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  • Let and <img alt="\tan \gamma ; \alpha ;\beta ;\gamma \neq \frac{\left ( 2n-1 \right )\pi }{2},n\; \epsilon \; N" src="/latex-image/?%5Ctan%20%5Cgamma%20%

Let \tan \alpha ,\tan \beta and \tan \gamma ; \alpha ;\beta ;\gamma \neq \frac{\left ( 2n-1 \right )\pi }{2},n\; \epsilon \; N be the slopes of three line segments OA, OB and OC, respectively, where O is origin.If circumcentre of \Delta ABC coincides with origin and its orthocentre lies on y-axis, then the value of \left ( \frac{\cos 3\alpha +\cos 3\beta +\cos 3\gamma }{\cos \alpha \cos \beta \cos \gamma } \right )^{2} is equal to ________.
Option: 1 81
Option: 2 64
Option: 3 144
Option: 4 256

Answers (1)

best_answer

\begin{aligned} &\text { Let } \mathrm{A}\left(\mathrm{x}_{1}, \mathrm{x}_{1} \tan \alpha\right), \mathrm{B}\left(\mathrm{x}_{2}, \mathrm{x}_{2} \tan \beta\right) \text { and } \mathrm{C}\left(\mathrm{x}_{3}, \mathrm{x}_{3} \tan \gamma\right)\\ &\text { circumcentre is } \mathrm{O}(0,0) \end{aligned}

Since orthocentre and circumcentre, both lies on the y-axis
⇒ Centroid also lies on y -axis

\\\text { also } \mathrm{OA}=\mathrm{OB}=\mathrm{OC}=\mathrm{R} \\ \mathrm{x}_{1} \sec \alpha=\mathrm{x}_{2} \sec \beta=\mathrm{x}_{3} \sec \gamma=\mathrm{R} \\ \mathrm{x}_{1}+\mathrm{x}_{2}+\mathrm{x}_{3}=0 \Rightarrow \mathrm{R}(\cos \alpha+\cos \beta+\cos \gamma)=0 \\ \Rightarrow \cos \alpha+\cos \beta+\cos \gamma=0 \\ \Rightarrow \cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma=3 \cos \alpha \cos \beta \cos \gamma

\\ \left(\frac{\cos 3 \alpha+\cos \beta+\cos 3 \gamma}{\cos \alpha \cos \beta \cos \gamma}\right)^{2}\\=\left\{\frac{\left.4\left(\cos ^{3} \alpha+\cos ^{3} \beta+\cos ^{3} \gamma\right)-3(\cos \alpha+\cos \beta+\cos \gamma)\right)}{\cos \alpha \cos \beta \cos \gamma}\right\} \\ =\left(\frac{4 \times 3 \cos \alpha \cos \beta \cos \gamma}{\cos \alpha \cos \beta \cos \gamma}\right)^{2}=144

Posted by

himanshu.meshram

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