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Question Number 10695 by okhema last updated on 23/Feb/17

prove that tanx−cotx=−2cot2x

$${prove}\:{that}\:{tanx}−{cotx}=−\mathrm{2}{cot}\mathrm{2}{x} \\ $$

Answered by nume1114 last updated on 23/Feb/17

tan x−cot x =((sin x)/(cos x))−((cos x)/(sin x)) =((sin^2 x−cos^2 x)/(sin xcos x)) =((−(cos^2 x−sin^2 x))/((1/2)×2sin xcos x)) =−2((cos 2x)/(sin 2x)) =−2cot 2x

$$\:\:\:\:\mathrm{tan}\:{x}−\mathrm{cot}\:{x} \\ $$$$=\frac{\mathrm{sin}\:{x}}{\mathrm{cos}\:{x}}−\frac{\mathrm{cos}\:{x}}{\mathrm{sin}\:{x}} \\ $$$$=\frac{\mathrm{sin}^{\mathrm{2}} {x}−\mathrm{cos}^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}\mathrm{cos}\:{x}} \\ $$$$=\frac{−\left(\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{sin}^{\mathrm{2}} {x}\right)}{\frac{\mathrm{1}}{\mathrm{2}}×\mathrm{2sin}\:{x}\mathrm{cos}\:{x}} \\ $$$$=−\mathrm{2}\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{sin}\:\mathrm{2}{x}} \\ $$$$=−\mathrm{2cot}\:\mathrm{2}{x} \\ $$

Commented by okhema last updated on 23/Feb/17

i dnt understand how you come about with the fourth line come down

$${i}\:{dnt}\:{understand}\:{how}\:{you}\:{come}\:{about}\:{with}\:{the}\:{fourth}\:{line}\:{come}\:{down} \\ $$

Commented by nume1114 last updated on 23/Feb/17

i used double angle fomulas: sin 2x=2sin xcos x cos 2x=cos^2 x−sin^2 x

$${i}\:{used}\:{double}\:{angle}\:{fomulas}: \\ $$$$\mathrm{sin}\:\mathrm{2}{x}=\mathrm{2sin}\:{x}\mathrm{cos}\:{x} \\ $$$$\mathrm{cos}\:\mathrm{2}{x}=\mathrm{cos}^{\mathrm{2}} {x}−\mathrm{sin}^{\mathrm{2}} {x} \\ $$$$ \\ $$

Answered by bar Jesús last updated on 24/Feb/17

((senx)/(cosx))−((cosx)/(senx))= −2cot2x ((sen^2 x−cos^2 x)/(senxcosx))= −2cot2x senxcosx=((sen2x)/2) −(cos^2 x−sen^2 x)= sen^2 x−cos^2 x= −cos2x ((−cos2x)/((sen2x)/2))= −2cot2x ((−2cos2x)/(sen2x))= −2cot2x −cot2x= −2cot2x

$$ \\ $$$$ \\ $$$$\frac{{senx}}{{cosx}}−\frac{{cosx}}{{senx}}=\:−\mathrm{2}{cot}\mathrm{2}{x} \\ $$$$ \\ $$$$\frac{{sen}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}}{{senxcosx}}=\:−\mathrm{2}{cot}\mathrm{2}{x} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:{senxcosx}=\frac{{sen}\mathrm{2}{x}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\: \\ $$$$−\left({cos}^{\mathrm{2}} {x}−{sen}^{\mathrm{2}} {x}\right)=\:{sen}^{\mathrm{2}} {x}−{cos}^{\mathrm{2}} {x}=\:−{cos}\mathrm{2}{x} \\ $$$$ \\ $$$$ \\ $$$$\frac{−{cos}\mathrm{2}{x}}{\frac{{sen}\mathrm{2}{x}}{\mathrm{2}}}=\:−\mathrm{2}{cot}\mathrm{2}{x} \\ $$$$ \\ $$$$\frac{−\mathrm{2}{cos}\mathrm{2}{x}}{{sen}\mathrm{2}{x}}=\:−\mathrm{2}{cot}\mathrm{2}{x} \\ $$$$ \\ $$$$−{cot}\mathrm{2}{x}=\:−\mathrm{2}{cot}\mathrm{2}{x} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by bar Jesús last updated on 24/Feb/17

jojojojo, excuse me −2cot2x= −2cot2x

$${jojojojo},\:{excuse}\:{me} \\ $$$$−\mathrm{2}{cot}\mathrm{2}{x}=\:−\mathrm{2}{cot}\mathrm{2}{x} \\ $$

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