Question

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

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

Answer 1

Answer: (D)

none of these

Given, $\frac{x^2-y^2}{x^2+y^2}$

Is always less than or equal to 1, and greater than or equal to -1 since

$x^2-y^2<x^2+y^2$ for all real numbers and for $x,y\in \left\{0\right\}$

Now,
$\frac{x^2-y^2}{x^2+y^2}\in [-1,1]$

Now the range of $cosec\theta$ is $(-\infty ,-1]\cup [1,\infty )$

Hence $cosec\theta \ne \frac{x^2-y^2}{x^2+y^2}$.