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Question
Differentiate the following w.r.t. x : `tan^-1[(1 + cos(x/3))/(sin(x/3))]`
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Solution
Let y = `tan^-1[(1 + cos(x/3))/(sin(x/3))]`
= `tan^-1[(2cos^2(x/6))/(2sin(x/6)cos(x/6))]`
= `tan^-1[cot(x/6)]`
= `tan^-1[tan(pi/2 - x/6)]`
= `pi/(2) - x/(6)`
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"(pi/2 - x/6)`
= `"d"/"dx"(pi/2) - (1)/(6)"d"/"dx"(x)`
= `0 - (1)/(6) xx 1`
= `-(1)/(6)`.
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