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

Explanation

`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)`

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