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### Answers (1)

Answer:

Correct option (c)

Hint:

Solve the determinant by applying column operations.

Given:

Given that,

$\left|\begin{array}{lll} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$

We have to find the real root of the given determinant in interval

$[\frac{-\pi }{4},\frac{\pi }{4}]$

Solution:

Here given that,

$\left|\begin{array}{lll} \sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x \end{array}\right|=0$

Applying C1 → C1 - C2; C2 → C2 - C3

$\Rightarrow \quad\left|\begin{array}{ccc} \sin x-\cos x & 0 & \cos x \\ \cos x-\sin x & \sin x-\cos x & \cos x \\ 0 & \cos x-\sin x & \sin x \end{array}\right|=0$

Taking common (sin x- cos x) from column 1 and 2 , we get

\begin{aligned} &\quad(\sin x-\cos x)^{2}\left|\begin{array}{ccc} 1 & 0 & \cos x \\ -1 & 1 & \cos x \\ 0 & -1 & \sin x \end{array}\right|=0 \\ &\Rightarrow (\sin x-\cos x)^{2}(1(\sin x+\cos x)+\cos x(1))=0 \\ &\Rightarrow (\sin x-\cos x)^{2}(\sin x+2 \cos x)=0 \\ &\Rightarrow (\sin x-\cos x)^{2}=0 \text { or } \sin x+2 \cos x=0 \\ &\Rightarrow \, \sin x-\cos x=0 \text { or } \frac{\sin x}{\cos x}=-2 \\ &\Rightarrow \frac{\sin x}{\cos x}=1 \: \text \: or \tan x=-2 \\ &\Rightarrow \tan x=1 \text { or } \tan x=-2 \notin\left[\frac{-\pi}{4}, \frac{\pi}{4}\right] \end{aligned}

$Since, \frac{-\pi }{4}\leq x\leq \frac{\pi }{4}$

Hence tan x=1  is the required answer.

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