A and B are positive acute angles satisfying the equations
`3cos^(2)A+2cos^(2)B=4` and `(3sinA)/(sinB)=(2cosB)/(cosA)`, then `A+2B` is equal to (a) `pi/4` (b) `pi/3` (c) `pi/6` (d)`pi/2`

Options

A. `(pi)/(3)`
B. `(pi)/(2)`
C. `(pi)/(6)`
D. `(pi)/(4)`

Answer

B

Explanation

We have,
`(3sinA)/(cosA)=(2cosB)/(cosA)`
`implies(3sinA)/(cosA)=(2cosB sinB)/(cos^(2)A)`
`impliestanA=1/3(sin2B)/(cos^(2)A)`
`impliestanA=1/3tan2B.(cos2B)/(cos^(2)A)`
`impliestanA=1/3(tan2B)/(cos^(2)A)(2cos^(2)B-1)`
`impliestanA=1/3(tan2B)/(cos^(2)A)(4-3cos^(2)A-1)" "[{:(,because2cos^(2)B),(,=4-3cos^(2)A):}]`
`impliestanA=tan2Btan^(2)A`
`impliestanA tan2B=1`
`impliestanA =cot2B`
`impliestanA=tan((pi)/(2)-2B)`
`impliesA=(pi)/(2)-2BimpliesA+2B=(pi)/(2)`