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Question
If x sin (a + y) + sin a cos (a + y) = 0, prove that `"dy"/"dx" = (sin^2("a" + y))/sin"a"`
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Solution
Given that: x sin (a + y) + sin a cos (a + y) = 0
⇒ x sin (a + y) = – sin a cos (a + y)
⇒ x = `(-sin"a" * cos("a" + y))/(sin ("a" + y))`
⇒ x = – sin a.cot (a + y)
Differentiating both sides w.r.t. y
⇒ `"dx"/"dy" = - sin"a"*"d"/"dy" cot("a" + y)`
⇒ `"dx"/"dy" = -sin"a"[-"cosec"^2("a" + y)`
⇒ `"dx"/"dy" = sin"a"/(sin^2("a" + y))`
∴ `"dy"/"dx" = 1/("dx"/"dy")`
= `1/(sin"a"/(sin^2("a" + y))`
Hence, `"dy"/"dx" = (sin^2("a" + y))/sin"a"`.
Hence proved.
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