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Question-140803




Question Number 140803 by jlewis last updated on 12/May/21
Answered by Ar Brandon last updated on 12/May/21
tany=((2x)/(x^2 −1)) , 1+tan^2 y=sec^2 y  ⇒cosy=±(1/( (√(1+tan^2 y))))=±(1/( (√(1+(((2x)/(x^2 −1)))^2 ))))
$$\mathrm{tany}=\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:,\:\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}=\mathrm{sec}^{\mathrm{2}} \mathrm{y} \\ $$$$\Rightarrow\mathrm{cosy}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{y}}}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\left(\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} }} \\ $$

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