Question

α+β+γ=90° then tanα+tanβ+tanα tanβ cotγ is?

Answer 1

Sol. As α+β+γ=90°
By using tan both sides ;
So, tan(α+β+γ) = tan 90°
Using, tan(α+β+γ) =[ tanα+tanβ+tanγ-tanαtanβtanγ]/[1-tanαtanβ -tanβtanγ-tanγtanα]

So using this we have,

1-tanαtanβ -tanβtanγ-tanγtanα = 0 (as tan90 = 1/0 or infinity)

1-tanβtanγ-tanγtanα = tanαtanβ

or tanγ[cotγ-tanβ-tabα] = tanαtanβ

or cotγ-tanβ-tabα = tanαtanβcotγ

or cotγ = tabα+tanβ+tanαtanβcotγ

so, tabα+tanβ+tanαtanβcotγ = cotγ (Ans)

Answer 2

I'm useing A,B,C instead of alfa , beta and gamma
A+B+C=90
A+B=90-C
Take bothsides tan
tan(A+B)=tan(90-C)
(tanA+tanB)/(1-tanAtanB)=cotC
tanA+tanB=cotC(1-tanAtanB)
tanA+tanB=cotC-tanAtanBcotC
tanA+tanB+tanAtanBcotC=cotC