Question

if sin y =(a+y) , show that dy/dx =sin²(a+y)/sin a​

Answer 1

$\bigstar \underline {\underline{\sf \bf CORRECT\ QUESTION:-}}$

❥ If sin y= x sin(a_y) ,Show that dy/dx = sin²(a+y)/sin a​.

$\bigstar \underline{\underline{\sf \bf ANSWER:-}}$

We need to know :

❥ Chain rule :

$\mapsto \sf \orange{\frac{dy}{dx} =\frac{dy}{du} \frac{du}{dx}}$

❥ Quotient rule :

$\mapsto \sf \orange{\frac{dy}{dx} =\frac{v(du/dx)-u(dv/dx)}{v^2}}$

---------------------------

Now,

Given that sin y = x sin(a+y)

=> x = siny / sin(a+y)

Differentiating w.r.t y ,

$:\implies \sf \frac{dx}{dy} = \frac{cosysin(a+y)-sinycos(a+y)}{sin^2(a+y)} \\\\\\:\implies \sf \frac{dx}{dy} = \frac{cosy(sinacosy+cosasiny)-siny(cosacosy-sinasiny)}{sin^2(a+y)} \\\\\\:\implies \sf \frac{dx}{dy} = \frac{cos^2ysina+cosasinycosy-sinycosacosy+sinasin^2y}{sin^2(a+y)} \\\\\\:\implies \sf \frac{dx}{dy} =\frac{cos^2ysina+sinasin^2y}{sin^2(a+y)} \\\\\\:\implies \sf \frac{dx}{dy} =\frac{sina(cos^2y+sin^2y)}{sin^2(a+y)}$

$\tt [cos^2 \theta+sin^2 \theta = 1]$

$:\implies \sf \frac{dx}{dy} =\frac{sina(1)}{sin^2a(a+y)}\\\\\\:\implies \sf \frac{dx}{dy} = \frac{sina}{sin^2a(a+y)}$

Therefore,

$\mapsto \boxed{\boxed{\bold{\frac{dy}{dx} =\frac{sin^2(a+y)}{sina}}}}$

Hence proved !

__________________

Answer 2

$\bigstar\huge\bf\underline{\underline{\red{Answer:-}}}$

$\bigstar\bf\underline\blue{Given:-}$

• $\bf sin \ y \ = \ x \ sin \ (a+y)$

$\bigstar\bf\underline\green{To \ prove:-}$

• $\bf sin \ y \ = \ x \ sin \ (a+y) \ = \frac{sin^{2}(a+y)}{sin \ a}$

$\bigstar\bf\underline\orange{Proof:-}$

$\bf Given,$

$\bf sin \ y \ = \ x \ sin \ (a+y)$

$\bf\implies x=\frac{sin \ y}{sin(a+y)}$

$\bf Differentiating \ w.r.t.x \ on \ both \ sides$

$\bf \frac{dx}{dy}=\frac{sin(a+y).cos \ y-sin \ y \ cos(a+y)}{sin^{2}(a+y)}$

$\bf \frac{dx}{dy}=\frac{sin(a+y-y)}{sin^{2}(a+y)}$

$\bf\implies\frac{dx}{dy}=\frac{sin \ a}{sin^{2}(a+y)}$

$\bf Hence,$

$\bf\boxed{\boxed{\purple{\frac{dx}{dy}=\frac{sin \ a}{sin^{2}(a+y)}}}}$

$\bf\underline\pink{Hence \ proved \ !!}$