Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 20 maths

Correct option (b)

Hint:

Simply first solve the determinant.

Given:

Here the distinct real roots of

$\left|\begin{array}{ccc} \cos e c x & \sec x & \sec x \\ \sec x & cos\, ecx & \sec x \\ \sec x & \sec x & \cos e c x \end{array}\right|=0$

lies in the interval

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

We have to find the number of roots of the given determinant.

Solution:

$Here\; \; \left|\begin{array}{ccc} \cos e c x & \sec x & \sec x \\ \sec x & cos\, ecx & \sec x \\ \sec x & \sec x & \cos e c x \end{array}\right|=0$

Applying C1→C1-C3; C2→C2-C3

Taking common cos sec x-sec sec x  from column 1 and 2, we get

$\Rightarrow(cos\, \, ecx-\sec x)^{2}\left|\begin{array}{ccc} 1 & 0 & \sec x \\ 0 & 1 & \sec x \\ -1 & -1 & \cos e c x \end{array}\right|=0$

Expanding along R1 , we get

$\Rightarrow(cos\, \, ecx-\sec x)^{2}[(cos\, ecx+sec\, x)+sec\, x(0+1)]=0$

$\Rightarrow(cos\, \, ecx-\sec x)^{2}(cos\, ecx+2sec\, x)=0$

$\Rightarrow(cos\, \, ecx-\sec x)^{2}=0\; \; or\; \; (cos\, ecx+2sec\, x)=0$

$\Rightarrow cos\, \, ecx-\sec x =0\; \; or\; \; cos\, ecx=-2sec\, x$

$\Rightarrow cos\, \,cos\, x-sin sin\, x =0\; \; or\; \; cos \, cos\, x=-2sin\, sin\, x$

$\Rightarrow sin\, x =cos\, x\; \; or\; \;\frac{cos\, cos\, x}{sin\, sin\, x}=-2$

$\Rightarrow \frac{cos\, cos\, x}{sin\, sin\, x}=1\; \;or\, \, cot\, cot\, x=-2$

$\Rightarrow cot\, cot\, x=1\; \;or\, \, cot\, cot\, x=-2$

$\Rightarrow tan\, tan\, x=1\; \;or\, \, tan\, tan\, x=-\frac{1}{2}$

Hence, there are two solutions.