α and β are the positive acute angles and satisfying equations 5sin⁡2β=3sin⁡2α and tan⁡β=3tan⁡α simultaneously. Then the value oftan⁡α+tan⁡β is

5×2tan⁡β1+tan2⁡β=3×2tan⁡α1+tan2⁡α or  5tan⁡β1+tan2⁡β=3tan⁡α1+tan2⁡α Substituting tan⁡β=3tan⁡α, we have     5×3tan⁡α1+9tan2⁡α=3tan⁡α1+tan2⁡α or     5+5tan2⁡α=1+9tan2⁡α or     4tan2⁡α=4 or      tan⁡α=1, i.e., tan⁡β=3∴     tan⁡α+tan⁡β=4