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

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}

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

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

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