Question

Tan inverse of acosx/2 +bsinx/2 divided by bcosx/2+asinx/2

Answer 1

Answer:

Step-by-step explanation:

y = tan^-1 [(a cos x - b sin x)/(b cos x + a sin x)]

Divide numerator & denominator by b cos x

y = tan^-1 [ (a/b - tan x)/ (1+ a/b tan x)]

Taking a/b as tan c we have

y = tan^-1 [ tan (c-x)]

y = c-x

So dy/dx = -1.

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