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Question Number 46573 by Rio Michael last updated on 28/Oct/18
Show that sin2x ≡((2tanx)/(1+tan^2 x))
$${Show}\:{that}\: \\ $$$${sin}\mathrm{2}{x}\:\equiv\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}} \\ $$
Commented by peter frank last updated on 28/Oct/18
sin2x=((2sinxcosx)/1) = ((2sinxcosx)/(cos^2 x+sin^2 x)) =(((2sinxcosx)/(cos^2 x))/(1+tan^2 x))=((2tanx)/(1+tan^2 x)) please recheck
$$\mathrm{sin2x}=\frac{\mathrm{2sinxcosx}}{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{2sinxcosx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}+\mathrm{sin}^{\mathrm{2}} \mathrm{x}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{\frac{\mathrm{2sinxcosx}}{\mathrm{cos}^{\mathrm{2}} \mathrm{x}}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}}=\frac{\mathrm{2tanx}}{\mathrm{1}+\mathrm{tan}^{\mathrm{2}} \mathrm{x}} \\ $$$$\mathrm{please}\:\mathrm{recheck} \\ $$
Commented by maxmathsup by imad last updated on 28/Oct/18
your answer is corrct peter but the Q.contain a error.
$${your}\:{answer}\:{is}\:{corrct}\:{peter}\:{but}\:{the}\:{Q}.{contain}\:{a}\:{error}. \\ $$
Commented by peter frank last updated on 28/Oct/18
thank you sir....I also had such doubt.
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}….\mathrm{I}\:\mathrm{also}\:\mathrm{had}\:\mathrm{such}\:\mathrm{doubt}. \\ $$
Commented by hknkrc46 last updated on 29/Nov/18
tan 2x=((2tan x)/(1−tan^2 x))⇒2tan x=tan 2x(1−tan^2 x) {1−tan^2 x=1−((sin^2 x)/(cos^2 x))=((cos^2 x−sin^2 x)/(cos^2 x))=((cos 2x)/(cos^2 x))} 2tan x=tan 2x∙((cos 2x)/(cos^2 x))=((sin 2x)/(cos 2x))∙((cos 2x)/(cos^2 x))=((sin 2x)/(cos^2 x)) {1+tan^2 x=1+((sin^2 x)/(cos^2 x))=((sin^2 x+cos^2 x)/(cos^2 x))=(1/(cos^2 x))} ((2tan x)/(1+tan^2 x))=(((sin 2x)/(cos^2 x))/(1/(cos^2 x)))=sin 2x
$$\mathrm{tan}\:\mathrm{2}{x}=\frac{\mathrm{2tan}\:{x}}{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}}\Rightarrow\mathrm{2tan}\:{x}=\mathrm{tan}\:\mathrm{2}{x}\left(\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}\right) \\ $$$$\left\{\mathrm{1}−\mathrm{tan}\:^{\mathrm{2}} {x}=\mathrm{1}−\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{cos}\:^{\mathrm{2}} {x}−\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\right\} \\ $$$$\mathrm{2tan}\:{x}=\mathrm{tan}\:\mathrm{2}{x}\centerdot\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:\mathrm{2}{x}}\centerdot\frac{\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}} \\ $$$$\left\{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}=\mathrm{1}+\frac{\mathrm{sin}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{sin}\:^{\mathrm{2}} {x}+\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}=\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}\right\} \\ $$$$\frac{\mathrm{2tan}\:{x}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {x}}=\frac{\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{cos}\:^{\mathrm{2}} {x}}}{\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} {x}}}=\mathrm{sin}\:\mathrm{2}{x} \\ $$
Answered by Nabraj Awasthi last updated on 30/Oct/18
Left side =sin2x =2sinx.cosx multiplying and diving it by sex^2 x, we have 2sinx.cosx×((sec^2 x)/(sec^2 x)) =((2sinx.secx)/(sec^2 x)) =((2tanx)/(1+tan^2 x))
$${Left}\:{side} \\ $$$$={sin}\mathrm{2}{x}\:\:\:\:=\mathrm{2}{sinx}.{cosx} \\ $$$${multiplying}\:{and}\:{diving}\:{it}\:{by}\:{sex}^{\mathrm{2}} {x},\:{we}\:{have} \\ $$$$\mathrm{2}{sinx}.{cosx}×\frac{{sec}^{\mathrm{2}} {x}}{{sec}^{\mathrm{2}} {x}}\:\:\:\:\:=\frac{\mathrm{2}{sinx}.{secx}}{{sec}^{\mathrm{2}} {x}}\:\:=\frac{\mathrm{2}{tanx}}{\mathrm{1}+{tan}^{\mathrm{2}} {x}} \\ $$

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