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Ierofanga [76]
3 years ago
8

What is the constant in the following equation? 12 = 3k

Mathematics
1 answer:
stiks02 [169]3 years ago
5 0

Answer:

k = 4

Step-by-step explanation:

12=3k\\\frac{12}{3}=\frac{3k}{3} \\4=k

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\rm\displaystyle  0,\pm\pi

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 <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>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 ) =  \frac{ \tan( \alpha )  +  \tan( \beta ) }{ \tan( \alpha )  \tan( \beta )    - 1}

multiply both numerator and denominator by-1 which yields:

\rm\displaystyle   \tan( \gamma ) =   -  \bigg(\frac{ \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 <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> 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  }

<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

\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!

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