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\operatorname { tan } ( 45 ^ { \circ } + \theta ) - \operatorname { tan } ( 45 ^ { \circ } - \theta ) = 2 \operatorname { tan } 2 \theta |
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Answer» tan (45°+x)=(1+tanx)/(1-tanx)........ (1)tan (45°-x)=(1-tanx)/(1+tanx)........ (2)subtracting eq.2 from1tan (45°+x)-tan (45°-x)=(1+tanx)/(1-tanx) - (1-tanx)/(1+tanx) =[(1+tanx)(1+tanx)-(1-tanx)(1-tanx)]/(1+tanx)(1-tanx) =[(1+tan^2 x +2tanx)-(1+tan^2 x -2tanx)]/(1-tan^2 x) =(2tanx + 2tanx)/(1-tan^2 x) =4tanx/(1-tan^2 x) =2×(2tanx/(1-tan^2 x) =2×tan2x =2tanx here i have used x for theta |
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