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# A quadratic equation whose roots are $cosec^2\theta$ and $\sec^2\theta$ can be

$\begin{array}{1 1}(a)\;x^2-2x+2=0&(b)\;x^2-3x+3=0\\(c)\;x^2-5x+5=0&(d)\;None\;of\;these\end{array}$

$\sec^2\theta+cosec^2\theta=\large\frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}$
$\qquad\qquad\qquad\;\;=\large\frac{\sin^2\theta+\cos^2\theta}{\cos^2\theta\sin^2\theta}$
$\sin^2\theta+\cos^2\theta=1$
$\sin 2\theta=2\sin\theta\cos\theta$
$\Rightarrow \sin\theta\cos\theta=\large\frac{1}{2}$$\sin2\theta \therefore \sec^2\theta+cosec^2\theta=\large\frac{4}{\sin^22\theta}$$\geq 4$
Also $\sec^2\theta.cosec^2\theta=\large\frac{4}{\sin^22\theta}$$\geq 4$
Thus required quadratic equation will be $x^2-tx+t=0$ where $t\geq 4$
Hence $x^2-5x+5=0$ can be the solution.