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Question Number 128336 by liberty last updated on 06/Jan/21

$$\mathrm{If}\frac{\sec \mathrm{x}+\tan \mathrm{x}}{\sec \mathrm{x}-\tan \mathrm{x}}=5\mathrm{then}\mathrm{what}\mathrm{is}\mathrm{the}$$$$\mathrm{value}\mathrm{of}\frac{3\cos 2\mathrm{x}+1}{3\cos 2\mathrm{x}-1}?$$

Answered by bramlexs22 last updated on 06/Jan/21

$$\Rightarrow \frac{\sec x+\tan x}{\sec x-\tan x}=5$$$$\Rightarrow \frac{\cos x(\sec x+\tan x)}{\cos x(\sec x-\tan x)}=5$$$$\Rightarrow \frac{1+\sin x}{1-\sin x}=5;1+\sin x=5-5\sin x$$$$\Rightarrow 6\sin x=4;\sin x=\frac{2}{3}$$$$then\frac{3\cos 2x+1}{3\cos 2x-1}=\frac{3(1-2\sin {}^{2}x)+1}{3(1-2\sin {}^{2}x)-1}$$$$\Rightarrow \frac{4-2\left(\frac{4}{9}\right)}{2-2\left(\frac{4}{9}\right)}=\frac{2-\frac{4}{9}}{1-\frac{4}{9}}=\frac{14}{5}.$$

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