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Question Number 187891 by Rupesh123 last updated on 23/Feb/23

Answered by mr W last updated on 24/Feb/23

Commented by mr W last updated on 25/Feb/23

$$\mathrm{\Delta }=areaof\mathrm{\Delta }ABC$$$$\frac{\sin (\beta +C)}{\sin \beta }=\frac{a}{\frac{b}{2}}=\frac{2a}{b}$$$$\cos C+\frac{\sin C}{\tan \beta }=\frac{2a}{b}$$$$\frac{a^2+b^2-c^2}{2ab}+\frac{2\mathrm{\Delta }}{ab\tan \beta }=\frac{2a}{b}$$$$\tan \beta =\frac{4\mathrm{\Delta }}{3a^2-b^2+c^2}$$$$similarly$$$$\tan \gamma =\frac{4\mathrm{\Delta }}{3a^2-c^2+b^2}$$$$\theta =\pi -(\beta +\gamma )$$$$\tan \theta =-\tan (\beta +\gamma )=-\frac{\tan \beta +\tan \gamma }{1-\tan \beta \tan \gamma }$$$$=\frac{\frac{4\mathrm{\Delta }}{3a^2-b^2+c^2}+\frac{4\mathrm{\Delta }}{3a^2-c^2+b^2}}{\frac{4\mathrm{\Delta }}{3a^2-b^2+c^2}\times \frac{4\mathrm{\Delta }}{3a^2-c^2+b^2}-1}$$$$=\frac{6a^2}{16{\mathrm{\Delta }}^2-(3a^2-b^2+c^2)(3a^2-c^2+b^2)}\times 4\mathrm{\Delta }$$$$=\frac{24a^2\mathrm{\Delta }}{16{\mathrm{\Delta }}^2-{\left(3a^2\right)}^2+{(b^2-c^2)}^2}$$$$=\frac{24a^2\mathrm{\Delta }}{2(a^2{b}^2+b^2{c}^2+c^2{a}^2)-(a^4+b^4+c^4)-9a^4+b^4+c^4-2b^2{c}^2}$$$$=\frac{12a^2\mathrm{\Delta }}{a^2{b}^2+c^2{a}^2-5a^4}$$$$=\frac{12\mathrm{\Delta }}{b^2+c^2-5a^2}✓$$

Commented by Rupesh123 last updated on 26/Feb/23

Excellent, sir!

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