Question

Given that
$f\left(\alpha \right)=\left|\begin{array}{ccc}\cos (x+\alpha )& -\sin (x+\alpha )& \cos 2\alpha \\ \sin x& \cos x& \sin \alpha \\ -\cos x& \sin x& \cos \alpha \end{array}\right|=A+B{\cos }^2\alpha$
then

• A. A=0, B=2
• B. A=1, B=1
• C. A=1, B=0
• D. none of these

Answer 1

The correct option is A A=0, B=2

$f\left(\alpha \right)=\left|\begin{array}{ccc}\cos (x+\alpha )& -\sin (x+\alpha )& \cos 2\alpha \\ \sin x& \cos x& \sin \alpha \\ -\cos x& \sin x& \cos \alpha \end{array}\right|$

$=\left|\begin{array}{ccc}\cos x\cos \alpha -\sin x\sin \alpha & -\sin x\cos \alpha -\cos x\sin \alpha & {\cos }^2\alpha -{\sin }^2\alpha \\ \sin x& \cos x& \sin \alpha \\ -\cos x& \sin x& \cos \alpha \end{array}\right|$

$=\left|\begin{array}{ccc}\cos x\cos \alpha & -\sin x\cos \alpha & {\cos }^2\alpha \\ \sin x& \cos x& \sin \alpha \\ -\cos x& \sin x& \cos \alpha \end{array}\right|-\left|\begin{array}{ccc}\sin x\sin \alpha & \cos x\sin \alpha & {\sin }^2\alpha \\ 0& 0& 0\\ 0& 0& 0\end{array}\right|$

$=\cos \alpha \left|\begin{array}{ccc}\cos x& -\sin x& \cos \alpha \\ \sin x& \cos x& \sin \alpha \\ -\cos x& \sin x& \cos \alpha \end{array}\right|$

Applying $R_3\to R_3+R_1$

$f\left(\alpha \right)=\cos \alpha \left|\begin{array}{ccc}\cos x& -\sin x& \cos \alpha \\ \sin x& \cos x& \sin \alpha \\ 0& 0& 2\cos \alpha \end{array}\right|=2{\cos }^2\alpha$

By comparing with given $A=0andB=2$