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Question Number 111382 by john santu last updated on 03/Sep/20

Commented by kaivan.ahmadi last updated on 03/Sep/20

$${sec}^{2}x=12+secx\Rightarrow {sec}^{2}x-secx-12=0\Rightarrow$$$$(secx-4)(secx+3)=0\Rightarrow secx=4,secx=-3$$$$byhypothesissecx>0sowehavesecx=4$$$$\Rightarrow cosx=\frac{1}{4}=\frac{1-{tan}^2\left(\frac{x}{2}\right)}{1+{tan}^2\left(\frac{x}{2}\right)}\Rightarrow$$$$1+{tan}^2\left(\frac{x}{2}\right)=4-4{tan}^2\left(\frac{x}{2}\right)\Rightarrow$$$$5{tan}^2\left(\frac{x}{2}\right)=3\Rightarrow {tan}^2\left(\frac{x}{2}\right)=\frac{3}{5}$$$$$$

Answered by bobhans last updated on 03/Sep/20

$$\mathrm{let}\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{...}}}}}=\mathrm{♭}$$$$\Rightarrow 12+\mathrm{♭}={\mathrm{♭}}^2;{\mathrm{♭}}^2-\mathrm{♭}-12=0,\mathrm{♭}=4$$$$\Rightarrow \sec \mathrm{x}=4,\cos \mathrm{x}=\frac{1}{4},2\cos {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5}{4}$$$$\cos {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{5}{8}\to \sin {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{3}{8}$$$$\mathrm{so}\tan {}^2\left(\frac{\mathrm{x}}{2}\right)=\frac{3}{8}\times \frac{8}{5}=\frac{3}{5}$$

Commented by peter frank last updated on 03/Sep/20

$$thankyouboth$$

Answered by Dwaipayan Shikari last updated on 03/Sep/20

$$\sqrt{12+\sqrt{12+\sqrt{12+....}}}=a$$$$12+a=a^2$$$$a^2-a-12=0\Rightarrow a=4$$$$secx=4$$$$cosx=\frac{1}{4}$$$$\frac{1-{tan}^2\left(\frac{x}{2}\right)}{1+{tan}^2\left(\frac{x}{2}\right)}=\frac{1}{4}$$$${tan}^2\left(\frac{x}{2}\right)=\frac{3}{5}$$

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