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If 3tanα+4tanβ+5tanγ=λ,then the minimum value of tan²α+tan²β+tan²γ is
a.λ²/9
b.λ²/16
c.λ²/25
d.none of these
Hi Menka,
Consider vectors:
A = 3i+4j+5k,
B = tanxi + tanyj + tanzk
A(dot)B = 3tanx+4tany+5tanz = λ.
Now A(dot)B ≤ |A| |B|.
So minimum value of the given expression would be λ2/50.
Hence (D).
Hope that helps.
Best Regards,
Ashwin (IIT Madras).
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