A curve is given by the equations  $x={\sec }^2\theta ,y=\cot \theta$,  If the tangent at P where   $\theta =\frac{\pi }{4}$ meets the curve again at $Q$ , then$\left[PQ\right]$ is, where [.] represents the greatest integer function______

$=-\frac{1}{2}{\cot }^3\theta =-\frac{1}{2}\text{at}\theta =\frac{\pi }{4}$

Also, the point P for   $\theta =\frac{\pi }{4} \text{is }(2,1)$

Equation of tangent is

$y-1=-\frac{1}{2}(x-2) \text{or }x+2y-4=0$

This meets the curve whosse Cartesian equation on  eliminating  $\theta$  by ${\sec }^2\theta -{\tan }^2\theta =1 \text{is }{y}^2=\frac{1}{x-1}$

Solving (1) and (2), we get  $y=1,-\frac{1}{2}$  $\therefore x=2,5$

Hence, P is (2,1) as given and Q is  $(5,-\frac{1}{2})$.

$\therefore PQ=\sqrt{\frac{45}{4}}=\frac{3\sqrt{5}}{2}$