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  • Need solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 27

Need solution for RD Sharma maths class 12 chapter Determinants exercise multiple choise question 27

Answers (1)

Answer:

Correct option (a)

Hint:

Use formula,

        \begin{aligned} &\cos (x+y)=\cos x \cos y-\sin x \sin y \\ &\sin (x+y)=\sin x \cos y+\cos x \sin y \end{aligned}

Given:

Here x, y \in R,

        \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0 \end{array}\right|

We have to find the value of \Delta.

Solution:

Here\; \; \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x+y) & -\sin (x+y) & 0 \end{array}\right|

Using formula,

        \begin{aligned} &\cos (x+y)=\cos x \cos y-\sin x \sin y \\ &\sin (x+y)=\sin x \cos y+\cos x \sin y \end{aligned}

        \Rightarrow \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \\cos x \cos y-\sin x \sin y & -\sin x \cos y+\cos x \sin y & 0 \end{array}\right|

Applying R3→R3 - R1 cos y + R2 sin y

        \Rightarrow \Delta=\left|\begin{array}{ccc} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ 0 & 0 & -\cos y+\sin y \end{array}\right|

Expanding along R3

        \begin{array}{ll} \Rightarrow \quad \Delta =(\sin y-\cos y)\left(\sin ^{2} x+\cos ^{2} x\right) \\ \Rightarrow \quad \Delta =(\sin y-\cos y)\end{array}                              [\because x + x=1]

Again, multiplying and dividing by \sqrt{2}, then we have

        \begin{aligned} &\Delta=\sqrt{2} \times \frac{1}{\sqrt{2}}(\sin y-\cos y) \\ &\Delta=\sqrt{2}\left(\frac{1}{\sqrt{2}} \sin y-\frac{1}{\sqrt{2}} \cos y\right) \\ &\Delta=\sqrt{2}\left(\cos \frac{\pi}{4} \sin y-\sin \frac{\pi}{4} \cos y\right) \\ &\Delta=\sqrt{2} \sin \left(y-\frac{\pi}{4}\right) \end{aligned}

Since \: \: -1\leq sin\, \theta \leq 1

        \begin{gathered} -1 \leq \sin \sin \left(y-\frac{\pi}{4}\right) \leq 1 \\ -\sqrt{2} \leq \sqrt{2} \sin \sin \left(y-\frac{\pi}{4}\right) \leq \sqrt{2} \end{gathered}

Hence,\: \: \sqrt{2}sin\: sin(y-\frac{\pi }{4})\in [-\sqrt{2}, \sqrt{2}]

Posted by

Gurleen Kaur

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