y= log squareroot of 1+tanx/1-tanx 
then prove that dy/dx=sec2x

Asked by mlipsita | 18th Jun, 2015, 10:25: AM

Expert Answer:

y equals log square root of fraction numerator 1 plus tan space x over denominator 1 minus tan space x end fraction end root y equals 1 half log space open square brackets fraction numerator 1 plus tan space x over denominator 1 minus tan space x end fraction close square brackets fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator 2 open square brackets fraction numerator 1 plus tan space x over denominator 1 minus tan space x end fraction close square brackets end fraction fraction numerator left parenthesis 1 minus tan space x right parenthesis s e c squared x plus left parenthesis 1 plus tan space x right parenthesis s e c squared x over denominator left parenthesis 1 minus tan space x right parenthesis squared end fraction fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator 2 open square brackets fraction numerator 1 plus tan space x over denominator 1 minus tan space x end fraction close square brackets end fraction fraction numerator s e c squared x minus tan space x. s e c squared x plus s e c squared x plus tan space x. s e c squared x over denominator left parenthesis 1 minus tan space x right parenthesis squared end fraction fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator 2 open square brackets fraction numerator 1 plus tan space x over denominator 1 minus tan space x end fraction close square brackets end fraction fraction numerator 2. s e c squared x over denominator left parenthesis 1 minus tan space x right parenthesis squared end fraction fraction numerator d y over denominator d x end fraction equals negative fraction numerator 1 over denominator left parenthesis 1 plus tan space x right parenthesis end fraction fraction numerator s e c squared x over denominator left parenthesis 1 minus tan space x right parenthesis end fraction fraction numerator d y over denominator d x end fraction equals fraction numerator s e c squared x over denominator 1 minus tan squared x end fraction fraction numerator d y over denominator d x end fraction equals fraction numerator 1 plus tan squared x over denominator 1 minus tan squared x end fraction fraction numerator d y over denominator d x end fraction equals fraction numerator 1 over denominator cos space 2 x end fraction fraction numerator d y over denominator d x end fraction equals s e c space 2 x

Answered by Prasenjit Paul | 19th Jun, 2015, 10:40: AM

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