1.

If sin³A + cos³B + 3sinAcosB = 1 then [2sinA + 3cosB] is (Where sinA+cosB not equal to 1)​

Answer»

-step explanation:ANSWERtanBtanA = 31 ⇒ cosAsinBsinAcosB = 31 PUT sinAcosB= 41 ∴cosAsinB= 43 ∴sin(A+B)= 41 + 43 =1orsinC=1=sin 2π ∴C= 2π Hence triangle is right ANGLED



Discussion

No Comment Found

Related InterviewSolutions