Question

Given that $\cos (A-B)=\cos A\cos B+\sin A\sin B$, the value of $\cos {15}^{\circ }=\frac{\sqrt{3}+a}{b\sqrt{2}}$. Then the value of $b-a$ is:

Answer 1

Putting $A={45}^{\circ }$ and $B={30}^{\circ }$ we get

$\cos ({45}^{\circ }-{30}^{\circ })=\cos {45}^{\circ }\cos {30}^{\circ }+\sin {45}^{\circ }\sin {30}^{\circ }$
$\Rightarrow \cos {15}^{\circ }=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}$
$\Rightarrow \cos {15}^{\circ }=\frac{\sqrt{3}+1}{2\sqrt{2}}$

Hence, $a=1,b=2$

$\Rightarrow b-a=2-1=1$