Algebra Questions

Question Number 147933 by mathdanisur last updated on 24/Jul/21

$${Simplify} \\$$ $$\left(\mathrm{1}+\:\frac{\mathrm{1}}{{cos}\mathrm{2}{x}}\right)\centerdot\left(\mathrm{1}+\frac{\mathrm{1}}{{cos}\mathrm{4}{x}}\right)\centerdot\left(\mathrm{1}+\frac{\mathrm{1}}{{cos}\mathrm{8}{x}}\right)=? \\$$

Answered by som(math1967) last updated on 24/Jul/21

$$\frac{{tanx}}{{tanx}}\left(\mathrm{1}+\frac{\mathrm{1}+{tan}^{\mathrm{2}} {x}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\right)\left(\mathrm{1}+\frac{\mathrm{1}+{tan}^{\mathrm{2}} \mathrm{2}{x}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{2}{x}}\right) \\$$ $$\:\:\:\:\:\:\:\:\:\:\:×\left(\mathrm{1}+\frac{\mathrm{1}+{tan}^{\mathrm{2}} \mathrm{4}{x}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{4}{x}}\right) \\$$ $$=\frac{\mathrm{1}}{{tanx}}\left(\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\right)\left(\frac{\mathrm{2}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{2}{x}}\right)\left(\frac{\mathrm{2}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{4}{x}}\right) \\$$ $$=\frac{\mathrm{1}}{{tanx}}×\frac{\mathrm{2}{tan}\mathrm{2}{x}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{2}{x}}×\frac{\mathrm{2}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{4}{x}} \\$$ $$=\frac{\mathrm{1}}{{tanx}}×\frac{\mathrm{2}{tan}\mathrm{4}{x}}{\mathrm{1}−{tan}^{\mathrm{2}} \mathrm{4}{x}}=\frac{{tan}\mathrm{8}{x}}{{tanx}} \\$$ $$\left[\because{tan}\mathrm{2}{x}=\frac{\mathrm{2}{tanx}}{\mathrm{1}−{tan}^{\mathrm{2}} {x}}\right] \\$$

Commented bymathdanisur last updated on 24/Jul/21

$${Thank}\:{you}\:{Ser} \\$$