2) } / [ 1-tan?/4 tan(x/2) ] )

Apply : { tan x+ tan y / [ 1-tan x * tany ] } = tan(x+y)

tan ^-1 ( tan( ?/4 + (x/2) ) )

Apply : tan ^-1 ( tan z) = z

?/4 + (x/2)

tan^-1 (cosx/1-sinx) = ?/4 + (x/2)