2tanB+cotB=tanA ,then prove that 2tan(A-B)=cotB
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2tan(A-B)=2{tanA-tanB/1+tanAtanB}........1Given:tanA=2tanB+cotB.......2tanA=tanB+tanB+1/tanBtanA-tanB={(tan^2B+1)/(tanB)}.........3From 2tanAtanB=2tan^2B+1.........4Substitute 3 and 4 in 1You get2tan(A-B)=2{tan^2B+1)/(tanB)/2tan^2B+2}2tan(A-B)={(tan^2B+1)/(tanB)/tan^2B+1}2tan(A-B)=(1/tanB)2tan(A-B)=cotB
2tan(A-B)=2{tanA-tanB/1+tanAtanB}........1Given:tanA=2tanB+cotB.......2tanA=tanB+tanB+1/tanBtanA-tanB={(tan^2B+1)/(tanB)}.........3From 2tanAtanB=2tan^2B+1.........4Substitute 3 and 4 in 1You get2tan(A-B)=2{tan^2B+1)/(tanB)/2tan^2B+2}2tan(A-B)={(tan^2B+1)/(tanB)/tan^2B+1}2tan(A-B)=(1/tanB)2tan(A-B)=cotB
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Answer:proved
Step-by-step explanation:explained
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