Question

if x sin cube theta + y cos cube theta = sin theta cos theta× sin theta = y cos theta then prove - x square + y square = 1

Answer 1

hope you understand

Answer 2

$\large{\underline{\bf{\green{Given:-}}}}$

✰ $\bf\:x\:sin^3\theta+ y \:cos^3\theta=sin\theta.cos\theta$

✰ $\bf\:x\:sin\:\theta=y\:cos\theta$

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to prove that $\bf\:x^2+y^2=1$

$\huge{\underline{\bf{\red{Solution:-}}}}$

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

$:\implies \:\:(x\:sin\theta)sin^2\theta+(y\:cos\theta)cos^2\theta=sin\theta.cos\theta$

$:\implies \:\:(x \:sin\theta)sin^2\theta+(x\:sin\theta)cos^2\theta=sin\theta.cos\theta$

$:\implies {\pink\:{\bf\:{y\:cos\theta=x\:sin\theta}}}$

$:\implies \:\:(x\:sin\theta)(sin^2\theta+cos^2\theta)=sin\theta.cos\theta$

$:\implies \:\:x\:sin\theta=sin\theta.cos\theta$

$:\implies \:\:{\pink{\bf{sin^2\theta+cos^2\theta=1}}}$

$:\implies \:\:x=cos\theta...................(i)$

Now, $\:x\:sin\theta=y\:cos\theta$

$:\implies \:\:cos\theta.sin\theta=y\:cos\theta$

$:\implies{\pink{\bf{x=cos\theta}}}$

$:\implies \:y=sin\theta\:..................(ii)$

squaring (i) and (ii) and adding both the equations ,

we get,

$:\implies \:\bf\:x^2+y^2=1$

Hence proved.

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