Question

acosφ=bcosθ

Show that

atanθ+btanφ=(a+b) tan (θ+φ) ÷2

Answer 1

i) atanӨ + btanΦ = a.sinӨ/cosӨ + b.sinΦ/cosΦ

= asinӨ/{acosΦ/b} + bsinΦ/cosΦ [From the given data, cosӨ = {acosΦ/b}]

= bsinӨ/cosΦ + bsinΦ/cosΦ

= b{sinӨ + sinΦ}/cosΦ

= b[2sin{(Ө + Φ)/2}cos{(Ө - Φ)/2}]/cosΦ [Application of Sum-Product rule of trigonometry identities]

= b[{sin{(Ө + Φ)/2}/cos{(Ө + Φ)/2}*2cos{(Ө - Φ)/2}cos{(Ө + Φ)/2}]/cos(Φ)

= b[tan{(Ө + Φ)/2}{cosӨ + cosΦ}]/cosΦ

= b[tan{(Ө + Φ)/2}{cosӨ/cosΦ + 1}

= b[tan{(Ө + Φ)/2}{a/b + 1} [From the given one, cos(Ө)/cos(Φ) = a/b]

= (a + b)tan{(Ө + Φ)/2} = Right side [Proved]