Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 19

Need solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 19

Answers (1)

Answer:

Correct option

Hint:

Here A+B+C=\pi, then we can write sin(A+B+C)=sin\pi

Given:

Here given that A+B+C=\pi

Here we have to find the value of

        \left|\begin{array}{ccc} \sin (A+B+C) & \sin (A+C) & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & \tan (B+C) & 0 \end{array}\right|

Solution:

        Let\: \: \Delta =\left|\begin{array}{ccc} \sin (A+B+C) & \sin (A+C) & \cos C \\ -\sin B & 0 & \tan A \\ \cos (A+B) & \tan (B+C) & 0 \end{array}\right|

Here A+B+C=\pi

        \Delta =\left | sin\, sin\, \pi \: sin\: sin(\pi -B)\: cos\: cos\, C-sin\, sin\, B\, 0\, tan\, tan\, A\, cos\, cos(\pi -C)\, tan\, tan(\pi -A)\, 0 \right |....(i)

[\because sin\, sin(\pi -B)=sin\, sin\, B, \, \, cos\, cos(\pi -c)=C,\, tan\, tan(\pi -A)=A]

\Delta =\left | 0\: sin\, sin\, B\: cos\, cos\, C-sin\, sin\, B\: 0\: tan\, tan\, A-cos\, cos\, C-tan\, tan\, A\, \, 0 \right |

Transposing the determinant, we get

        \Delta =\begin{vmatrix} 0 &-sinB &-cosC \\ sinB &0 &-tanA \\ cosC &tanA &0 \end{vmatrix}                            ....(ii)

Adding (i) and (ii) we get

        2\Delta =0\Rightarrow \Delta =0

This is the required answer.

Posted by

Gurleen Kaur

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

CBSE Results 2024: Marks verification from 4th day of result declaration; board issues schedule
anam-khan

CBSE 10th, 12th results 2024 ‘shortly’; board issues Digilocker access code for students
anam-khan

CBSE Class 10, 12 results 2024 likely after May 20, says result portal

Latest Question