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

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Correct option


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


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|


        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.

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