The greater of the two angles `A=2tan^(-1)(2sqrt(2)-1)` and `B=3sin^(-1)(1/3)+sin^(-1)(3/5)i s` ____.

Answer

`A`

Explanation

Given, `A = 2 tan ^(-1) (2 sqrt2 -1)`
and `B = 3 sin ^(-1) ((1)/(3)) + sin ^(-1) ((3)/(5))`
Here, ` A = 2 tan ^(-1) ( 2sqrt2 - 1)`
`" " = 2 tan ^(-1) ( 2xx 1.414 - 1)`
`" " 2 tan ^(-1) (1.828)`
`therefore " " A gt 2 tan ^(-1) (sqrt3) = 2*(pi)/(3) = ( 2pi)/(3) `
To find the value of B, we first say
`sin ^(-1) ""(1)/(3) lt sin ^(-1)""(1)/(2) = (pi)/(6)`
so that `0 lt 3 sin ^(-1) ""(1)/(3) lt (pi)/(2)`
Now, `3 sin ^(-1)""(1)/(3) = sin ^(-1) (3 *(1)/(3) - 4 *(1)/(27))`
`" " = sin ^(-1)((23)/(27))`
`sin ^(-1)(0.851) lt sin^(-1) ((sqrt3)/(2)) = (pi)/(3)`
`sin ^(-1) ((3)/(5)) = sin ^(-1) (0.6) lt sin ^(-1)((sqrt3)/(2)) = (pi)/(3)`
`therefore B lt (pi)/(3) + (pi)/(3) = ( 2pi)/(3)`
Thus, `A gt ( 2pi)/(3) and B lt ( 2pi)/(3)`
Hence, greater angle is A.