1tan@=t
nw
u get a quadratic in terms of t
t^2(b^2-a^2)+2act+b^2-c^2=0
tan α+tan β=-2ac/(b^2-a^2)
tan α*tan β =b^2-c^2/(b^2-a^2)
nw u can find tan(α+β) =tanα+tanβ/(1-tanαtanβ)
nw
u can easily find cos(α+β) ...........also
if no then do knck
1
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·2009-04-20 11:15:05
answer is
tan(α+β)=2ac/(a^2-c^2 )
1cos(α+β)==a^2-b^2/a^2+b^2
its corrct???????
11cos(α+β)==a^2-b^2/a^2+b^2
its corrct???????
nope dharun
check ur answer again
