Please help me with C-6 sum
correct option is A

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Given} \mathrm{point} \mathrm{P}\left(\sqrt{2}\mathrm{sec\theta },\sqrt{2}\mathrm{tan\theta }\right) \\ \mathrm{Distance} \mathrm{from} \mathrm{origin} \mathrm{O}=\sqrt{6} \\ \mathrm{OP}=\sqrt{6} \\ \sqrt{{\left(\sqrt{2}\mathrm{sec\theta }-0\right)}^2+{\left(\sqrt{2}\mathrm{tan\theta }-0\right)}^2}=\sqrt{6} \\ \Rightarrow 2{\sec }^2\mathrm{\theta }+2{\tan }^2\mathrm{\theta }=6 \\ \mathrm{now}, \\ 1+{\tan }^2\mathrm{\theta }={\sec }^2\mathrm{\theta }, \mathrm{so} \\ \Rightarrow 2\left(1+{\tan }^2\mathrm{\theta }\right)+2{\tan }^2\mathrm{\theta }=6 \\ 1+{\tan }^2\mathrm{\theta }+{\tan }^2\mathrm{\theta }=3 \\ 2{\tan }^2\mathrm{\theta }=2 \\ {\tan }^2\mathrm{\theta }=1 \\ \mathrm{so}, \\ \mathrm{tan\theta }=1 \mathrm{or} \mathrm{tan\theta }=-1 \\ \mathrm{since} \mathrm{P} \mathrm{lies} \mathrm{in} \mathrm{first} \mathrm{quadrant}, \mathrm{tan\theta }>0 \\ \mathrm{hence} \\ \mathrm{tan\theta }=1 \\ \mathrm{\theta }=\frac{\mathrm{\pi }}{4} \\ \mathbf{Ans}\mathbf{ }\mathbf{is}\mathbf{ }\mathbf{A} \end{gathered}$$

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