#### Provide Solution For  R. D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 2 Sub Question 13 Maths Textbook Solution.

Answer:$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|=0$

Hint: We will try to do any two column or row equal

Given:$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|$

Solution:$\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|$

\begin{aligned} &\text { On applying } \mathrm{C}_{2} \rightarrow \cos \delta \cdot \mathrm{C}_{2}-\sin \delta . C_{1} \\ &\left|\begin{array}{lll} \sin \alpha & \cos \alpha \cdot \cos \delta-\sin \alpha \cdot \sin \delta & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta \cdot \cos \delta-\sin \beta \cdot \sin \delta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma \cdot \cos \delta-\sin \gamma \cdot \sin \delta & \cos (\gamma+\delta) \end{array}\right| \\ &\{\because \cos (\alpha+\delta)=\cos \alpha \cos \delta-\sin \alpha \sin \delta\} \\ &=\left|\begin{array}{lll} \sin \alpha & \cos (\alpha+\delta) & \cos (\alpha+\delta) \\ \sin \beta & \cos (\beta+\delta) & \cos (\beta+\delta) \\ \sin \gamma & \cos (\gamma+\delta) & \cos (\gamma+\delta) \end{array}\right| \end{aligned}

If any two rows or columns of a determinant are identical.

The value of the determinant is zero

$=0 \quad\left(\because C_{2}=C_{3}\right)$

Hence $\left|\begin{array}{lll} \sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta) \end{array}\right|=0$