If (sec⁡A−tan⁡A)(sec⁡B−tan⁡B)(sec⁡C−tan⁡C) =(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C) then each side is equal to

We have,(sec⁡A−tan⁡A)(sec⁡B−tan⁡B)(sec⁡C−tan⁡C)=(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C)⇒sec2⁡A−tan2⁡Asec2⁡B−tan2⁡Bsec2⁡C−tan2⁡C={(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(scc⁡C+tan⁡C)}2⇒1={(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C)}2⇒(sec⁡A+tan⁡A)(sec⁡B+tan⁡B)(sec⁡C+tan⁡C)=±1Hence, LHS=RHS=±1.