If 3tanα+4tanβ+5tanγ=λ,then the minimum value of tan²α+tan²β+tan²γ isa.λ²/9b.λ²/16c.λ²/25d.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).