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If sinA=1/2,cosB=(sqrt(3))/2, w here pi/...

If `sinA=1/2,cosB=(sqrt(3))/2, w here pi/2 lt A lt pi and 0ltBltpi/2` find the following: `tan(A-B)dot`

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Given,`sinA=1/2,cosB=(sqrt(3))/2`
We know that `sin^2A+cos^2A=1`
Then,`cosA=-sqrt(3)/2`
`sinB=1/2`
`tan (A – B) = (tan A – tan B)/(1 + tan A tan B)`
= `((-1/sqrt(3)) – (1/sqrt(3)))/(1 + (-1/sqrt(3)) × (1/sqrt(3))) `
=` ((-2/sqrt(3))/(1 – 1/3)) `
...
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