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-

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

If \tan^{2} x - 3 \tan x = 1 has the roots

are x_{1} and x_{2} then find the value

of ∣ \cos x_{1} . \cos x_{2} ∣ .

Commented by liberty last updated on 12/Jan/21

\tan^{2} x - 3 \tan x - 1 = 0

\tan x = \frac{3 \pm \sqrt{13}}{2} \to {\begin{cases} \tan x = \frac{3 + \sqrt{13}}{2} \\ \tan x = \frac{3 - \sqrt{13}}{2} \end{cases}

{\begin{cases} \cos x = \frac{2}{\sqrt{26 + 6 \sqrt{13}}} \\ \cos x = \frac{2}{\sqrt{26 - 6 \sqrt{13}}} \end{cases} then ∣ \cos x_{1} . \cos x_{2} ∣= \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} x - 3 tanx + {(- \frac{3}{2})}^{2} = \frac{13}{4}

{(tanx - \frac{3}{2})}^{2} = \frac{13}{4} \Rightarrow \tan (x) = \frac{3 \pm \sqrt{13}}{2}

x_{1} = \tan^{- 1} (\frac{3 + \sqrt{13}}{2}) \Rightarrow \cos (x_{1}) = \frac{2}{\sqrt{26 + 6 \sqrt{13}}}

x_{2} = \tan^{- 1} (\frac{3 - \sqrt{13}}{2}) \Rightarrow \cos (x_{2}) = \frac{2}{\sqrt{26 - 6 \sqrt{13}}}

Hence, ∣ \cos (x_{1}) \cos (x_{2}) ∣ = \frac{1}{\sqrt{13}}

Commented by liberty last updated on 12/Jan/21

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

Commented by Lordose last updated on 12/Jan/21

you are right

I have found an error in my solution

Thanks

Answered by MJS_new last updated on 12/Jan/21

\tan x = t

t^{2} - 3 t - 1 = 0

(t - u) (t - v) = 0 \Rightarrow u + v = 3 \land u v = - 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} - 2 u v + 1}} =

= \frac{1}{\sqrt{{(- 1)}^{2} + 3^{2} - 2 (- 1) + 1}} = \frac{1}{\sqrt{13}}

Facebook Tweet Pin