Question

please answer ASAP !!!!!!!!

Answer 1

atanα + btanβ = ( a + b)tan(α + β)/2

atanα + btanβ = atan(α +β)/2 + btan(α +β)/2

atanα − atan(α +β)/2 = btan(α +β)/2 −btanβ

a( sinα.cos(α+β)/2 −sin(α+β)/2.cosα)/cosα.cos(α+β)/2 = b(sin(α +β)/2.cosβ−sinβ.cos(α+β)/2)/cosβ.cos(α +β)/2

a sin(α−α/2 −β/2)/cosα = bsin(β/2 +α/2−β)/cosβ

asin(α−β)/2/cosα = bsin(α−β)/2/cosβ

a/cosα = b/cosβ

acosβ =bcosα
hence. proved /

Answer 2

Answer:

atana + btanß = ( a + b)tan(a + B)/2

atana + btanß = atan(a +B)/2 + btan(a +ß)/ 2

atana - atan(a +ß)/2 = btan(a +ß)/2 -btanß

al sina.cos(a+b)/2-sin(a+b)/2.cosa)/ cosa.cos(a+b)/2 = b(sin(a +ß)/ 2.cosß-sinß.cos(a+b)/2)/cosß.cos(a +ß)/2

a sin(a-a/2 -B/2)/cosa = bsin(B/2 +a/2-B)/ cos

asin(a-ß)/2/cosa = bsin(a-B)/2/cosß

a/cosa = b/cosß

acosß =bcosa hence. proved /