Question

If $ f(\theta )=\left| \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta & {{\sin }^{2}}\theta & \cos \theta \\ \sin \theta & -\cos \theta & 0 \\ \end{matrix} \right| $ Then, for all $ \theta $

• A. $ f(\theta )=0 $
• B. $ f(\theta )=1 $
• C. $ f(\theta )=-1 $
• D. None of these

Answer 1

Answer: (B)

Given, $ f(\theta )=\left| \begin{matrix} {{\cos }^{2}}\theta & \cos \theta \sin \theta & -\sin \theta \\ \cos \theta \sin \theta & {{\sin }^{2}}\theta & \cos \theta \\ \sin \theta & -\cos \theta & 0 \\ \end{matrix} \right| $
$ ={{\cos }^{2}}\theta (0+{{\cos }^{2}}\theta )-\cos \theta $
$ .\sin \theta (0-\sin \theta .\cos \theta ) $
$ -\sin \theta (-{{\cos }^{2}}\theta .\sin \theta -{{\sin }^{3}}\theta ) $
$ ={{\cos }^{4}}\theta +2{{\sin }^{2}}\theta .{{\cos }^{2}}\theta +{{\sin }^{4}}\theta $ $ ={{({{\cos }^{2}}\theta +{{\sin }^{2}}\theta )}^{2}} $
$ =1 $
$ \therefore $ For all $ \theta ,f(\theta )=1 $