VectorFundament120
Answered

Looking for answers? Westonci.ca is your go-to Q&A platform, offering quick, trustworthy responses from a community of experts. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

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)
Show your work too. Thank you!​

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

===========================

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.

We appreciate your time on our site. Don't hesitate to return whenever you have more questions or need further clarification. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.