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प्रश्न
If cos y = x cos (a + y), with cos a ≠ ± 1, prove that `dy/dx = cos^2(a+y)/(sin a)`
उत्तर
cos y = x cos (a + y)
`therefore x = (cos y)/(cos (a + y))`
On differentiating with respect to y,
`cos (a + y) d/dy cos y - cos y d/dy`
`therefore dx/dy = (cos (a + y))/(cos^2 (a + y))`
`= (- sin y cos (a + y) + cos y sin (a + y))/(cos^2 (a + y))`
`= (sin (a + y) cos y - cos (a + y) sin y)/(cos^2 (a + y))`
`= (sin (a + y - y))/(cos^2 (a + y))` ... [∵ sin (A-B) = sin A cos B - cos A sin B]
`= (sin a)/(cos^2 (a + y))`
`therefore dy/dx = (cos^2 (a + y))/(sin a)`
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