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18 November, 17:04

Prove identity: tanx-1/tanx+1 = 1-cotx/1+cotx

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  1. 18 November, 17:18
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    to answer the question above, take the LHS.

    [ (tan x - 1) / (tan x + 1) ] =

    Remember that tan x = 1 / cot x.

    {[ (1 / cot x) - 1] / [ (1 / cot x) + 1]} =

    The LCD is cot x. Multiply as needed to get the common denominator for all terms.

    {[ (1 / cot x) - 1 (cot x / cot x) ] / [ (1 / cot x) + 1 (cot x / cot x) ]} =

    {[ (1 / cot x) - (cot x / cot x) ] / [ (1 / cot x) + (cot x / cot x) ]} =

    Then Simplify.

    [ (1 - cot x) / cot x] / [ (1 + cot x) / cot x] =

    Remember that (a / b) / (c / d) = (a / b) * (d / c).

    [ (1 - cot x) / cot x] * [cot x / (1 + cot x) ] =

    [ (1 - cot x) / (1 + cot x) ] =

    RHS

    The answer is

    [ (tan x - 1) / (tan x + 1) ] = [ (1 - cot x) / (1 + cot x) ]
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