Question

In ∆leABC, 3sinA+4cosB=6, 4sinB+3cosA=2 then show that the maximum value of S=(1-a)(1-b)+(1-alpha)(1-beta)is 8​

Answer 1

square the first and second equation..

9sin^2A+16cos^2+12SinAcosB=36

16sin^2B+9cos^B+12SinBCosA=4

add both terms

9(cos^A+sin^A)+16(cos^A+sin^B)+12(SinAcosB+CosASinB)=40..........1)

we know that

Sin^A+cos^A=1

also,SinACosB+Cos ASinB=Sin(A+B)

put this value in equation 1)

9+16+12Sin(A+B)=40

26+12Sin(A+B)=40

12Sin(A+B)=14....2)

also,we know,ABC=180

so

angle(A+B+C)=180

A+B=180-C......3)

Put this value in equation 2)

12Sin(180-C)=14

12sinC=14

sinc=7/6

using this take out A,B and C and put this in given eqaution...

u will get 8 as answer