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Show that ∣ ∣ ∣1cos(βα)cos(γα)cos(αβ)1cos(γβ)cos(αγ)cos(βγ)1∣ ∣ ∣=

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Solution





L.H.S=∣ ∣ ∣1cos(βα)cos(γα)cos(αβ)1cos(γβ)cos(αγ)cos(βγ)1∣ ∣ ∣=∣ ∣cosαsinα0cosβsinβ0cosγsinγ0∣ ∣×∣ ∣cosαsinα0cosβsinβ0cosγsinγ0∣ ∣(RowbyRow)=0=R.H.S
232038_133145_ans_aa89a280478a46d1a43c0f20bc761071.jpg

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