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

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 (A) Δ is independent of θ (B) .( d Δ/ d θ)]θ=π / 2=0 (C) Δ is a constant (D) None of these

Tardigrade Admin · Accepted Answer

Applying C 1 arrow C 1-( cot φ) C 2, we get Δ=|0 sin θ sin φ cos θ 0 cos θ sin φ - sin θ - sin θ / sin φ sin θ cos φ 0| On expanding along C 1, we get Δ=-( sin θ/ sin φ)[ sin φ sin 2 θ- cos 2 θ sin φ] = sin θ, This is independent of φ. Also, .( d Δ/ d θ)= cos θ ⇒ ( d Δ/ d θ)]θ=π / 2= cos ((π/2))=0

Q. Let $Δ = ∣ ∣ sin θ cos ϕ cos θ cos ϕ - sin θ sin ϕ sin θ sin ϕ cos θ sin ϕ sin θ cos ϕ cos θ - sin θ 0 ∣ ∣$ , then

$Δ$ is independent of $θ$

$\frac{d Δ}{d θ}]_{θ = π /2} = 0$

$Δ$ is a constant

None of these

Q. Let Δ=∣∣​sinθcosϕcosθcosϕ−sinθsinϕ​sinθsinϕcosθsinϕsinθcosϕ​cosθ−sinθ0​∣∣​, then

Δ is independent of θ

dθdΔ​]θ=π/2​=0

Δ is a constant

None of these

Q. Let $Δ = ∣ ∣ sin θ cos ϕ cos θ cos ϕ - sin θ sin ϕ sin θ sin ϕ cos θ sin ϕ sin θ cos ϕ cos θ - sin θ 0 ∣ ∣$ , then

$Δ$ is independent of $θ$

$\frac{d Δ}{d θ}]_{θ = π /2} = 0$

$Δ$ is a constant