Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 128995 by bramlexs22 last updated on 12/Jan/21

If tan^2 x −3tan x=1 has the roots are x_1 and x_2 then find the value of ∣ cos x_1 .cos x_2 ∣.

$$\mathrm{If}\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}\:−\mathrm{3tan}\:\mathrm{x}=\mathrm{1}\:\mathrm{has}\:\mathrm{the}\:\mathrm{roots} \\ $$$$\mathrm{are}\:\mathrm{x}_{\mathrm{1}} \:\mathrm{and}\:\mathrm{x}_{\mathrm{2}} \:\mathrm{then}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value} \\ $$$$\mathrm{of}\:\mid\:\mathrm{cos}\:\mathrm{x}_{\mathrm{1}} .\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:\mid.\: \\ $$

Commented by liberty last updated on 12/Jan/21

tan^2 x−3tan x−1=0 tan x = ((3 ± (√(13)))/2) → { ((tan x=((3+(√(13)))/2))),((tan x=((3−(√(13)))/2))) :} { ((cos x = (2/( (√(26+6(√(13)))))))),((cos x = (2/( (√(26−6(√(13)))))))) :} then ∣ cos x_1 .cos x_2 ∣=(4/( (√(676−468))))=(4/(4(√(13)))) =(1/( (√(13))))

$$\:\mathrm{tan}\:^{\mathrm{2}} \mathrm{x}−\mathrm{3tan}\:\mathrm{x}−\mathrm{1}=\mathrm{0} \\ $$$$\:\mathrm{tan}\:\mathrm{x}\:=\:\frac{\mathrm{3}\:\pm\:\sqrt{\mathrm{13}}}{\mathrm{2}}\:\rightarrow\begin{cases}{\mathrm{tan}\:\mathrm{x}=\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}}\\{\mathrm{tan}\:\mathrm{x}=\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}}\end{cases} \\ $$$$\begin{cases}{\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{26}+\mathrm{6}\sqrt{\mathrm{13}}}}}\\{\mathrm{cos}\:\mathrm{x}\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{26}−\mathrm{6}\sqrt{\mathrm{13}}}}}\end{cases}\:\mathrm{then}\:\mid\:\mathrm{cos}\:\mathrm{x}_{\mathrm{1}} .\mathrm{cos}\:\mathrm{x}_{\mathrm{2}} \:\mid=\frac{\mathrm{4}}{\:\sqrt{\mathrm{676}−\mathrm{468}}}=\frac{\mathrm{4}}{\mathrm{4}\sqrt{\mathrm{13}}}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}} \\ $$$$\: \\ $$

Answered by Lordose last updated on 12/Jan/21

tan^2 x − 3tanx + (−(3/2))^2 = ((13)/4) (tanx − (3/2))^2 = ((13)/4) ⇒ tan(x) = ((3±(√(13)))/2) x_1 = tan^(−1) (((3+(√(13)))/2)) ⇒ cos(x_1 ) = (2/( (√(26+6(√(13)))))) x_2 = tan^(−1) (((3−(√(13)))/2)) ⇒ cos(x_2 ) = (2/( (√(26−6(√(13)))))) Hence, ∣cos(x_1 )cos(x_2 )∣ = (1/( (√(13))))

$$\mathrm{tan}^{\mathrm{2}} \mathrm{x}\:−\:\mathrm{3tanx}\:+\:\left(−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{13}}{\mathrm{4}} \\ $$$$\left(\mathrm{tanx}\:−\:\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{13}}{\mathrm{4}}\:\Rightarrow\:\mathrm{tan}\left(\mathrm{x}\right)\:=\:\frac{\mathrm{3}\pm\sqrt{\mathrm{13}}}{\mathrm{2}} \\ $$$$\mathrm{x}_{\mathrm{1}} \:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}+\sqrt{\mathrm{13}}}{\mathrm{2}}\right)\:\Rightarrow\:\mathrm{cos}\left(\mathrm{x}_{\mathrm{1}} \right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{26}+\mathrm{6}\sqrt{\mathrm{13}}}}\:\:\: \\ $$$$\mathrm{x}_{\mathrm{2}} \:=\:\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{3}−\sqrt{\mathrm{13}}}{\mathrm{2}}\right)\:\Rightarrow\:\mathrm{cos}\left(\mathrm{x}_{\mathrm{2}} \right)\:=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{26}−\mathrm{6}\sqrt{\mathrm{13}}}}\:\:\:\:\: \\ $$$$\mathrm{Hence},\:\mid\mathrm{cos}\left(\mathrm{x}_{\mathrm{1}} \right)\mathrm{cos}\left(\mathrm{x}_{\mathrm{2}} \right)\mid\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}} \\ $$$$ \\ $$$$ \\ $$

Commented by liberty last updated on 12/Jan/21

i think it (1/( (√(13))))

$$\mathrm{i}\:\mathrm{think}\:\mathrm{it}\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}}\: \\ $$

Commented by Lordose last updated on 12/Jan/21

you are right I have found an error in my solution Thanks

$$\mathrm{you}\:\mathrm{are}\:\mathrm{right} \\ $$$$\mathrm{I}\:\mathrm{have}\:\mathrm{found}\:\mathrm{an}\:\mathrm{error}\:\mathrm{in}\:\mathrm{my}\:\mathrm{solution} \\ $$$$\mathrm{Thanks} \\ $$

Answered by MJS_new last updated on 12/Jan/21

tan x =t t^2 −3t−1=0 (t−u)(t−v)=0 ⇒ u+v=3∧uv=−1 cos x =(1/( (√(t^2 +1)))) cos x_1 cos x_2 =(1/( (√((u^2 +1)(v^2 +1)))))= =(1/( (√(u^2 v^2 +u^2 +v^2 +1))))=(1/( (√(u^2 v^2 +(u+v)^2 −2uv+1))))= =(1/( (√((−1)^2 +3^2 −2(−1)+1))))=(1/( (√(13))))

$$\mathrm{tan}\:{x}\:={t} \\ $$$${t}^{\mathrm{2}} −\mathrm{3}{t}−\mathrm{1}=\mathrm{0} \\ $$$$\left({t}−{u}\right)\left({t}−{v}\right)=\mathrm{0}\:\Rightarrow\:{u}+{v}=\mathrm{3}\wedge{uv}=−\mathrm{1} \\ $$$$\mathrm{cos}\:{x}\:=\frac{\mathrm{1}}{\:\sqrt{{t}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\mathrm{cos}\:{x}_{\mathrm{1}} \:\mathrm{cos}\:{x}_{\mathrm{2}} \:=\frac{\mathrm{1}}{\:\sqrt{\left({u}^{\mathrm{2}} +\mathrm{1}\right)\left({v}^{\mathrm{2}} +\mathrm{1}\right)}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} {v}^{\mathrm{2}} +{u}^{\mathrm{2}} +{v}^{\mathrm{2}} +\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{{u}^{\mathrm{2}} {v}^{\mathrm{2}} +\left({u}+{v}\right)^{\mathrm{2}} −\mathrm{2}{uv}+\mathrm{1}}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\left(−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} −\mathrm{2}\left(−\mathrm{1}\right)+\mathrm{1}}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{13}}} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com