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Question Number 129181 by liberty last updated on 13/Jan/21

$$\mathrm{Given}\tan {}^2\mathrm{x}-5\tan \mathrm{x}=3\mathrm{has}\mathrm{the}\mathrm{roots}$$$$\mathrm{are}{\mathrm{x}}_1\mathrm{and}{\mathrm{x}}_2.\mathrm{Find}\mid \cos {\mathrm{x}}_1.\cos {\mathrm{x}}_2\mid .$$

Commented by MJS_new last updated on 13/Jan/21

$$\mathrm{see}\mathrm{my}\mathrm{solution}\mathrm{to}\mathrm{question}128995$$

Answered by bramlexs22 last updated on 13/Jan/21

$$\mathrm{Consider}\mathrm{the}\mathrm{sum}\mathrm{and}\mathrm{product}\mathrm{of}$$$$\mathrm{the}\mathrm{equation}\tan {}^2\mathrm{x}-5\tan \mathrm{x}-3=0$$$$\{\begin{array}{l}\tan {\mathrm{x}}_1+\tan {\mathrm{x}}_2=5\\ \tan {\mathrm{x}}_1.\tan {\mathrm{x}}_2=-3\end{array}$$$$\mathrm{we}\mathrm{get}\tan ({\mathrm{x}}_1+{\mathrm{x}}_2)=\frac{5}{1-(-3)}=\frac{5}{4}$$$$\mathrm{and}\sin ({\mathrm{x}}_1+{\mathrm{x}}_2)=\pm \frac{5}{\sqrt{41}}$$$$\mathrm{from}\tan {\mathrm{x}}_1+\tan {\mathrm{x}}_2=5\mathrm{we}\mathrm{get}$$$$5\cos {\mathrm{x}}_1.\cos {\mathrm{x}}_2=\sin ({\mathrm{x}}_1+{\mathrm{x}}_2)$$$$\mid \cos {\mathrm{x}}_1.\cos {\mathrm{x}}_2\mid =\frac{1}{\sqrt{41}}$$$$$$

Commented by liberty last updated on 13/Jan/21

$$\mathrm{nice}...$$

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