Question

If sin A = `(sqrt3)/(2)` and cos B = `(sqrt3)/(2)` , find the value of : `(tan"A" – tan"B")/(1+tan"A" tan"B")`

Answer 1

Consider the diagram below :
'in A = `(sqrt3)/(2)`

i.e.`"perpendicular"/"hypotenuse" = (sqrt3)/(2) ⇒"BC"/"AC" = (sqrt3)/(2)`

Therefore if length of BC = `sqrt3x`, length of AC = 2x

Since
AB² + BC² = AC²    ...[ Using Pythagoras Therorm]

`(sqrt3x)^2 + AB^2 = (2x)^2`

AB² = x²

∴ AB = x (base)

Consider the diagram below :

cos B = `(sqrt3)/(2)`

i.e.`"base"/"perpendicular" = (sqrt3)/(2) ⇒ "AB"/"BC" = (sqrt3)/(2)`

Therefore if length of AB = `sqrt3x` , length of  BC = 2x

Since
AB² + AC² = BC²  ...[ Using Pythagoras Theorem ]

AC² + `(sqrt3x)^2 = (2x)^2`

AC² = x²

∴ AC = x(perpendicular)

Now

tan A = `"perpendicular"/"base" = (sqrt3x)/(x) = sqrt3`

tan B = `"perpendicular"/"base" = (x)/(sqrt3x)  = 1/(sqrt3)`

Therefore

`(tan A – tan B)/(1 + tan A tan B) = (sqrt3 - 1/(sqrt3))/(1+sqrt3 1/(sqrt3)`

= `((2)/(sqrt3))/(2)`

= `(1)/(sqrt3)`