If the two determinants D1 and D2 given below be equal, then prove that cos2θ+cos2ϕ+cos2ψ=1 where D1=∣∣
∣∣1cosθcosϕcosθ1cosψcosϕcosψ1∣∣
∣∣ D2= ∣∣
∣∣0cosθcosϕcosθ0cosψcosϕcosψ0∣∣
∣∣
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Solution
Expand both D1=1−∑cos2θ+2cosθcosϕcosψ D2=2cosθcosϕcosψ D1=D2 implies the result.