यदि `n * sin (A +2B)= sin A `सिद्ध कीजिए कि
` tan (A+B) = (1+n)/(1-n)*tan B.`

Explanation

` n sin (A+2B)= sinA`
` rArr n = (sinA)/(sin(A+2B))`
` :. R.H.S = (1+n)/(1-n)* tanB = 1+((sinA)/(sin(A+2B)))/(1-(sin A)/(sin(A+2B)))* tan B`
` = ((sin (A+2B)+sinA))/(sin(A+2B)-sinA)* tanB`
` = (2 sin. (A+2B+A)/2 cos. (A+2B-A)/2 )/(2cos. (A+2B+A)/2sin. (A+2B-A)/2)* tan B`
` (sin (A+B)cosB sin B)/(cos(A+B)sinBcosB)`
` = tan (A+B) = L.H.S`