Question

The number of distinct real roots of $\left|\begin{array}{ccc}\mathrm{cosec}x& \sec x& \sec x\\ \sec x& \mathrm{cosec}x& \sec x\\ \sec x& \sec x& \mathrm{cosec}x\end{array}\right|=0$ lies in the interval $-\frac{\pi }{4}\le x\le \frac{\pi }{4}$ is
(a) 1
(b) 2
(c) 3
(d) 0

Answer 1

(b) 2
$$\begin{gathered} \mathrm{Let}∆=\left|\mathrm{cosec}x\sec x\sec x \\ \sec x\mathrm{cosec}x\sec x \\ \sec x\sec x\mathrm{cosec}x\right| \\ ={\left(\mathrm{cosec}x\right)}^3\left|1\frac{\sec x}{\mathrm{cosec}x}\frac{\sec x}{\mathrm{cosec}x} \\ \frac{\sec x}{\mathrm{cosec}x}1\frac{\sec x}{\mathrm{cosec}x} \\ \frac{\sec x}{\mathrm{cosec}x}\frac{\sec x}{\mathrm{cosec}x}1\right| \\ ={\left(\mathrm{cosec}x\right)}^3\left|1\tan x\tan x \\ \tan x1\tan x \\ \tan x\tan x1\right| \\ ={\left(c\mathrm{os}ecx\right)}^3\left|1-\tan x\tan x-10 \\ 01-\tan x\tan x-1 \\ \tan x\tan x1\right|\left[\mathrm{Applying}{R}_1\to R_1-R_2,R_2\to R_2-R_3\right] \\ ={\left(c\mathrm{os}ecx\right)}^3{\left(1-\tan x\right)}^2\left|1-10 \\ 01-1 \\ \tan x\tan x1\right|\left[\mathrm{Taking}\mathrm{out}\left(1-\tan x\right)\mathrm{common}\mathrm{from}{R}_1\mathrm{and}{R}_2\right] \\ ={\left(c\mathrm{os}ecx\right)}^3{\left(1-\tan x\right)}^2\left\{1\left|\begin{array}{cc}1& -1\\ \tan x& 1\end{array}\right|+\tan x\left|\begin{array}{cc}-1& 0\\ 1& -1\end{array}\right|\right\}\left[\mathrm{Expanding}\mathrm{along}{C}_1\right] \\ ={\left(c\mathrm{os}ecx\right)}^3{\left(1-\tan x\right)}^2\left\{1+\tan x+\tan x\right\} \\ ={\left(c\mathrm{os}ecx\right)}^3{\left(1-\tan x\right)}^2\left\{1+2\tan x\right\} \\ ∆=0 \\ {\left(c\mathrm{os}ecx\right)}^3{\left(1-\tan x\right)}^2\left(1+2\tan x\right)=0 \\ \Rightarrow \left(1-\tan x\right)=0,{\left(\mathrm{cosec}x\right)}^3=0\mathrm{and}\left(1+2\tan x\right)=0 \\ \mathrm{or} \\ \tan x=1,\mathrm{cosec}x=0\mathrm{and}\tan x=\frac{-1}{2} \\ \Rightarrow -\frac{\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{4}\left[\tan x=1,\tan x=\frac{-1}{2}\mathrm{are}2\mathrm{real}\mathrm{roots}\mathrm{as}\mathrm{cosec}x=0\mathrm{has}\mathrm{no}\mathrm{solution}\right] \\ \mathrm{Thus},\mathrm{there}\mathrm{are}2\mathrm{solutions}. \end{gathered}$$