Question

If $x\sin \theta =y\cos \theta =\frac{2z\tan \theta }{1-\tan {}^2\theta },$ then $z^2(x^2+y^2)$ is equal to

• A. ${(x^2+y^2)}^3$
• B. ${\left(x^2{y}^2\right)}^3$
• C. ${(x^2-y^2)}^2$
• D. ${(x^2+y^2)}^2$

Answer 1

The correct option is D ${(x^2+y^2)}^2$

$x\sin \theta =\frac{2z\tan \theta }{1-{\tan }^2\theta }$

$y\cos \theta =\frac{2z\tan \theta }{1-{\tan }^2\theta }$

$x^2{\sin }^2\theta =4z^2\frac{{\tan }^2\theta }{{(1-{\tan }^2\theta )}^2}$

$x^2{\sin }^2\theta =z^2\left({\tan }^{2}2\theta \right)$

$x^2=\frac{z^2{\tan }^{2}2\theta }{{\sin }^2\theta }$

Similarly for $y^2=\frac{z^2{\tan }^{2}2\theta }{{\cos }^2\theta }$

$4z^2(x^2+y^2)=4z^2(\frac{z^2{\tan }^{2}2\theta }{{\sin }^2\theta }+\frac{z^2{\tan }^{2}2\theta }{{\cos }^2\theta })$

$=4z^2\left(z^2{\tan }^{2}2\theta \right)(\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta })$

$=4z^2{\tan }^{2}2\theta \left(\frac{4}{4{\sin }^2\theta {\cos }^2\theta }4\right)$

$=\frac{16z^2{\tan }^{2}2\theta }{{\sin }^{2}2\theta }=\frac{16z^2}{{\cos }^2\theta }$

$={(x^2+y^2)}^2$

So option $\left(d\right)$ is correct.