Question

If $x\sin^{3}\alpha + y\cos^{3}\alpha = \sin\alpha\cos\alpha$ and $x\sin\alpha - y\cos\alpha = 0,$ then $x^{2} + y^{2} =$

• A. – 1
• B. ±1
• C. 1
• D. None of these

Answer 1

Answer: (C)

1

Answer 2

We have $x\sin^{3}\alpha + y\cos^{3}\alpha = \sin\alpha\cos\alpha$ …..(i)

and $x\sin\alpha - y\cos\alpha = 0$ …..(ii)

Now from (ii), $x\sin\alpha = y\cos\alpha$

Putting in (i), we get

$\Rightarrow y\cos\alpha\sin^{2}\alpha + y\cos^{3}\alpha = \sin\alpha\cos\alpha$

$\Rightarrow y\cos\alpha\left\{ \sin^{2}\alpha + \cos^{2}\alpha \right\} = \sin\alpha\cos\alpha$

$\Rightarrow y\cos\alpha = \sin\alpha\cos\alpha \Rightarrow y = \sin\alpha$ and $x = \cos\alpha$

Hence, $x^{2} + y^{2} = \sin^{2}\alpha + \cos^{2}\alpha = 1.$