`y=tan^(-1)((acosx-bsinx)/(bcosx+asinx)), `where `-pi/2ltxltpi`and `a/btanx gt -1`

y=tan^(-1)((acosx-bsinx)/(bcosx+asinx)),w h e r e-pi/2 -1

tan^(-1)((a+bcosx)/(b-acosx))

Differentiate tan^(-1)((acosx-bsinx)/(bcosx+asinx))-(pi)/(2)ltxlt(pi)/(2)" and "(a)/(b)tanxgt-1 . Firstly, convert the given inverse trigonometic function into the simplest form and the differentiate it.

Write the simplest form of tan^(-1)[(acosx-bsinx)/(bcosx + asinx)] , if a/b tanx gt -1.

Simplify: tan^(-1)[(acosx-bsinx)/(b cosx+a sinx)]

If y = tan^-1((acosx-bsinx)/(bcosx+asinx)) , then find (dy)/(dx)

Simplify tan^(-1)[(acosx-bsinx)/(bcosx+asinx)] , if a/btanxgt-1 .

Simplify tan^-1[(acosx-bsinx)/(bcosx+asinx)] if a/b tanx>-1