You can use the identity $\sin \left(A\+B\right)\=\sin \left(A\right)\cos \left(B\right)\+\cos \left(A\right)\sin \left(B\right)$ to determine the value of sin(75°):

First, note that 45° + 30° = 75°. This is important, as these angles form one of the special triangles:

From this, you can substitute the two angles into the identity for $\sin \left(A\+B\right)$:

$\sin \left(A\+B\right)\=\sin \left(A\right)\cos \left(B\right)\+\cos \left(A\right)\sin \left(B\right)$

$\sin \left(30°\+45°\right)\=\sin \left(30°\right)\cos \left(45°\right)\+\cos \left(30°\right)\sin \left(45°\right)$

$\sin \left(75°\right)\=\left(\frac{1}{2}\right)\left(\frac{1}{\sqrt{2}}\right)\+\left(\frac{\sqrt{3}}{2}\right)\left(\frac{1}{\sqrt{2}}\right)$

$\sin \left(75°\right)\=\frac{\sqrt{2}}{4}\+\frac{\sqrt{6}}{4}$

$\sin \left(75°\right)\=\frac{\sqrt{2}\+\sqrt{6}}{4}$

First, note that:

$\tan \left(A\+B\right)\=\frac{\sin \left(A\+B\right)}{\cos \left(A\+B\right)}$

Keeping this in mind, you can substitute the trig identities for $\sin \left(A\+B\right)$ and $\cos \left(A\+B\right)$ and get:

$\frac{\sin \left(A\right)\cos \left(B\right)\+\cos \left(A\right)\sin \left(B\right)}{\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)}$

Now, divide every term by $\cos \left(A\right)\cos \left(B\right)$:

$\frac{\frac{\sin \left(A\right)\cos \left(B\right)}{\cos \left(A\right)\cos \left(B\right)}\+\frac{\cos \left(A\right)\sin \left(B\right)}{\cos \left(A\right)\cos \left(B\right)}}{\frac{\cos \left(A\right)\cos \left(B\right)}{\cos \left(A\right)\cos \left(B\right)}-\frac{\sin \left(A\right)\sin \left(B\right)}{\cos \left(A\right)\cos \left(B\right)}}$

Simplify this expression as follows:

$\frac{\frac{\sin \left(A\right)}{\cos \left(A\right)}\+\frac{\sin \left(B\right)}{\cos \left(B\right)}}{1-\frac{\sin \left(A\right)\sin \left(B\right)}{\cos \left(A\right)\cos \left(B\right)}}$

Finally, you can simplify this expression further by recalling that $\tan \left(Q\right)\=\frac{\sin \left(Q\right)}{\cos \left(Q\right)}$:

$\frac{\tan \left(A\right)\+\tan \left(B\right)}{1-\tan \left(A\right)\tan \left(B\right)}$