#### Need solution for RD Sharma maths Class 12 Chapter 5 Determinants Exercise VSQ Question 55 maths textbook solution.

Answer:  $x = -2$

Hint: Here we use basic concept of determinant of matrix

Given: $\left|\begin{array}{ccc} x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x \end{array}\right|=8$

Solution :

\begin{aligned} &x\left|\begin{array}{cc} -x & 1 \\ 1 & x \end{array}\right|-\sin \theta\left|\begin{array}{cc} -\sin \theta & 1 \\ \cos \theta & x \end{array}\right|+\cos \theta\left|\begin{array}{cc} -\sin \theta & -x \\ \cos \theta & 1 \end{array}\right|=8 \\ &x\left(-x^{2}-1\right)-\sin \theta(-\sin \theta(x)-\cos \theta)+\cos \theta(-\sin \theta-(-x) \cos \theta)=8 \\ &-x^{3}-x+x \sin ^{2} \theta+\sin \theta \cos \theta-\sin \theta \cos \theta+x \cos ^{2} \theta=8 \\ &-x^{3}-x+x\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=8 \\ &-x^{3}-x+x=8 \ldots\left\{\sin ^{2} \theta+\cos ^{2} \theta=1\right\} \end{aligned}

\begin{aligned} &x^{3}=-8 \\ &x=-2 \end{aligned}