If α+β+γ=2π, then
tanα2+tanβ2+tanγ2=tanα2tanβ2tanγ2
tanα2tanβ2+tanβ2tanγ2+tanγ2tanα2=1
tanα2+tanβ2+tanγ2=-tanα2tanβ2tanγ2
None of these
Explanation for the correct option:
Given, α+β+γ=2π
⇒ (α+β+γ)2=π
Take “tan” on both sides,
tanα2+β2+γ2=tanπ
⇒tanα2+tanβ2+tanγ2–tanα2tanβ2tanγ21–tanα2tanβ2–tanβ2tanγ2–tanγ2tanα2=0
⇒ tanα2+tanβ2+tanγ2–tanα2tanβ2tanγ2=0
∴tanα2+tanβ2+tanγ2=tanα2tanβ2tanγ2
Hence, Option ‘A’ is Correct.
If cos−1α+cos−1β
+cos−1γ=3π, then
α(β+γ)+β(γ+α)+γ(α+β) equals