Let Δ= | sin θ cos φ sin θ sin φ cos θ cos θ cos φ cos θ sin φ -sin θ -sin θ sin φ sin θ cos φ 0 | . Then

Question

MARKS Content Team · Accepted Answer

Correct answer: (B) Δ is independent of φ; (D) ((d Δ)/(d θ) )_θ-(π)/2=0. Δ= | sin θ cos φ sin θ sin φ cos θ cos θ cos φ cos θ sin φ -sin θ -sin θ sin φ sin θ cos φ 0 | C_1 longrightarrow C_1- ((cos φ)/(sin φ) ) C_2 ⇒ Δ= | 0 sin θ sin φ cos θ 0 cos θ sin φ -sin θ -(sin θ)/(sin φ) sin θ cos φ θ | =(sin θ)/(sin φ) · sin φ (cos^2 θ+sin^2 θ )= sin θ arrow Independent of φ (d Δ)/(d θ)= .cos θ ⇒ (d Δ)/(d θ) |_θ-(π)/2=0

Let $\Delta=\left|\begin{array}{ccc}\sin \theta \cos \phi & \sin \theta \sin \phi & \cos \theta \\ \cos \theta \cos \phi & \cos \theta \sin \phi & -\sin \theta \\ -\sin \theta \sin \phi & \sin \theta \cos \phi & 0\end{array}\right|$. Then

Solution

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