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प्रश्न
The number of distinct real roots of \[\begin{vmatrix}cosec x & \sec x & \sec x \\ \sec x & cosec x & \sec x \\ \sec x & \sec x & cosec x\end{vmatrix} = 0\] lies in the interval
\[- \frac{\pi}{4} \leq x \leq \frac{\pi}{4}\]
पर्याय
1
2
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0
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उत्तर
\[\text{ Let }∆ = \begin{vmatrix} cosec x & \sec x & \sec x\\\sec x & cosec x & \sec x\\\sec x & \sec x & cosec x \end{vmatrix}\]
\[ = \left( cosec x \right)^3 \begin{vmatrix} 1 &\frac{\sec x}{cosec x} & \frac{\sec x}{cosec x}\\\frac{\sec x}{cosec x} & 1 & \frac{\sec x}{cosec x}\\\frac{\sec x}{cosec x} &\frac{\sec x}{cosec x} & 1 \end{vmatrix}\]
\[ = \left( cosec x \right)^3 \begin{vmatrix} 1 & \tan x & \tan x \\\tan x & 1 & \tan x\\\tan x & \tan x & 1 \end{vmatrix}\]
\[ = \left( cosec x \right)^3 \begin{vmatrix} 1 - \tan x & \tan x - 1 & 0 \\ 0 & 1 - \tan x & \tan x - 1\\\tan x & \tan x & 1 \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_2 , R_2 \to R_2 - R_3 \right]\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \begin{vmatrix} 1 & - 1 & 0 \\ 0 & 1 & - 1\\\tan x & \tan x & 1 \end{vmatrix} \left[\text{ Taking out }\left( 1 - \tan x \right)\text{ common from }R_1\text{ and }R_2 \right]\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left\{ 1\begin{vmatrix}1 & - 1 \\ \tan x & 1\end{vmatrix} + \tan x\begin{vmatrix}- 1 & 0 \\ 1 & - 1\end{vmatrix} \right\} \left[ \text{ Expanding along }C_1 \right]\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left\{ 1 + \tan x + \tan x \right\}\]
\[ = \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left\{ 1 + 2 \tan x \right\}\]
\[ ∆ = 0\]
\[ \left( cosec x \right)^3 \left( 1 - \tan x \right)^2 \left( 1 + 2 \tan x \right) = 0\]
\[ \Rightarrow \left( 1 - \tan x \right) = 0, \left( cosec x \right)^3 = 0\text{ and }\left( 1 + 2 \tan x \right) = 0\]
or
\[\tan x = 1, cosec x = 0\text{ and }\tan x = \frac{- 1}{2}\]
\[ \Rightarrow - \frac{\pi}{4} \leq x \leq \frac{\pi}{4} \left[ \tan x = 1, \tan x = \frac{- 1}{2}\text{ are 2 real roots as cosec x = 0 has no solution }\right]\]
Thus, there are 2 solutions .
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