If sin y - x sin a - y and dy - dx - A -1- x 2-2 x cos a - then the value of A isA- 2B- cos aC- sin aD- None of these

The correct option is C sin ax = sin y/sin(a + y).....(i) ⇒dx/dy = sin(a)/sin^2 (a + y) ∴dy/dx = sin^2 (a + y)/sin(a) = A/1 + x^2 2x cos a Put x = 0, y = 0, ...

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Standard XII

Mathematics

Implicit Differentiation

If sin y = x ...

Question

If sin y = x sin (a + y) and $\frac{d y}{d x} = \frac{A}{1 + x^{2} - 2 x c o s a^{'}},$ then the value of A is

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cos a

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sin a

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None of these

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Solution

The correct option is C sin a
$x = \frac{s i n y}{s i n (a + y)} . . . . . (i)$
$\Rightarrow \frac{d x}{d y} = \frac{s i n (a)}{s i n^{2} (a + y)}$
$∴ \frac{d y}{d x} = \frac{s i n^{2} (a + y)}{s i n (a)} = \frac{A}{1 + x^{2} - 2 x c o s a}$
$P u t x = 0, y = 0,$
$t h e n A = s i n a$

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If sin y = x sin (a + y) and dydx=A1+x2−2xcos a′, then the value of A is

The correct option is C sin a x=sin ysin(a+y).....(i) ⇒dxdy=sin(a)sin2(a+y) ∴dydx=sin2(a+y)sin(a)=A1+x2−2x cos a Put x=0,y=0, then A=sin a

If sin y = x sin (a + y) and $\frac{d y}{d x} = \frac{A}{1 + x^{2} - 2 x c o s a^{'}},$ then the value of A is

The correct option is C sin a
$x = \frac{s i n y}{s i n (a + y)} . . . . . (i)$
$\Rightarrow \frac{d x}{d y} = \frac{s i n (a)}{s i n^{2} (a + y)}$
$∴ \frac{d y}{d x} = \frac{s i n^{2} (a + y)}{s i n (a)} = \frac{A}{1 + x^{2} - 2 x c o s a}$
$P u t x = 0, y = 0,$
$t h e n A = s i n a$