1.

If `tan(A-B)=1a n dsec(A+B)=2/(sqrt(3))`, then the smallest positive values of A and B, respectively, are`(25pi)/(24),(19pi)/(24)`(b) `(19pi)/(24),(25pi)/(24)``(31pi)/(24),(31pi)/(24)`(d) `(13pi)/(24),(31pi)/(24)`

Answer» `tan(A-B) = 1 and sec(A+B) = 2/sqrt3`
`:. A-B = npi+pi/4 and A+B = 2npi+-pi/6`
As, we have to find the smallest positive values of `A` and `B`,
`:. A-B = pi/4->(1)`
Now, `A+B` can not be `pi/6` as it becmes less than `A-B` as `A` and `b` are positive.
So, with smallest positive values for `A` and `B`,
`A+B= 2(1)pi-pi/6 = (11pi)/6`
`:.A+B = (11pi)/6->(2)`
Now, adding (1) and (2),
`=>2A = pi/4+(11pi)/6 = (25pi)/12`
`=>A = (25pi)/24`
`=>B = A-pi/4 = (25pi)/24 - pi/4 = (19pi)/24.`


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