Answer:
[tex]\rm\displaystyle 0,\pm\pi [/tex]
Step-by-step explanation:
please note that to find but α+β+γ in other words the sum of α,β and γ not α,β and γ individually so it's not an equation
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we want to find all possible values of α+β+γ when tanα+tanβ+tanγ = tanαtanβtanγ to do so we can use algebra and trigonometric skills first
cancel tanγ from both sides which yields:
[tex] \rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma ) [/tex]
factor out tanγ:
[tex]\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)[/tex]
divide both sides by tanαtanβ-1 and that yields:
[tex]\rm\displaystyle \tan( \gamma ) = \frac{ \tan( \alpha ) + \tan( \beta ) }{ \tan( \alpha ) \tan( \beta ) - 1}[/tex]
multiply both numerator and denominator by-1 which yields:
[tex]\rm\displaystyle \tan( \gamma ) = - \bigg(\frac{ \tan( \alpha ) + \tan( \beta ) }{ 1 - \tan( \alpha ) \tan( \beta ) } \bigg)[/tex]
recall angle sum indentity of tan:
[tex]\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta ) [/tex]
let α+β be t and transform:
[tex]\rm\displaystyle \tan( \gamma ) = - \tan( t) [/tex]
remember that tan(t)=tan(t±kπ) so
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi ) [/tex]
therefore when k is 1 we obtain:
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi ) [/tex]
remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus
[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi ) [/tex]
recall that if we have common trigonometric function in both sides then the angle must equal which yields:
[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm \pi [/tex]
isolate -α-β to left hand side and change its sign:
[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }[/tex]
when is 0:
[tex]\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 ) [/tex]
likewise by Opposite Angle Identity we obtain:
[tex]\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 ) [/tex]
recall that if we have common trigonometric function in both sides then the angle must equal therefore:
[tex]\rm\displaystyle \gamma = - \alpha - \beta \pm 0 [/tex]
isolate -α-β to left hand side and change its sign:
[tex]\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }[/tex]
and we're done!
Answer:
-π, 0, and π
Step-by-step explanation:
You can solve for tan y :
tan y (tan a + tan B - 1) = tan a + tan y
Assuming tan a + tan B ≠ 1, we obtain
[tex]tan/y/=-\frac{tan/a/+tan/B/}{1-tan/a/tan/B/} =-tan(a+B)[/tex]
which implies that
y = -a - B + kπ
for some integer k. Thus
a + B + y = kπ
With the stated limitations, we can only have k = 0, k = 1 or k = -1. All cases are possible: we get k = 0 for a = B = y = 0; we get k = 1 when a, B, y are the angles of an acute triangle; and k = - 1 by taking the negatives of the previous cases.
It remains to analyze the case when "tan "a" tan B = 1, which is the same as saying that tan B = cot a = tan(π/2 - a), so
[tex]B=\frac{\pi }{2} - a + k\pi[/tex]
but with the given limitation we must have k = 0, because 0 < π/2 - a < π.
On the other hand we also need "tan "a" + tan B = 0, so B = - a + kπ, but again
k = 0, so we obtain
[tex]\frac{\pi }{2} - a=-a[/tex]
a contradiction.
