Question

if x sin ^3 theta +y cos^3 theta =sin theta cos theta and x sin theta =y cos theta ,then show that x^2 + y^2 =1

Answer 1

Answer:

 

$\large \text{ x^2+y^2=1 \ proved}$

Step-by-step explanation:

Given :

$\large \text{ x \ sin^3\theta+y \ cos^3\theta=sin \theta. cos \theta and x \ sin \theta=y \ cos\theta }$

We have prove

$\large x^2+y^2 =1$

We know that

$\large tan=\dfrac{sin}{cos}$

So we have

$\large x \ sin \theta=y \ cos\theta}\\\\\\\large tan \theta=\dfrac{y}{x} \ ...( i)$

We also have

$\large x \ sin^3\theta+y \ cos^3\theta=sin \theta. cos \theta$

Now divide it by $cos^3 \theta$ we get

$\large x \ \dfrac{sin^3\theta}{cos^3 \theta} +y \ \dfrac{cos^3\theta}{cos^3 \theta} =\dfrac{sin \theta. cos \theta}{cos^3 \theta} \\\\\\\large x \ tan^3 \theta + y = \dfrac{tan \theta}{cos \theta}$

Now put value of $tan \theta=\dfrac{y}{x}$

$\large \text{ x \ (\dfrac{y}{x})^{3} \theta + y = \dfrac{y\sqrt{x^2+y^2} }{x\times x} }}\\\\\\\large \text{Here we directly write cos \theta=\sqrt{1+tan^2 \theta} }\\\\\\\large y^3+x^2y=y\sqrt{x^2+y^2}\\\\\\\large x^2+y^2=\sqrt{x^2+y^2}\\\\\\\large 1=\sqrt{x^2+y^2}\\\\\\\large x^2+y^2=1$

One step it write directly you can see in picture.

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

xsin³∅ + ycos³∅ = sin∅. cos∅

x.sin³∅ + cos²∅.(ycos∅) = sin∅.cos∅

use xsin∅= ycos∅ in above

xsin³∅ + cos²∅.xsin∅ = sin∅.cos∅

xsin∅( sin²∅ + cos²∅) = sin∅.cos∅

xsin∅ = sin∅.cos∅

x = cos∅

so, y = sin∅

now we know,

sin²∅ + cos²∅ = 1

put sin∅ = y

cos∅ = x

hence, x² + y² = 1

Hope thse helps u