If atanα+a2−1tanβ+a2+1tanγ=2a, where a is constant and α,β,γ are variable angles then the least value of
3(tan2α+tan2β+tan2γ)
4.00
1
2
3
Let A¯=ai¯+a2−1j¯+a2+1k¯
B→=tanαi→+tanβj→+tanγk¯⇒A¯.B¯=2a
⇒|A¯|2|B¯|2cos2θ=4a2
3[tan2α+tan2β+tan2γ]≥4