Question Number 128010 by Study last updated on 03/Jan/21
Commented by Study last updated on 03/Jan/21
$${help}\:{me}\:{and}\:{show}\:{practice} \\ $$
Commented by som(math1967) last updated on 03/Jan/21
$$\frac{\mathrm{1}}{{x}} \\ $$
Commented by som(math1967) last updated on 03/Jan/21
$$\frac{\mathrm{1}}{{x}}\left\{\frac{{x}}{{cosocos}\mathrm{1}}+\frac{{x}}{{cos}\mathrm{1}{cos}\mathrm{2}}+…+\frac{{x}}{{cos}\mathrm{44}{cos}\mathrm{45}}\right\} \\ $$$$=\frac{\mathrm{1}}{{x}}\left\{\frac{{sin}\left(\mathrm{1}−\mathrm{0}\right)}{{cos}\mathrm{0}{cos}\mathrm{1}}+\frac{{sin}\left(\mathrm{2}−\mathrm{1}\right)}{{cos}\mathrm{1}{cos}\mathrm{2}}+..\frac{{sin}\left(\mathrm{45}−\mathrm{44}\right)}{{cos}\mathrm{44}{cos}\mathrm{45}}\right\} \\ $$$$=\frac{\mathrm{1}}{{x}}\left\{\frac{{sin}\mathrm{1}}{{cos}\mathrm{1}}−\frac{{sin}\mathrm{0}}{{cos}\mathrm{0}}+\frac{{sin}\mathrm{2}}{{cos}\mathrm{2}}−\frac{{sin}\mathrm{1}}{{cos}\mathrm{1}}\right. \\ $$$$\left.\:\:\:\:\:\:\:……+\frac{{sin}\mathrm{45}}{{cos}\mathrm{45}}−\frac{{sin}\mathrm{44}}{{cos}\mathrm{44}}\right\} \\ $$$$\frac{\mathrm{1}}{{x}}×\mathrm{1}=\frac{\mathrm{1}}{{x}} \\ $$
Commented by Study last updated on 03/Jan/21
$${good}!!\:\:\:{thanks}\:{sir} \\ $$