Question

Select the correct option from the given alternatives :

Let 0 < A, B < `pi/2` satisfying the equation 3 sin²A + 2 sin²B = 1 and 3 sin 2A − 2 sin 2B = 0 then A + 2B is equal to ______

Options

• π

• `pi/2`

• `pi/4`

• 2π

Answer 1

Let 0 < A, B < `pi/2` satisfying the equation 3 sin²A + 2 sin²B = 1 and 3 sin 2A − 2 sin 2B = 0 then A + 2B is equal to `underline(pi/2)`

Explanation:

3 sin 2A − 2 sin 2B = 0

∴ sin 2B = `3/2 sin 2"A"` ......(i)

3 sin²A + 2 sin²B = 1

∴ 3 sin²A = 1 − 2 sin²B

∴ 3 sin²A = cos 2B  ...(ii)

cos (A + 2B) = cos A cos 2B – sin A sin 2B

= `cos "A"(3 sin^2"A") – sin"A"(3/2 sin2"A")` ...[From (i) and (ii)]

= `3 cos "A" sin^2 "A" - 3/2 (sin"A") (2 sin "A" cos "A")`

= 3 cos A sin²A – 3 sin²A cos A

= 0

= `cos  pi/2`

∴ A + 2B = `pi/2   ...[because 0 < "A" + 2"B" < (3pi)/2]`