Trigonometric elements MCQ Quiz in मल्याळम - Objective Question with Answer for Trigonometric elements - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation

\(\Delta = \begin{bmatrix} \sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0 \end{bmatrix}\)

Expanding through R₃ we get

\(= -\sin \theta \sin \phi (-\sin^2 \theta \sin \phi - \cos^2 \theta \sin \phi) - \sin \theta \cos \phi (-\sin^2 \theta \cos \phi - \cos^2 \theta \cos \phi)\)

\(= \sin \theta \sin^2 \phi (\sin^2 \theta + \cos^2 \theta) + \sin \theta \cos^2 \phi (\sin^2 \theta + \cos^2 \theta)\)

\(= \sin \theta \sin^2 \phi + \sin \theta \cos^2 \phi = \sin \theta (\sin^2 \phi + \cos^2 \phi)\)

\(= \sin \theta\)

Independent of \(\phi\)

Also,

\(\frac{d\Delta}{d\theta} = \cos \theta\)

Therefore,

\(\frac{d\Delta}{d\theta}\) at \(\theta = \frac{\pi}{2}\) \(= \cos(\frac{\pi}{2}) = 0\)

Hence option 2 and 4 are correct

The correct option is: 90 

Explanation: We start with the matrix ( A ) and the given equation ( A + A^T - 2I = 0 ).

Given: \([ A = \begin{pmatrix} \sin θ & \cos θ \ \cos θ & \sin θ \end{pmatrix} ]\)

• First, determine the transpose of ( A ):\( [ A^T = \begin{pmatrix} \sin θ & \cos θ \ \cos θ & \sin θ \end{pmatrix} ]\)
• Next, add ( A ) and ( A^T ): [ A + A^T = \(\begin{pmatrix} \sin θ & \cos θ \ \cos θ & \sin θ \end{pmatrix} + \begin{pmatrix} \sin θ & \cos θ \ \cos θ & \sin θ \end{pmatrix} = \begin{pmatrix} 2\sin θ & 2\cos θ \ 2\cos θ & 2\sin θ \end{pmatrix} ]\)
• Now, subtract ( 2I ) from ( A + A^T ): \([ 2I = \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix} ] [ A + A^T - 2I = \begin{pmatrix} 2\sin θ & 2\cos θ \ 2\cos θ & 2\sin θ \end{pmatrix} - \begin{pmatrix} 2 & 0 \ 0 & 2 \end{pmatrix} = \begin{pmatrix} 2\sin θ - 2 & 2\cos θ \ 2\cos θ & 2\sin θ - 2 \end{pmatrix} ]\)
• Set ( A + A^T - 2I ) equal to the zero matrix: \([ \begin{pmatrix} 2\sin θ - 2 & 2\cos θ \ 2\cos θ & 2\sin θ - 2 \end{pmatrix} = \begin{pmatrix} 0 & 0 \ 0 & 0 \end{pmatrix} ]\)
• This results in the following equations: \([ 2\sin θ - 2 = 0 ] [ 2\cos θ = 0 ]\)
• Solving these gives: \([ 2\sin θ - 2 = 0 \rightarrow \sin θ = 1 ] [ 2\cos θ = 0 \rightarrow \cos θ = 0 ]\)
• The only angle that satisfies both equations is \((θ = 90^\circ).\)

Thus, the value of (θ) is 90.