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Question Number 176785 by Ar Brandon last updated on 26/Sep/22

$$\frac{6}{1+c^x}+\frac{1}{1+c^{-x}}=y$$$$\mathrm{Express}\frac{11}{1+c^x}+\frac{1}{1+c^{-x}}\mathrm{in}\mathrm{terms}\mathrm{of}y$$

Commented by Ar Brandon last updated on 26/Sep/22

$$\mathrm{Ans}.2y-1$$

Answered by a.lgnaoui last updated on 26/Sep/22

$$y=6\frac{1}{1+c^x}+\frac{1}{1+\frac{1}{c^x}}=6\frac{1}{1+c^x}+\frac{c^x}{1+c^x}$$$$y=6\frac{1}{1+c^x}+\frac{1+c^x-1}{1+c^x}=5\frac{1}{1+c^x}+1$$$$\frac{11}{1+c^x}+\frac{1}{1+c^{-x}}=\frac{11}{1+c^x}+1-\frac{1}{1+c^x}=10\frac{1}{1+c^x}+1$$$$\frac{11}{1+c^x}+\frac{1}{1+c^{-x}}=2y-1$$

Commented by Ar Brandon last updated on 26/Sep/22

Thank you, Sir!

Answered by mr W last updated on 26/Sep/22

$$\frac{6+c^x}{1+c^x}=y$$$$\frac{5}{1+c^x}=y-1$$$$$$$$\frac{11}{1+c^x}+\frac{1}{1+c^{-x}}$$$$=\frac{6}{1+c^x}+\frac{1}{1+c^{-x}}+\frac{5}{1+c^x}$$$$=y+y-1$$$$=2y-1$$

Commented by Tawa11 last updated on 26/Sep/22

$$\mathrm{Great}\mathrm{sirs}$$

Commented by Ar Brandon last updated on 26/Sep/22

Thank you, Sir!

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