If siny -x sin a-y- price that d y-d x-sin 2a-y-sin a

Since, siny = xsin (a + y) we have: x=siny/sin(a+y) Deriving both sides we have: dx=cosy/sin (a+y) sin(y)cos(a+y)/sin^2 (a+y)dy dx/dy=cos(y)sin(a+y) sin(y)co ...

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Byju's Answer

Standard XII

Mathematics

Implicit Differentiation

If siny =x si...

Question

If $s i n y = x s i n (a + y)$ , price that $\frac{d y}{d x} = s i n^{2} \frac{(a + y)}{s i n a}$

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Solution

Since, $s i n y = x s i n (a + y)$ we have:
$x = \frac{s i n y}{s i n (a + y)}$
Deriving both sides we have:
$d x = \frac{c o s y}{s i n (a + y)} - \frac{s i n (y) c o s (a + y)}{s i n^{2} (a + y)} d y$
$\frac{d x}{d y} = \frac{c o s (y) s i n (a + y) - s i n (y) c o s (a + y)}{s i n^{2} (a + y)}$
Using $s i n (a + b) = s i n (a) c o s (b) + s i n (b) c o s (a)$ , we get that:
$\frac{d x}{d y} = \frac{s i n (a + y - y)}{s i n^{2} (a + y)} = s i n (a) s i n^{2} (a + y)$
So, $\frac{d x}{d y} = \frac{s i n^{2} (a + y)}{s i n (a)}$

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If siny=xsin(a+y), price that dydx=sin2(a+y)sina

Since, siny=xsin(a+y) we have: x=sinysin(a+y) Deriving both sides we have: dx=cosysin(a+y)−sin(y)cos(a+y)sin2(a+y)dy dxdy=cos(y)sin(a+y)−sin(y)cos(a+y)sin2(a+y) Using sin(a+b)=sin(a)cos(b)+sin(b)cos(a), we get that: dxdy=sin(a+y−y)sin2(a+y)=sin(a)sin2(a+y) So, dxdy=sin2(a+y)sin(a)

If $s i n y = x s i n (a + y)$ , price that $\frac{d y}{d x} = s i n^{2} \frac{(a + y)}{s i n a}$