if A =! B and atanA + btanB =
(a + b)tan{(A+B)/2} , then prove
cosA / cosB = a/b
Answers
\underline{\text{Proof :}}Proof :
\text{Given,}Given,
\mathrm{a\:tan\alpha+b\:tan\beta=(a+b)\:tan\frac{\alpha+\beta}{2}}atanα+btanβ=(a+b)tan2α+β
\to \mathrm{a\frac{sin\alpha}{cos\alpha}+b\frac{sin\beta}{cos\beta}=a\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}}+b\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}}}→acosαsinα+bcosβsinβ=acos2α+βsin2α+β+bcos2α+βsin2α+β
\to \small{\mathrm{a(\frac{sin\alpha}{cos\alpha}-\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}})+b (\frac{sin\beta}{cos\beta}-\frac{sin\frac{\alpha+\beta}{2}}{cos\frac{\alpha+\beta}{2}})=0}}→a(cosαsinα−cos2α+βsin2α+β)+b(cosβsinβ−cos2α+βsin2α+β)=0
\to \tiny{\mathrm{a\frac{sin\alpha\:cos\frac{\alpha+\beta}{2}-cos\alpha\:sin\frac{\alpha+\beta}{2}}{cos\alpha\:cos\frac{\alpha+\beta}{2}}+b\frac{sin\beta\:cos\frac{\alpha+\beta}{2}-cos\beta\:sin\frac{\alpha+\beta}{2}}{cos\beta\:cos\frac{\alpha+\beta}{2}}=0}}→acosαcos2α+βsinαcos2α+β−cosαsin2α+β+bcosβcos2α+βsinβcos2α+β−cosβsin2α+β=0
\to \mathrm{a\frac{sin(\alpha-\frac{\alpha+\beta}{2})}{cos\alpha}+b\frac{sin(\beta-\frac{\alpha+\beta}{2})}{cos\beta}=0}→acosαsin(α−2α+β)+bcosβsin(β−2α+β)=0
\to \mathrm{a\frac{sin\frac{\alpha-\beta}{2}}{cos\alpha}+b\frac{sin\frac{\beta-\alpha}{2}}{cos\beta}=0}→acosαsin2α−β+bcosβsin2β−α=0
\to \mathrm{a\frac{sin\frac{\alpha-\beta}{2}}{cos\alpha}-b\frac{sin\frac{\alpha-\beta}{2}}{cos\beta}=0}→acosαsin2α−β−bcosβsin2α−β=0
\to \mathrm{\frac{a}{cos\alpha}-\frac{b}{cos\beta}=0}→