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Question
If x sin(a + y) + sin a cos(a + y) = 0 then show that `("d"y)/("d"x) = (sin^2("a" + y))/(sin"a")`
Solution
x sin(a + y) + sin a cos(a + y) = 0 .......(i)
`x.cos("a" + y)*"d"/("d"x)("a" + y) + sin("a" + y)*"d"/("d"x)(x) + sin"a"[-sin("a" + y)]*"d"/("d"x)("a" + y)` = 0
∴ `xcos("a" + y )("d"y)/("d"x) + sin("a" + y)(1) - sin("a" + y)("d"y)/("d"x)` = 0
∴ `[x cos("a" + y) - sin "a" sin("a" + y)]("d"y)/("d"x)` = − sin(a + y) .......(ii)
From (i), we get
x = `(-sin"a"cos("a" + y))/(sin("a" + y))`
Substituting the value of x in (ii), we get
`[(-sin"a"cos("a" + y))/(sin("a" + y))*cos("a" + y) - sin"a"sin("a" + y)]("d"y)/("d"x)` = − sin(a + y)
∴ `-sin"a"[(cos^2("a" + y))/(sin("a" + y)) + sin("a" + y)]("d"y)/("d"x)` = − sin(a + y)
∴ `(-sin"a"[cos^2("a" + y) + sin^2("a" + y)])/(sin("a" + y))("d"y)/("d"x)` − sin(a + y)
∴ `-(sin"a"(1))/(sin("a" + y))*("d"y)/("d"x)` = − sin(a + y)
∴ `("d"y)/("d"x) = sin("a" + y) [(sin("a" + y))/(sin"a")]`
∴ `("d"y)/("d"x) = (sin^2("a" + y))/(sin"a")`
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