Using the formula :

$\cos \left(\mathrm{A}+\mathrm{B}\right)=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\mathrm{sinB},$ prove that the value of $\cos {75}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}.$

$\left(\mathbf{Hint}\mathbf{:}\mathrm{Take}\mathrm{A}={45}^{\circ },\mathrm{B}={30}^{\circ }\right).$

Step 1:-Verify that L.H.S and R.H.S

$\cos \left(\mathrm{A}+\mathrm{B}\right)=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\mathrm{sinB},$

$$\begin{gathered} \angle \mathrm{A}={45}^{\circ } \\ \mathrm{and}\angle \mathrm{B}={30}^{\circ } \end{gathered}$$

Step 2:- Prove that, $\cos {75}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$

$\therefore \cos \left(\mathrm{A}+\mathrm{B}\right)=\cos \mathrm{A}\cos \mathrm{B}-\sin \mathrm{A}\mathrm{sinB},$

$\cos \left({45}^{\circ }+{30}^{\circ }\right)=\cos {45}^{\circ }\cos {30}^{\circ }-\sin {45}^{\circ }\sin {30}^{\circ }$

$\cos {75}^{\circ }=\left(\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}\right)-\left(\frac{1}{\sqrt{2}}\times \frac{1}{2}\right)$

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

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

Hence, $\cos {75}^{\circ }=\frac{\sqrt{3}-1}{2\sqrt{2}}$proved.