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Question
If x sin (a + y) + sin a . cos (a + y) = 0, then show that `"dy"/"dx" = (sin^2(a + y))/(sina)`.
Solution 1
x sin (a + y) + sin a . cos (a + y) = 0 ...(1)
Differentiating w.r.t. x, we get
`x"d"/"dx"[sin(a + y)] + sin(a + y)."d"/"dx"(x) + (sina)."d"/"dx"[cos(a + y)]= 0`
`∴ xcos(a + y)."d"/"dx"(a + y) + sin(a + y) xx 1 + (sina) [-sin(a + y)]."d"/"dx"(a + y) = 0`
`∴ xcos(a + y).(0 + dy/dx) + sin(a + y) - sina.sin(a + y)(0 + dy/dx) = 0`
`∴ xcos(a + y)"dy"/"dx" + sin(a + y) - sina.sin(a + y)"dy"/"dx" = 0`
`∴ sina.sin(a + y)"dy"/"dx" - xcos(a + y)"dy"/"dx" = sin(a + y)`
`∴ [sina.sin(a + y) - xcos(a + y)]"dy"/"dx" = sin(a + y)`
`∴ "dy"/"dx" = (sin(a + y))/(sina.sin(a + y) - xcos(a + y)`
From (1),
`x = (-sina.cos(a + y))/(sin(a + y)`
`∴ "dy"/"dx" = (sin(a + y))/(sina.sin(a + y) + (sina.cos(a + y))/(sin(a + y)).cos(a + y)`
`= (sin^2(a + y))/(sina.sin^2(a + y) + sina.cos^2(a + y)`
`= (sin^2(a + y))/(sina[sin^2(a + y) + cos^2(a + y)]`
`∴ "dy"/"dx" = (sin^2(a + y))/(sina)`.
Solution 2
x sin (a + y) + sin a . cos (a + y) = 0
∴ x sin (a + y) = – sin a. cos (a + y)
∴ x = `-sina.(cos(a + y))/(sin(a + y)`
∴ x = – sin a . cot (a + y)
Differentiating both sides w.r.t. y, we get
`"dx"/"dy" = -sina."d"/"dx"[cot(a + y)]`
= `-sina.[-"cosec"^2(a + y)]."d"/"dx"(a + y)`
= sin a . cosec2(a + y) . (0 + 1)
= `(sina)/(sin^2(a + y)`
∴ `"dy"/"dx" = (1)/((dx/dy)`
= `(sin^2(a + y))/(sina)`.
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