if x sin ^3 theta +y cos^3 theta =sin theta cos theta and x sin theta

1.	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» 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 $\large \text{$ x^2+y^2 =1$}$ We KNOW that $\large \text{$tan=\dfrac{sin}{cos} $}$ So we have $\large \text{ $x \ sin \theta=y \ cos\theta$}}\\\\\\\large \text{$tan \theta=\dfrac{y}{x} \ ...( i)$}$ We ALSO have $\large \text{$x \ sin^3\theta+y \ cos^3\theta=sin \theta. cos \theta$}$ Now divide it by $cos^3 \theta$ we get $\large \text{$x \ \dfrac{sin^3\theta}{cos^3 \theta} +y \ \dfrac{cos^3\theta}{cos^3 \theta} =\dfrac{sin \theta. cos \theta}{cos^3 \theta} $}\\\\\\\large \text{$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} $}\\\\\\\large \text{$y^3+x^2y=y\sqrt{x^2+y^2}$}\\\\\\\large \text{$x^2+y^2=\sqrt{x^2+y^2}$}\\\\\\\large \text{$1=\sqrt{x^2+y^2}$}\\\\\\\large \text{$x^2+y^2=1$}$ One step it write directly you can see in picture. Hence proved.

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»

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

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

We KNOW that

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

So we have

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

We ALSO have

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

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

$\large \text{$x \ \dfrac{sin^3\theta}{cos^3 \theta} +y \ \dfrac{cos^3\theta}{cos^3 \theta} =\dfrac{sin \theta. cos \theta}{cos^3 \theta} $}\\\\\\\large \text{$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} $}\\\\\\\large \text{$y^3+x^2y=y\sqrt{x^2+y^2}$}\\\\\\\large \text{$x^2+y^2=\sqrt{x^2+y^2}$}\\\\\\\large \text{$1=\sqrt{x^2+y^2}$}\\\\\\\large \text{$x^2+y^2=1$}$