Question

If cos A = `(1)/(2)` and sin B = `(1)/(sqrt2)`, find the value of: `(tan"A"  –  tan"B")/(1+tan"A" tan"B")`.

Are angles A and B from the same triangle? Explain.

Answer 1

Consider the diagram below: 

cos A = `(1)/(2)`

i.e.`"base"/"hypotenuse"= (1)/(2)`

⇒ `"AB"/"AC" = (1)/(2)`

Therefore if length of AB = x, length of AC = 2x

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

(x)² + BC² = (2x)²

BC² = 4x² – x² = 3x²

∴ BC = `sqrt3x`     ...(perpendicular)

Consider the diagram below: 

sin B = `(1)/(sqrt2)`

i.e.`"perpendicular"/"hypotenuse" = (1)/(sqrt2)`

⇒ `"AC"/"BC" = (1)/(sqrt2)`

Therefore if length of AC = x, length of BC = `sqrt2`

Since

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

AB² + x² = `(sqrt2x)^2`

AB² = 2x² – x² = x²

∴ AB = x (base)

Now

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

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

Thererfore

`(tan"A"  –  tan"B")/(1+tan"A"tan"B")`

= `(sqrt3  –  1)/(1+sqrt3)`

= `(sqrt3  –  1)/(1+sqrt3) xx (1-sqrt3) /(1-sqrt3)`

= `((sqrt3 - 1) (1-sqrt3))/(1-3) `

= `(sqrt3 - 3 - 1 + sqrt3) / 2`

= `(2sqrt3-4)/2`

= `(cancel2 (sqrt3-2))/cancel2`

= `sqrt3 -2`