If α,β,γ,δ are the solutions of the equation tan⁡θ+π4= 3 tan3θ, no two of which have equal tangents.The value of tan⁡α+tan⁡β+tan⁡γ+tan⁡δ isThe value of tan⁡α tan⁡β tan⁡γ tan⁡δ is The value of 1tan⁡α+1tan⁡β+1tan⁡γ+1tan⁡δ is

We have tan⁡θ+π4=3tan⁡3θor 1+tan⁡θ1−tan⁡θ=3×3tan⁡θ−tan3⁡θ1−3tan2⁡θ⇒1+t1−t=33t−t31−3t2 (putting t=tan⁡θ )or 3t4-6t2+8t- 1= 0Hence,S1 : sum of roots =t1+t2+t3+t4=0S2 : sum of product of roots taken two at a time = - 2S3 : sum of product of roots taken three at time = - 8 /3S4 : product of all roots = - 1/31t1+1t2+1t3+1t4=∑t1t2t3t1t2t3t4=−8/3−1/3=8