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Find all possible values of α+
β+γ when tanα+tanβ+tanγ = tanαtanβtanγ (-π/2<α<π/2 , -π/2<β<π/2 , -π/2<γ<π/2)
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Sagot :

Nayefx

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.

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