If `Aa n dB` are acute positive angles satisfying the equations `3sin^2A+2sin^2B=1a n d3sin2A-2sin2B=0,t h e nA+2B` is equal to `pi` (b) `pi/2` (c) `pi/4` (d) `pi/6`

Options

A. `pi`
B. `(pi)/(2)`
C. `(pi)/(4)`
D. `(pi)/(6)`

Answer

B

Explanation

`3sin^(2)A+2sin^(@)B=1`
or `3sin^(2)A=cos2B`
Also `3sin2A-2sin2B=0`
or `sin2B=(3)/(2)sin2A`
Now, `cos(A+2B)=cos A cos 2B-sinA sin2B`
`=cos A3 sin^(2)A-sinA((3)/(2)) sin2A`
`=3sin^(2)AcosA-3sin^(2)Acos A=0`
`therefore A+2B=(pi)/(2)`