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  • 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

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