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Question Number 126405 by amns last updated on 20/Dec/20

If tanA = (√3) , prove it: (√3) sinA cosA = (3/4)

$$\boldsymbol{\mathrm{If}}\:{tanA}\:=\:\sqrt{\mathrm{3}}\:,\:\boldsymbol{\mathrm{prove}}\:\boldsymbol{\mathrm{it}}:\:\sqrt{\mathrm{3}}\:{sinA}\:{cosA}\:=\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Commented by amns last updated on 20/Dec/20

help me, plz...

$$\mathrm{help}\:\mathrm{me},\:\mathrm{plz}... \\ $$

Commented by som(math1967) last updated on 20/Dec/20

tanA=(√3) cosA=(√(1/(1+tan^2 A)))=(1/2) tanA=(√3) sinA=tanAcosA=((√3)/2) (√3)sinAcosA =(√3)×((√3)/2)×(1/2)=(3/4)

$$\boldsymbol{{tanA}}=\sqrt{\mathrm{3}} \\ $$$${c}\boldsymbol{{osA}}=\sqrt{\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {A}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\boldsymbol{{tanA}}=\sqrt{\mathrm{3}} \\ $$$$\boldsymbol{{sinA}}=\boldsymbol{{tanAcosA}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\sqrt{\mathrm{3}}\boldsymbol{{sinAcosA}} \\ $$$$=\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$

Answered by benjo_mathlover last updated on 20/Dec/20

if A in 1^(st) −quadrant ⇒tan A=(√3) ⇒ A=60° ∧ (√3) sin A.cos A=(√3) sin 60°.cos 60°=(√3)×((√3)/2)×(1/2)=(3/4) if A in 3^(rd) −quadrant ⇒tan A=(√3) A=240° then (√3) ×(−((√3)/2))×(−(1/2))=(3/4)

$${if}\:{A}\:{in}\:\mathrm{1}^{{st}} −{quadrant}\:\Rightarrow\mathrm{tan}\:{A}=\sqrt{\mathrm{3}} \\ $$$$\Rightarrow\:{A}=\mathrm{60}°\:\wedge\:\sqrt{\mathrm{3}}\:\mathrm{sin}\:{A}.\mathrm{cos}\:{A}=\sqrt{\mathrm{3}}\:\mathrm{sin}\:\mathrm{60}°.\mathrm{cos}\:\mathrm{60}°=\sqrt{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$${if}\:{A}\:{in}\:\mathrm{3}^{{rd}} −{quadrant}\:\Rightarrow\mathrm{tan}\:{A}=\sqrt{\mathrm{3}} \\ $$$${A}=\mathrm{240}°\:{then}\:\sqrt{\mathrm{3}}\:×\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$

Answered by TITA last updated on 20/Dec/20

tan A=(√3) we show (√3)sin Acos A=(3/4) (√3)cos A=sin A ⇒ let t=(√3)sin Acos A=sin^2 A ⇒ t=sin^2 A=((tan^2 A)/(1+tan^2 A))=(3/(1+3))=(3/4)

$$\mathrm{tan}\:\mathrm{A}=\sqrt{\mathrm{3}}\:\:\:\mathrm{we}\:\mathrm{show}\:\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{Acos}\:\mathrm{A}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{A}=\mathrm{sin}\:\mathrm{A}\:\:\Rightarrow\:\mathrm{let}\:\mathrm{t}=\sqrt{\mathrm{3}}\mathrm{sin}\:\mathrm{Acos}\:\mathrm{A}=\mathrm{sin}\:^{\mathrm{2}} \mathrm{A} \\ $$$$\Rightarrow\:\mathrm{t}=\mathrm{sin}\:^{\mathrm{2}} \mathrm{A}=\frac{\mathrm{tan}\:^{\mathrm{2}} \mathrm{A}}{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \mathrm{A}}=\frac{\mathrm{3}}{\mathrm{1}+\mathrm{3}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$ \\ $$

Answered by MJS_new last updated on 20/Dec/20

t=tan A ⇒ sin A =(t/( (√(t^2 +1))))∧cos A =(1/( (√(t^2 +1)))) (√3)sin A cos A =(((√3)t)/(t^2 +1))=^((t=(√3))) (3/4)

$${t}=\mathrm{tan}\:{A}\:\Rightarrow\:\mathrm{sin}\:{A}\:=\frac{{t}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}}\wedge\mathrm{cos}\:{A}\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\sqrt{\mathrm{3}}\mathrm{sin}\:{A}\:\mathrm{cos}\:{A}\:=\frac{\sqrt{\mathrm{3}}{t}}{{t}^{\mathrm{2}} +\mathrm{1}}\overset{\left({t}=\sqrt{\mathrm{3}}\right)} {=}\frac{\mathrm{3}}{\mathrm{4}} \\ $$

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