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If y = cos^-1(3cosx +4sinx/5) π/2<x <π then dY/dx is equal to

If y = cos^-1(3cosx +4sinx/5) π/2<x <π then dY/dx is equal to

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A)
Cos¯¹ (1/5) [ 3 Cos x - 4 Sin x ] = y (say)

=> Cos y = (1/5) [ 3 Cos x - 4 Sin x ]

=> ...= (3/5) Cos x - (4/5) Sin x

angle  of ' A ' such that Cos A = 3/5 and Sin A = 4/5

both Sin A and Cos A are constant quantities .

=>..= Cos A. Cos x - Sin A. Sin x

Using formula Cos A . Cos B - Sin A. Sin B = Cos ( A + B ) we get

=> Cos y = Cos ( A + x )

Hence y = A + x

Differentiating both the sides w.r.t x , taking care that ' A ' is a constant we get

=> y' = 0 + 1

dy/dx = 1
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