Question Number 131230 by Fikret last updated on 02/Feb/21
![((sin5)/(cos0.cos5))+((sin5)/(cos5.cos10))+ . . . +((sin5)/(cos55.cos60)) =?](https://www.tinkutara.com/question/Q131230.png)
$$\frac{{sin}\mathrm{5}}{{cos}\mathrm{0}.{cos}\mathrm{5}}+\frac{{sin}\mathrm{5}}{{cos}\mathrm{5}.{cos}\mathrm{10}}+\:\:.\:\:.\:\:.\:\:+\frac{{sin}\mathrm{5}}{{cos}\mathrm{55}.{cos}\mathrm{60}}\:=? \\ $$
Answered by mr W last updated on 02/Feb/21
![example: sin 5=sin (10−5)=sin 10 cos 5−cos 10 sin 5 ((sin 5)/(cos 5 cos 10))=tan 10−tan 5 ((sin5)/(cos0.cos5))+((sin5)/(cos5.cos10))+ . . . +((sin5)/(cos55.cos60)) =(tan 5−tan 0)+(tan 10−tan 5)+...+(tan 60−tan 55) =tan 60−tan 0 =tan 60 =(√3)](https://www.tinkutara.com/question/Q131234.png)
$${example}: \\ $$$$\mathrm{sin}\:\mathrm{5}=\mathrm{sin}\:\left(\mathrm{10}−\mathrm{5}\right)=\mathrm{sin}\:\mathrm{10}\:\mathrm{cos}\:\mathrm{5}−\mathrm{cos}\:\mathrm{10}\:\mathrm{sin}\:\mathrm{5} \\ $$$$\frac{\mathrm{sin}\:\mathrm{5}}{\mathrm{cos}\:\mathrm{5}\:\mathrm{cos}\:\mathrm{10}}=\mathrm{tan}\:\mathrm{10}−\mathrm{tan}\:\mathrm{5} \\ $$$$ \\ $$$$\frac{{sin}\mathrm{5}}{{cos}\mathrm{0}.{cos}\mathrm{5}}+\frac{{sin}\mathrm{5}}{{cos}\mathrm{5}.{cos}\mathrm{10}}+\:\:.\:\:.\:\:.\:\:+\frac{{sin}\mathrm{5}}{{cos}\mathrm{55}.{cos}\mathrm{60}} \\ $$$$=\left(\mathrm{tan}\:\mathrm{5}−\mathrm{tan}\:\mathrm{0}\right)+\left(\mathrm{tan}\:\mathrm{10}−\mathrm{tan}\:\mathrm{5}\right)+…+\left(\mathrm{tan}\:\mathrm{60}−\mathrm{tan}\:\mathrm{55}\right) \\ $$$$=\mathrm{tan}\:\mathrm{60}−\mathrm{tan}\:\mathrm{0} \\ $$$$=\mathrm{tan}\:\mathrm{60} \\ $$$$=\sqrt{\mathrm{3}} \\ $$
Commented by malwan last updated on 02/Feb/21
![great sir thank you](https://www.tinkutara.com/question/Q131240.png)
$${great}\:{sir} \\ $$$${thank}\:{you}\: \\ $$
Commented by PRITHWISH SEN 2 last updated on 03/Feb/21
![Beautiful as always !](https://www.tinkutara.com/question/Q131267.png)
$$\boldsymbol{\mathrm{Beautiful}}\:\boldsymbol{\mathrm{as}}\:\boldsymbol{\mathrm{always}}\:! \\ $$