Question
Sat May 26, 2012 By: Mahesh Meghwal
 

how should I simplify :- tan^-1[(cos x)/(1-sin x)].

Expert Reply
Mon May 28, 2012

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)

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