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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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:

factor out tanγ:

divide both sides by tanαtanβ-1 and that yields:

multiply both numerator and denominator by-1 which yields:

recall angle sum indentity of tan:

let α+β be t and transform:

remember that tan(t)=tan(t±kπ) so,

we know that -tanx=tan(-x) therefore we obtain:

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

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

and we're done!

additional comment:—

in fact it's an identity known as triple tangent indentity it states that if α+β+γ=π then tanα+tanβ+tanγ = tanαtanβtanγ that's being said there's no value except π that's equal to α+β+γ for the given condition

step-by-step explanation:

5×5×5+2(8)÷2(2)+3(3)

15+16÷4+9

31÷13

do 31÷13 and u will get ur answer