# What is the derivative of #y=(3(1-sinx))/(2cosx)#?

So, the derivative is

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To find the derivative of ( y = \frac{3(1-\sin x)}{2\cos x} ), use the quotient rule:

[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} ]

Where ( u = 3(1-\sin x) ) and ( v = 2\cos x ). Now, differentiate ( u ) and ( v ) with respect to ( x ):

[ u' = 3(-\cos x) ] [ v' = -2\sin x ]

Now, substitute into the quotient rule formula:

[ \frac{d}{dx}\left(\frac{3(1-\sin x)}{2\cos x}\right) = \frac{(3(-\cos x))(2\cos x) - (3(1-\sin x))(-2\sin x)}{(2\cos x)^2} ]

Simplify:

[ = \frac{-6\cos^2 x + 6\sin x - 6\cos x + 6\sin x\cos x}{4\cos^2 x} ]

[ = \frac{-6\cos^2 x - 6\cos x + 12\sin x\cos x}{4\cos^2 x} ]

[ = \frac{-3\cos^2 x - 3\cos x + 6\sin x\cos x}{2\cos^2 x} ]

[ = \frac{-3(\cos^2 x + \cos x - 2\sin x\cos x)}{2\cos^2 x} ]

[ = \frac{-3(\cos x(\cos x + 1 - 2\sin x))}{2\cos^2 x} ]

[ = \frac{-3\cos x(1 - \sin x)}{2\cos^2 x} ]

[ = \frac{-3(1-\sin x)}{2\cos x} ]

So, the derivative of ( y = \frac{3(1-\sin x)}{2\cos x} ) is ( \frac{-3(1-\sin x)}{2\cos x} ).

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