Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Explain solution for RD Sharma Class 12 Chapter 5 Determinants Exercise Very short Answers Question 20 maths textbook solution.

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

 

Answers (1)

Answer : 1

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}

Posted by

infoexpert23

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

Board Exam Time Table 2025 Live: UP Board exams from February 24; CBSE, BSEB date sheet soon
anam-khan

CBSE Date Sheet 2025 Live: Class 10, 12 board exam dates soon; syllabus, sample papers
anam-khan

Extra time in CBSE exams for students with type 1 diabetes: Kerala SHRC seeks report on plea

Latest Question