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Question: The points (1, 1); \((0,\sec^{2}\theta);(co\text{se}\text{c}^{2}\theta,0)\) are collinear for...

The points (1, 1); (0,sec2θ);(cosec2θ,0)(0,\sec^{2}\theta);(co\text{se}\text{c}^{2}\theta,0) are collinear for

A

θ=nπ/2\theta = n\pi/2

B

θnπ/2\theta \neq n\pi/2

C

θ=nπ\theta = n\pi

D

None of these

Answer

θnπ/2\theta \neq n\pi/2

Explanation

Solution

The given points are collinear, if Area of ∆

1 & 1 & 1 \\ 0 & \sec^{2}\theta & 1 \\ c\text{ose}\text{c}^{2}\theta & 0 & 1 \end{matrix} \right| = 0$$ ⇒ $1(\sec^{2}\theta) + 1(\text{cose}\text{c}^{2}\theta) + 1( - \text{cose}\text{c}^{2}\theta.\sec^{2}\theta) = 0$ ⇒ $\frac{1}{\cos^{2}\theta} + \frac{1}{\sin^{2}\theta} - \frac{1}{\sin^{2}\theta.\cos^{2}\theta} = 0$ ⇒$\frac{1}{\sin^{2}\theta.\cos^{2}\theta} - \frac{1}{\sin^{2}\theta.\cos^{2}\theta} = 0 \Rightarrow 0 = 0$ Therefore the points are collinear for all value of $\theta$, except only $\theta = \frac{n\pi}{2}$because at $\theta = \frac{n\pi}{2}$,$\sec^{2}\theta = \infty$ (Not defined)