Question

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

Answer 1

Step-by-step explanation:

ANSWER

tanB

tanA

=

3

1

⇒

cosAsinB

sinAcosB

=

3

1

Put sinAcosB=

4

1

∴cosAsinB=

4

3

∴sin(A+B)=

4

1

+

4

3

=1orsinC=1=sin

2

π

∴C=

2

π

Hence triangle is right angled