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write sin(cos-1x-tan-1y) in terms of x and y only where -1
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Recall the identity, \[\sin(x\pm y)=\sin x\cos y\pm\sin y\cos x\]
This means \[\sin(\cos^{-1}x-\tan^{-1}y)=\sin(\cos^{-1}x)\cos(\tan^{-1}y)-\sin(\tan^{-1}y)\cos(\cos^{-1}x)\]
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