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Question Number 31017 by 78987 last updated on 02/Mar/18

solve (√(1+tan^2 x/1+cot^2 x=   tanx))

$${solve}\:\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {x}/\mathrm{1}+{cot}^{\mathrm{2}} {x}=\:\:\:{tanx}} \\ $$

Answered by iv@0uja last updated on 02/Mar/18

1+tan^2 x=((cos^2 x)/(cos^2 x))+((sin^2 x)/(cos^2 x))=(1/(cos^2 x))  1+cot^2 x=((sin^2 x)/(sin^2 x))+((cos^2 x)/(sin^2 x))=(1/(sin^2 x))  hence  (√((1+tan^2 x)/(1+cot^2 x)))=(√((sin^2 x)/(cos^2 x)))=(√(tan^2 x))=tan x  so, this equation is always true  as long as tan x is defined.

$$\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}=\frac{\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}} {x}}+\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}} {x}}=\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}} {x}} \\ $$$$\mathrm{1}+\mathrm{cot}^{\mathrm{2}} {x}=\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{sin}^{\mathrm{2}} {x}}+\frac{\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}^{\mathrm{2}} {x}}=\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} {x}} \\ $$$${hence} \\ $$$$\sqrt{\frac{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} {x}}{\mathrm{1}+\mathrm{cot}^{\mathrm{2}} {x}}}=\sqrt{\frac{\mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{2}} {x}}}=\sqrt{\mathrm{tan}^{\mathrm{2}} {x}}=\mathrm{tan}\:{x} \\ $$$${so},\:{this}\:{equation}\:{is}\:{always}\:{true} \\ $$$${as}\:{long}\:{as}\:\mathrm{tan}\:{x}\:{is}\:{defined}. \\ $$

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