Question

$\text{cosec}\theta =\frac{x^2-y^2}{x^2+y^2}$, where $x\in R,$ $y\in R,$ gives real $\theta$ if and only if

• A. $x=y\ne 0$
• B. $\left|x\right|=\left|y\right|\ne 0$
• C. $x+y=0,x\ne 0$
• D. none of these.

Answer 1

The correct option is D none of these.

$cosec\theta =\frac{x^2-y^2}{x^2+y^2}$
$\because cosec\theta \in (-\infty ,\infty )-\left\{0\right\}$
$\Rightarrow cosec\theta \ne 0$
$\frac{x^2-y^2}{x^2+y^2}\ne 0$
$\Rightarrow x^2-y^2\ne 0$
$\therefore \left|x\right|\ne \left|y\right|$
Hence, option 'D' is correct.