Question Number 68390 by TawaTawa last updated on 10/Sep/19
Answered by som(math1967) last updated on 10/Sep/19
$${tan}\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}\:\right)+{tan}\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} {tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left(\pi−\theta\right)+{tan}\theta \\ $$$$=−{tan}\theta+{tan}\theta=\mathrm{0} \\ $$
Commented by TawaTawa last updated on 10/Sep/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$