Question

If cosy=xcos(a+y) with cosa≠±1 prove that dy/dx=cos^2(a+y) / sina

Answer 1

cosy = xcos(a + y) --------(1)
differentiate with respect to x,
d(cosy)/dx = d[xcos(a + y)]/dx
-siny.dy/dx = x. d{cos(a + y)}/dx + cos(a + y).dx/dx
-siny.dy/dx = x[-sin(a + y). dy/dx] + cos(a + y)
-siny. dy/dx = -xsin(a + y).dy/dx + cos(a + y)
dy/dx [xsin(a + y) - siny ] = cos(a + y)
dy/dx = cos(a + y)/[xsin(a + y) - siny]

now, from equation (1),
x = cosy/cos(a + y)

so, dy/dx = cos(a + y)/[cosy.sin(a + y)/cos(a + y) - siny]
= cos²(a + y)/[cosy.sin(a + y) - siny.cos(a + y)]
= cos²(a + y)/[sin(a + y - y)]
[ use formula, sin(C - D) = sinC.cosD - cosC.sinD]
= cos² (a + y)/sina

hence, $\bf{\frac{dy}{dx}=\frac{cos^2(a+y)}{sina}}$

Answer 2

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