If tan alpha=√3 tan beta=1/√3 find (aplbha+beta) plx give me and fast
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tanα=√3, tanβ=1/√3
tan(α+β)=(tanα+tanβ)/(1-tanα.tanβ)
or, tan(α+β)=(√3+1/√3)/(1-√3.1/√3)
or, tan(α+β)=(√3+1/√3)/(1-1)
or, tan(α+β)=undefined
or, α+β=tan-1(undefined)
or, α+β=π/2 or 90°
tan(α+β)=(tanα+tanβ)/(1-tanα.tanβ)
or, tan(α+β)=(√3+1/√3)/(1-√3.1/√3)
or, tan(α+β)=(√3+1/√3)/(1-1)
or, tan(α+β)=undefined
or, α+β=tan-1(undefined)
or, α+β=π/2 or 90°
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