Question Number 181592 by mathlove last updated on 27/Nov/22
$${prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{cos}\mathrm{0}\:{cos}\mathrm{1}\:}+\frac{\mathrm{1}}{{cos}\mathrm{1}\:{cos}\mathrm{2}}+......+\frac{\mathrm{1}}{{cos}\mathrm{88}\:{cos}\mathrm{89}}=\frac{{cos}\mathrm{1}}{{sin}^{\mathrm{2}} \mathrm{1}} \\ $$
Answered by som(math1967) last updated on 27/Nov/22
![(1/(sin1))[((sin(1−0))/(cos0cos1))+((sin(2−1))/(cos1cos2))+ ...+((sin(89−88))/(cos88cos89))] =(1/(sin1))[((sin1cos0)/(cos1cos0)) +((sin0cos1)/(cos0cos1)) +((sin2cos1)/(cos2cos1))+((sin1cos2)/(cos1cos2))+...+((sin89sin88)/(cos88cos89))] =(1/(sin1))[tan1−tan0+tan2−tan1 +...+tan88−tan87+tan89−tan88 ] =(1/(sin1))×tan89 =(1/(sin1))×((sin89)/(cos89)) =(1/(sin1))×((cos(90−89))/(sin(90−89))) =(1/(sin1))×((cos1)/(sin1))=((cos1)/(sin^2 1)) [proved]](Q181610.png)
$$\frac{\mathrm{1}}{{sin}\mathrm{1}}\left[\frac{{sin}\left(\mathrm{1}−\mathrm{0}\right)}{{cos}\mathrm{0}{cos}\mathrm{1}}+\frac{{sin}\left(\mathrm{2}−\mathrm{1}\right)}{{cos}\mathrm{1}{cos}\mathrm{2}}+\right. \\ $$$$\left....+\frac{{sin}\left(\mathrm{89}−\mathrm{88}\right)}{{cos}\mathrm{88}{cos}\mathrm{89}}\right] \\ $$$$=\frac{\mathrm{1}}{{sin}\mathrm{1}}\left[\frac{{sin}\mathrm{1}{cos}\mathrm{0}}{{cos}\mathrm{1}{cos}\mathrm{0}}\:+\frac{{sin}\mathrm{0}{cos}\mathrm{1}}{{cos}\mathrm{0}{cos}\mathrm{1}}\right. \\ $$$$\left.+\frac{{sin}\mathrm{2}{cos}\mathrm{1}}{{cos}\mathrm{2}{cos}\mathrm{1}}+\frac{{sin}\mathrm{1}{cos}\mathrm{2}}{{cos}\mathrm{1}{cos}\mathrm{2}}+...+\frac{{sin}\mathrm{89}{sin}\mathrm{88}}{{cos}\mathrm{88}{cos}\mathrm{89}}\right] \\ $$$$=\frac{\mathrm{1}}{{sin}\mathrm{1}}\left[\cancel{{tan}\mathrm{1}}−{tan}\mathrm{0}+\cancel{{tan}\mathrm{2}}−\cancel{{tan}\mathrm{1}}\right. \\ $$$$\left.+...+\cancel{{tan}\mathrm{88}}−\cancel{{tan}\mathrm{87}}+{tan}\mathrm{89}−\cancel{{tan}\mathrm{88}}\:\right] \\ $$$$=\frac{\mathrm{1}}{{sin}\mathrm{1}}×{tan}\mathrm{89} \\ $$$$=\frac{\mathrm{1}}{{sin}\mathrm{1}}×\frac{{sin}\mathrm{89}}{{cos}\mathrm{89}} \\ $$$$=\frac{\mathrm{1}}{{sin}\mathrm{1}}×\frac{{cos}\left(\mathrm{90}−\mathrm{89}\right)}{{sin}\left(\mathrm{90}−\mathrm{89}\right)} \\ $$$$=\frac{\mathrm{1}}{{sin}\mathrm{1}}×\frac{{cos}\mathrm{1}}{{sin}\mathrm{1}}=\frac{{cos}\mathrm{1}}{{sin}^{\mathrm{2}} \mathrm{1}}\:\:\left[{proved}\right] \\ $$
Commented by mathlove last updated on 27/Nov/22
$${thanks} \\ $$