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Question Number 13194 by Tinkutara last updated on 16/May/17

Answered by ajfour last updated on 16/May/17

$$\left(i\right)$$$$letx=\tan \alpha ,y=\tan \beta ,z=\tan \gamma$$$$\tan (\alpha +\beta +\gamma )=\frac{\mathrm{\Sigma }\tan \alpha -\mathrm{\Pi }\tan \alpha }{1-\mathrm{\Sigma }\tan \alpha \tan \beta }$$$$x+y+z=xyz....\left(given\right)$$$$\Rightarrow \mathrm{\Sigma }\tan \alpha =\mathrm{\Pi }\tan \alpha$$$$so,\tan (\alpha +\beta +\gamma )=0$$$$\Rightarrow \alpha +\beta +\gamma =n\pi$$$$\Rightarrow 2\alpha +2\beta +2\gamma =2n\pi$$$$\tan (2\alpha +2\beta +2\gamma )=0$$$$\Rightarrow \mathrm{\Sigma }\tan \left(2\alpha \right)=\mathrm{\Pi }\tan \left(2\alpha \right)$$$$or,$$$$\mathrm{\Sigma }\frac{2\tan \alpha }{1-\tan {}^2\alpha }=\mathrm{\Pi }\frac{2\tan \alpha }{1-\tan {}^2\alpha }$$$$or$$$$\frac{2x}{1-x^2}+\frac{2y}{1-y^2}+\frac{2z}{1-z^2}=\frac{8xyz}{(1-x^2)(1-y^2)(1-z^2)}$$$$$$$$\left(ii\right)$$$$\tan (3\alpha +3\beta +3\gamma )=0$$$$\Rightarrow \mathrm{\Sigma }\tan \left(3\alpha \right)=\mathrm{\Pi }\tan \left(3\alpha \right)$$$$\mathrm{\Sigma }\frac{3x-x^2}{1-3x^2}=\mathrm{\Pi }\frac{3x-x^2}{1-3x^2}.$$

Commented by Tinkutara last updated on 16/May/17

$$\mathrm{Thanks}\mathrm{a}\mathrm{lot}!$$

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