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, ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

cos a

No worries! We‘ve got your back. Try BYJU‘S free classes today!

sin a

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

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

Suggest Corrections

Similar questions

Q. If cos y = x cos(a+y) and

\frac{d y}{d x} = \frac{k}{1 + x^{2} - 2 x c o s a}

then the value of k is

Q. If

x sin y = sin (y + a)

and

\frac{d y}{d x} = \frac{A}{1 + x^{2} - 2 x cos a}

then the value of

A

If siny=xsin(a+y), then dy/dx=

Q. If

sin y = x sin (a + y),

then show that

\frac{d y}{d x} = \frac{{sin}^{2} (a + y)}{sin a}

Q. If

sin y = x sin (a + y)

, then prove that

\frac{d y}{d x} = \frac{{sin}^{2} (a + y)}{sin a}

Join BYJU'S Learning Program

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