#### Explain solution for RD Sharma Class 12 Chapter 5 Determinants Exercise Very short Answers Question 20 maths textbook solution.

Hint: Here we use basic concept of determinant of matrix

Given: $\left[\begin{array}{cc} \sin 20^{\circ} & -\cos 20^{\circ} \\ \sin 70^{\circ} & \cos 70^{\circ} \end{array}\right]$

Solution :

So let's find determinate

$\left|\begin{array}{cc} \sin 20^{0} & -\cos 20^{\circ} \\ \sin 70^{0} & \cos 70^{0} \end{array}\right|$

\begin{aligned} &=\sin 20^{\circ} \times \cos 70^{\circ}-\left(-\cos 20^{\circ} \times \sin 70^{\circ}\right) \\ &=\left(\sin 20^{\circ} \times \cos 70^{\circ}\right)+\left(\cos 20^{\circ} \times \sin 70^{\circ}\right) \end{aligned}

$\rightarrow$ using the formula

\begin{aligned} &\sin (A+B)=\sin A \cos B+\cos A \sin B \\ &\text { Here } \mathrm{A}=20^{\circ} \\ &\mathrm{B}=70^{\circ} \end{aligned}

So,

\begin{aligned} &\sin (20+70)^{0}=\left(\sin 20^{\circ} \times \cos 70^{\circ}\right)+\left(\cos 20^{\circ} \times \sin 70^{\circ}\right) \\ &=\sin 90^{\circ} \\ &=1 \end{aligned}