If tan⁡A⋅tan⁡B=12, then (5−3cos⁡2A)(5−3cos⁡2B)=

tan⁡A⋅tan⁡B=12⇒ tan⁡A1=cot⁡B2=λ( say )⇒ tan⁡A=λ and tan⁡B=12λ∴ (5−3cos⁡2A)(5−3cos⁡2B) =5−31−λ21+λ2×5−3⋅1−14λ21+14λ2=16