Question

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!​

Answer 1

-π, 0, and π

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

`tan/y/=-(tan/a/+tan/B/)/(1-tan/a/tan/B/) =-tan(a+B)`

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

`B=(\pi )/(2) - a + k\pi`

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

`(\pi )/(2) - a=-a`

a contradiction.

Answer 2

`\rm\displaystyle 0,\pm\pi`

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:

`\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \alpha ) \tan( \beta ) \tan( \gamma ) - \tan( \gamma )`

factor out tanγ:

`\rm\displaystyle \tan( \alpha ) + \tan( \beta ) = \tan( \gamma ) (\tan( \alpha ) \tan( \beta ) - 1)`

divide both sides by tanαtanβ-1 and that yields:

`\rm\displaystyle \tan( \gamma ) = ( \tan( \alpha ) + \tan( \beta ) )/( \tan( \alpha ) \tan( \beta ) - 1)`

multiply both numerator and denominator by-1 which yields:

`\rm\displaystyle \tan( \gamma ) = - \bigg(( \tan( \alpha ) + \tan( \beta ) )/( 1 - \tan( \alpha ) \tan( \beta ) ) \bigg)`

recall angle sum indentity of tan:

`\rm\displaystyle \tan( \gamma ) = - \tan( \alpha + \beta )`

let α+β be t and transform:

`\rm\displaystyle \tan( \gamma ) = - \tan( t)`

remember that tan(t)=tan(t±kπ) so

`\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm k\pi )`

therefore when k is 1 we obtain:

`\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta\pm \pi )`

remember Opposite Angle identity of tan function i.e -tan(x)=tan(-x) thus

`\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm \pi )`

recall that if we have common trigonometric function in both sides then the angle must equal which yields:

`\rm\displaystyle \gamma = - \alpha - \beta \pm \pi`

isolate -α-β to left hand side and change its sign:

`\rm\displaystyle \alpha + \beta + \gamma = \boxed{ \pm \pi }`

when is 0:

`\rm\displaystyle \tan( \gamma ) = -\tan( \alpha +\beta \pm 0 )`

likewise by Opposite Angle Identity we obtain:

`\rm\displaystyle \tan( \gamma ) = \tan( -\alpha -\beta\pm 0 )`

recall that if we have common trigonometric function in both sides then the angle must equal therefore:

`\rm\displaystyle \gamma = - \alpha - \beta \pm 0`

isolate -α-β to left hand side and change its sign:

`\rm\displaystyle \alpha + \beta + \gamma = \boxed{ 0 }`

and we're done!