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Please solve RD Sharma class 12 chapter 5 Determinants exercise Fill in the blanks question 21 maths textbook solution

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Answer: y

Hint: Here, we use basic concept of determinant of matrix

Given:  \left[\begin{array}{ccc} \cos (x+y) & -\sin (x+y) & \cos 2 y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & \sin y \end{array}\right]

Solution: Formula expanding,

                \begin{aligned} &{\left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y & -\sin x \cos y+\cos x \sin y & \cos ^{2} y-\sin ^{2} y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & \sin y \end{array}\right]} \\ &R_{1} \rightarrow R_{1}+R_{2}(\sin y)+R_{3}(\cos y) & \end{aligned}

                \left[\begin{array}{ccc} \cos x \cos y-\sin x \sin y+\sin y \sin x-\cos y \cos x & -\sin x \cos y-\cos x \sin y+\sin y \cos x+\cos y \sin x & \cos ^{2} y-\sin ^{2} y+\sin ^{2} y-\cos ^{2} y \\ \sin x & \cos x & \sin y \\ -\cos x & \sin x & -\sin y \end{array}\right]\text { So }\left[\begin{array}{ccc} 0 & 0 & 0 \\ \sin x & \cos x & \sin x \\ -\cos x & \sin x & -\cos x \end{array}\right]=0

Here, we get answer is zero which is obtained by multiplying 2nd row with sin y and 3rd row with cosy.

So, Value of determinant depends on y.

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