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

Answer 1

The correct option is C

${\left(x^2-y^2\right)}^2$

Explanation for the correct option.

Step 1: Simplify $\frac{2z\tan \theta }{1–{\tan }^2\theta }$

$\begin{array}{rcl}\frac{2z\tan \theta }{1–{\tan }^2\theta }& =& \frac{2z{\displaystyle \frac{\sin \theta }{\cos \theta }}}{1–{\displaystyle \frac{{\sin }^2\theta }{{\cos }^2\theta }}}\\ & =& \frac{{\displaystyle \frac{2z\sin \theta }{\cos \theta }}}{{\displaystyle \frac{{\cos }^2\theta -{\sin }^2\theta }{{\cos }^2\theta }}}\\ & =& \frac{2z\sin \theta \cos \theta }{\cos 2\theta }\left[by\cos 2x={\cos }^{2}x-{\sin }^{2}x\right]\end{array}$

Step 2: Simplify $x\sin \theta =y\cos \theta =\frac{2z\tan \theta }{1–{\tan }^2\theta }$

$\begin{array}{rcl}x\sin \theta & =& \frac{2z\sin \theta \cos \theta }{\cos 2\theta }\\ \Rightarrow x& =& \frac{2z\cos \theta }{\cos 2\theta }\end{array}$

Similarly

$\begin{array}{rcl}y\cos \theta & =& \frac{2z\sin \theta \cos \theta }{\cos 2\theta }\\ \Rightarrow y& =& \frac{2z\sin \theta }{\cos 2\theta }\end{array}$

Step 3: Find the value of $(x^2-y^2)$

$\begin{array}{rcl}{x}^2-y^2& =& \frac{4z^2{\cos }^2\theta -4z^2{\sin }^2\theta }{{\cos }^{2}2\theta }\\ & =& 4z^2\left({\cos }^2\theta -{\sin }^2\theta \right)\times \frac{1}{{\cos }^{2}2\theta }\\ & =& 4z^2\left(\cos 2\theta \right)\times \frac{1}{{\cos }^{2}2\theta }\\ & =& \frac{4z^2}{\cos 2\theta }........\left(1\right)\end{array}$

Step 4: Find the value of $4z^2(x^2+y^2)$

$\begin{array}{rcl}4z^2(x^2+y^2)& =& 4z^2\left[{\left(\frac{2z\cos \theta }{\cos 2\theta }\right)}^2+{\left(\frac{2z\sin \theta }{\cos 2\theta }\right)}^2\right]\\ & =& 4z^2\left(\frac{4z^2{\cos }^2\theta +4z^2{\sin }^2\theta }{{\cos }^{2}2\theta }\right)\\ & =& 4z^2\left(\frac{4z^2}{{\cos }^{2}2\theta }\right)\\ & =& \frac{16z^2}{{\cos }^{2}2\theta }\\ & =& {\left(x^2-y^2\right)}^2\left(\mathrm{from}\left(1\right)\right)\end{array}$

Hence, option C is correct