Question

If $x\sin \theta =y\cos \theta =\frac{2z\tan \theta }{1-{\tan }^2\theta },$ then $4z^2(x^2+y^2)$ is equal to

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

Answer 1

Answer: (C)

Given, $\frac{x}{\text{cosec }\theta }=\frac{y}{\sec \theta }=\frac{z}{\cot 2\theta }=k(say)$
$\therefore$ $4z^2(x^2+y^2)=4k^2{\cot }^{2}2\theta (k^2\text{cose}{\text{c}}^2\theta$
$+k^2{\sec }^2\theta )$
$=4k^4{\cot }^{2}2\theta \left(\frac{1}{{\sin }^2\theta }+\frac{1}{{\cos }^2\theta }\right)$
$=4k^2{\cot }^{2}2\theta \left(\frac{1}{{\sin }^2\theta {\cos }^2\theta }\right)$
$=\frac{4k^4}{4}\left(\frac{{\cos }^2\theta -{\sin }^2\theta }{{\sin }^2\theta {\cos }^2\theta }\right)$
$={(k^2\text{cose}{\text{c}}^2\theta -k^2{\sec }^2\theta )}^2$
$={(x^2-y^2)}^2$