What Two Numbers Adds Up To 75?

Relevant wiki: Sum and Difference Trigonometric Formulas - Problem Solving

$\large \displaystyle \begin{aligned} \tan (75^{\circ}) + \cot (75^{\circ}) &= \tan (75^{\circ}) + \frac{1}{\tan (75^{\circ})}\\ \large \displaystyle &= \frac{\tan ^2 (75^{\circ}) + 1}{\tan (75^{\circ})}\\ \large \displaystyle &= \frac{\sec^2 (75^{\circ})}{\tan (75^{\circ})}\\ \large \displaystyle &= \frac{\cos (75^{\circ})}{\sin (75^{\circ}) \times \cos^2 (75^{\circ})}\\ \large \displaystyle &= \frac{1}{\sin (75^{\circ}) \times \cos (75^{\circ})}\\ \large \displaystyle &= \frac{2}{2 \times \sin (75^{\circ}) \times \cos (75^{\circ})} \\ \large \displaystyle &= \frac{2}{\sin 2(75^{\circ})} \\ \large \displaystyle &= \frac{2}{\sin (150^{\circ})} = \frac{2}{\sin (90+60)}\\ \large \displaystyle &= \frac{2}{\cos 60^{\circ}} = \frac{2}{\frac{1}{2}} = 2 \times 2 = \color{#3D99F6}{\boxed{4}} \end{aligned}.$

$\,$ $\,$ $\large \displaystyle \begin{aligned} \text{Formula Used : } & \sin (2 \theta) = 2 \times \sin \theta \times \cos \theta\\ \large \displaystyle & \tan ^2 \theta + 1 = \sec ^2 \theta\\ \large \displaystyle & \tan \theta = \frac{1}{\cot \theta} \end{aligned}$

$\tan 75^{\circ} + \cot 75^{\circ}\\ =\dfrac{\sin 75^{\circ}}{\cos75^{\circ}} + \dfrac{\cos 75^{\circ}}{\sin 75^{\circ}}\\ =\dfrac{\sin^2 75^{\circ} + \cos^2 75^{\circ}}{\sin 75^{\circ} \cos 75^{\circ}}\\ =\dfrac{1}{\sin 75^{\circ} \cos 75^{\circ}}\\ =\dfrac{2}{2 \sin 75^{\circ}\cos 75^{\circ}}\\ =\dfrac{2}{\sin 150^{\circ}}\\ =\dfrac{2}{\sin 30^{\circ}}\\ =\dfrac{2}{\left(\frac{1}{2}\right)}\\ =\boxed{4}$

$\tan (75^{\circ}) + \cot (75^{\circ})$

$= \tan (75^{\circ}) + \tan (15^{\circ})$

$= \tan (45^{\circ} + 30^{\circ}) + \tan (45^{\circ} - 30^{\circ})$

$= \frac{ \tan (45^{\circ}) + \tan (30^{\circ}) }{ 1 - \tan (45^{\circ}) \tan (30^{\circ}) } + \frac{\tan (45^{\circ})-\tan (30^{\circ})}{1+\tan (45^{\circ})\tan (30^{\circ}) }$

$= \frac{ 1+{\sqrt{3}}/{3} }{ 1-{\sqrt{3}}/{3} } + \frac{ 1-{\sqrt{3}}/{3} }{ 1+{\sqrt{3}}/{3} }$

$= \large \color{#20A900}{\boxed{4}}$

Heres my method. $\tan(75^\circ) + \cot(75^\circ)$

Converting and writing in terms of $\sin$ and $\cos$ we get $\frac{\sin(75 ^\circ )}{\cos(75 ^\circ )} +\frac{\cos(75)}{\sin(75)}$ . Using the property $\sin^2 \theta + \cos^2 \theta =1$ , we get $\frac{1}{(\sin 75 ^\circ \cos 75 ^\circ )} (1)$

Now we make use of property $\sin 2 \theta = 2 \sin \theta \cos \theta$ . Multiplying and dividing equation $1$ by $2$ we get $\frac{2}{2 \times (\sin 75 ^\circ \cos 75 ^\circ )}$ .

Using the above stated property we get $\frac{2}{\sin 150 ^\circ }$ . Now $\sin 150 ^\circ$ can be written as $\sin(180 ^\circ -30 ^\circ )$ . We know by trigonometry that $\sin(180 ^\circ - \theta)=\sin \theta$ . Therefore we get $\frac{2}{\sin 30 ^\circ }$ .

We know that $\sin 30 ^\circ = \frac{1}{2}$ . Therefore the answer is $2 \times 2=4$

Sorry it was little informal. I'm new to posting solutions.