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Question Number 189201 by Shrinava last updated on 13/Mar/23

$$\mathrm{In}△\mathrm{ABC}\mathrm{holds}:$$$$\sqrt{2}\mathrm{a}\cos \frac{\mathrm{B}}{2}\cos \frac{\mathrm{C}}{2}=\mathrm{s}$$$$\Rightarrow \sec \left(2\mathrm{B}\right)+\tan \left(2\mathrm{B}\right)=\frac{\mathrm{c}+\mathrm{b}}{\mathrm{c}-\mathrm{b}}$$

Answered by som(math1967) last updated on 13/Mar/23

$$\sqrt{2}acos\frac{B}{2}cos\frac{C}{2}=s$$$$\sqrt{2}a\sqrt{\frac{s(s-b)}{ca}}\times \sqrt{\frac{s(s-c)}{ab}}=s$$$$\sqrt{2}a\frac{s}{a}\times \sqrt{\frac{(s-b)(s-c)}{bc}}=s$$$$\sqrt{2}sin\frac{A}{2}=1$$$$\therefore sin\frac{A}{2}=\frac{1}{\sqrt{2}}\Rightarrow \frac{A}{2}=\frac{\pi }{4}$$$$\therefore A=\frac{\pi }{2}\therefore a^2=b^2+c^2$$$$nowsec2B+tan2B$$$$=\frac{1}{cos2B}+\frac{sin2B}{cos2B}$$$$=\frac{1+sin2B}{cos2B}$$$$=\frac{{(cosB+sinB)}^2}{{cos}^{2}B-{sin}^{2}B}$$$$=\frac{cosB+sinB}{cosB-sinB}$$$$=\frac{cosB+sin(\frac{\pi }{2}-C)}{cosB-sin(\frac{\pi }{2}-C)}[A=\frac{\pi }{2}\therefore B+C=\frac{\pi }{2}]$$$$=\frac{cosB+cosC}{cosB-cosC}$$$$=\frac{\frac{c^2+a^2-b^2}{2ca}+\frac{a^2+b^2-c^2}{2ab}}{\frac{c^2+a^2-b^2}{2ca}-\frac{a^2+b^2-c^2}{2ab}}$$$$=\frac{\frac{2c^2}{2ca}+\frac{2b^2}{2ab}}{\frac{2c^2}{2ca}-\frac{2b^2}{2ab}}[a^2=b^2+c^2]$$$$=\frac{c+b}{c-b}$$

Commented by Shrinava last updated on 13/Mar/23

$$\mathrm{thankyou}\mathrm{dearSer}\mathrm{cool}$$

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