#### Explain solution RD Sharma class 12 chapter Determinants exercise 5.1 question 7 maths

$1$

Hint:

Determinant matrix must be square (i.e. same number of rows and columns)

Given:

$\left|\begin{array}{ccc} \cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha \end{array}\right|$

Solution:

\begin{aligned} &\Delta=\mathrm{a}_{11} \mathrm{C}_{11}+\mathrm{a}_{21} \mathrm{C}_{21}+\mathrm{a}_{31} \mathrm{C}_{31} \\ &=(-1)^{1+1} \cos \alpha \cos \beta(\cos \alpha \cos \beta-0)+(-1)^{1+2} \cos \alpha \sin \beta(-\sin \beta \cos \alpha-0)+(-\text { 1) }^{1+3}(-\sin \alpha)\left(-\sin \alpha \sin ^{2} \beta-\sin \alpha \cos ^{2} \beta\right) \\ &=\cos \alpha \cos \beta(\cos \alpha \cos \beta-0)-\cos \alpha \sin \beta(-\sin \beta \cos \alpha-0)+(-\sin \alpha)\left(-\sin \alpha \sin ^{2} \beta-\right. \left.\sin \alpha \cos ^{2} \beta\right) \\ &=\cos ^{2} \alpha \cos ^{2} \beta+\cos ^{2} \alpha \sin ^{2} \beta+\sin ^{2} \alpha \sin ^{2} \beta+\sin ^{2} \alpha \cos ^{2} \beta \\ &=\cos ^{2} \alpha\left(\cos ^{2} \beta+\sin ^{2} \beta\right)+\sin ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right) \quad \quad \quad \quad \quad \quad \quad\left[\cos ^{2} \theta+\sin ^{2} \theta=1\right] \\ &=\cos ^{2} \alpha+\sin ^{2} \alpha \\ &\Delta=1 \end{aligned}