Question

$\tan {\left(7\frac{1}{2}\right)}^{°}=?$

Answer 1

$\tan {\left(7\frac{1}{2}\right)}^{°}=\frac{\sin {\left(7\frac{1}{2}\right)}^{°}}{\cos {\left(7\frac{1}{2}\right)}^{°}}$

$=\frac{{\sin }^2{\left(7\frac{1}{2}\right)}^{°}}{\sin {\left(7\frac{1}{2}\right)}^{°}\cos {\left(7\frac{1}{2}\right)}^{°}}$

$=\frac{2{\sin }^2{\left(7\frac{1}{2}\right)}^{°}}{2\sin {\left(7\frac{1}{2}\right)}^{°}\cos {\left(7\frac{1}{2}\right)}^{°}}$

$=\frac{1-\cos 15°}{\sin 15°}$ $[\because 1-\cos 2A=2{\sin }^{2}A,2\sin A\cos A=\sin 2A]$

$=\frac{1-\cos (45°-30°)}{\sin \left(45°-30°\right)}$

$=\frac{1-\cos 45°\cos 30°-\sin 45°\sin 30°}{\sin 45°\cos 30°-\cos 45°\sin 30°}$

$=\frac{1-{\displaystyle \frac{1}{\sqrt{2}}}\times {\displaystyle \frac{\sqrt{3}}{2}}-\frac{1}{\sqrt{2}}\times {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{\sqrt{2}}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2}}$

$=\frac{1-\left({\displaystyle \frac{\sqrt{3}+1}{2\sqrt{2}}}\right)}{{\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}}}$

$=\frac{{\displaystyle \frac{2\sqrt{2}-\sqrt{3}-1}{2\sqrt{2}}}}{{\displaystyle \frac{\sqrt{3}-1}{2\sqrt{2}}}}$

$=\frac{2\sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1}$

$=\frac{2\sqrt{2}-\sqrt{3}-1}{\sqrt{3}-1}\times \frac{\sqrt{3}+1}{\sqrt{3}+1}$

$=\frac{2\sqrt{6}-3-\sqrt{3}+2\sqrt{2}-\sqrt{3}-1}{{\left(\sqrt{3}\right)}^2-{\left(1\right)}^2}$

$=\frac{2\sqrt{6}-2\sqrt{3}+2\sqrt{2}-4}{3-1}$

$=\frac{2\left(\sqrt{6}-\sqrt{3}+\sqrt{2}-2\right)}{2}$

$=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

Hence, $\tan {\left(7\frac{1}{2}\right)}^{°}=\sqrt{6}-\sqrt{3}+\sqrt{2}-2$