If a tan⁡α+a2−1tan⁡β+a2+1tan⁡γ=2a where a is constant and  α,β,γ are variable angles. Then the least value of 2727 tan2⁡α+tan2⁡β+tan2⁡γ must be

we have , atan⁡β−a2−1tan⁡α2+a2−1tan⁡γ−a2+1tan⁡β2+a2+1tan⁡α−atan⁡γ2≥0⇒a2+a2−1+a2+1tan2⁡α+tan2⁡β+tan2⁡γ−atan⁡α+a2−1tan⁡β+a2+1tan⁡γ2≥0   [using Lagrange's identity]⇒3a2tan2⁡α+tan2⁡β+tan2⁡γ−(2a)2≥0∴ 3tan2⁡α+tan2⁡β+tan2⁡γ≥4Hence 2727tan2⁡α+tan2⁡β+tan2⁡γ≥3636Least value is 3636.