Question

There exists some $\theta$ such that the points $(1,1)$; $(0,{\sec }^2\theta )$; and $(cose{c}^2\theta ,0)$ are collinear.

• A. True
• B. False

Answer 1

The correct option is B False

$\because Collinear,\therefore Slopemustbesame.$

$⟹\frac{\sec {}^2\theta -1}{0-1}=\frac{0-\sec {}^2\theta }{\csc {}^2\theta -0}$

$⟹-\tan {}^2\theta =-\tan {}^2\theta$

$Slopessame.$

$letA->(1,1)b->(0,\sec {}^2\theta )$

$c->(\csc {}^2\theta ,0)$

$AB:y-\sec {}^2\theta =\frac{\sec {}^2\theta -1}{0-1}(x-0)$

$⟹y-\sec {}^2\theta =-\tan {}^2\theta \left(x\right)$

$⟹y+x\tan {}^2\theta =\sec {}^2\theta$

$BC:y=0=\frac{-\sec {}^2\theta }{\csc {}^2\theta }(x-\csc {}^2\theta )$

$⟹y=-\tan {}^2\theta (x-\csc {}^2\theta )$

$⟹y=-x\tan {}^2\theta +\sec {}^2\theta$

$\because A,B,Carecollinear.$

$⟹EquationAB=\lambda BC$

$⟹y+x\tan {}^2\theta +\sec {}^2\theta =\lambda (y+x\tan {}^2\theta -\sec {}^2\theta )$

$⟹\lambda =1$

$and\sec {}^2\theta =-\sec {}^2\theta$

$⟹\sec \theta =0$

$\sec \theta ϵ(-\infty ,-1]\cup [1,\infty )$

$\therefore \sec \theta =0$ is not true for any value of$\theta .$