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

$$\mathrm{If}\tan {}^2\mathrm{x}-3\tan \mathrm{x}=1\mathrm{has}\mathrm{the}\mathrm{roots}$$$$\mathrm{are}{\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}$$$$\mathrm{of}\mid \cos {\mathrm{x}}_1.\cos {\mathrm{x}}_2\mid .$$

Commented by liberty last updated on 12/Jan/21

$$\tan {}^2\mathrm{x}-3\tan \mathrm{x}-1=0$$$$\tan \mathrm{x}=\frac{3\pm \sqrt{13}}{2}\to \{\begin{array}{l}\tan \mathrm{x}=\frac{3+\sqrt{13}}{2}\\ \tan \mathrm{x}=\frac{3-\sqrt{13}}{2}\end{array}$$$$\{\begin{array}{l}\cos \mathrm{x}=\frac{2}{\sqrt{26+6\sqrt{13}}}\\ \cos \mathrm{x}=\frac{2}{\sqrt{26-6\sqrt{13}}}\end{array}\mathrm{then}\mid \cos {\mathrm{x}}_1.\cos {\mathrm{x}}_2\mid =\frac{4}{\sqrt{676-468}}=\frac{4}{4\sqrt{13}}=\frac{1}{\sqrt{13}}$$$$$$

Answered by Lordose last updated on 12/Jan/21

$${\tan }^2\mathrm{x}-3\mathrm{tanx}+{(-\frac{3}{2})}^2=\frac{13}{4}$$$${(\mathrm{tanx}-\frac{3}{2})}^2=\frac{13}{4}\Rightarrow \tan \left(\mathrm{x}\right)=\frac{3\pm \sqrt{13}}{2}$$$${\mathrm{x}}_1={\tan }^{-1}\left(\frac{3+\sqrt{13}}{2}\right)\Rightarrow \cos \left({\mathrm{x}}_1\right)=\frac{2}{\sqrt{26+6\sqrt{13}}}$$$${\mathrm{x}}_2={\tan }^{-1}\left(\frac{3-\sqrt{13}}{2}\right)\Rightarrow \cos \left({\mathrm{x}}_2\right)=\frac{2}{\sqrt{26-6\sqrt{13}}}$$$$\mathrm{Hence},\mid \cos \left({\mathrm{x}}_1\right)\cos \left({\mathrm{x}}_2\right)\mid =\frac{1}{\sqrt{13}}$$$$$$$$$$

Commented by liberty last updated on 12/Jan/21

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

Commented by Lordose last updated on 12/Jan/21

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

Answered by MJS_new last updated on 12/Jan/21

$$\tan x=t$$$$t^2-3t-1=0$$$$(t-u)(t-v)=0\Rightarrow u+v=3\wedge uv=-1$$$$\cos x=\frac{1}{\sqrt{t^2+1}}$$$$\cos x_1\cos x_2=\frac{1}{\sqrt{(u^2+1)(v^2+1)}}=$$$$=\frac{1}{\sqrt{u^2{v}^2+u^2+v^2+1}}=\frac{1}{\sqrt{u^2{v}^2+{(u+v)}^2-2uv+1}}=$$$$=\frac{1}{\sqrt{{(-1)}^2+3^2-2(-1)+1}}=\frac{1}{\sqrt{13}}$$

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