Question

Given sin=1/2 and sin b =1/underoot 2 find tan a -tan b/1+tan a tan b

Answer 1

if
Sina = 1/2
and
sinb = 1/√2

then
tana = 1/√3
tanb = 1

so
[tana - tanb]/[ 1 + tana tanb ]
= [1/√3 - 1]÷[ 1 + 1/√3 ]
= [1/√3-1]²÷[1/3-1]
= 3/2[1/√3 - 1]²
= 3[1+1/3-2/√3]/2
= 3(4/3 - 2/√3)/2