Question

$\left(i\right)$Assume that $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$. Find $\tan {75}^{\circ }$ when $A={45}^{\circ },B={30}^{\circ }$
$\left(ii\right)$ Assume $\cos (A-B)=\cos A\cos B+\sin A\sin B$. Find $\cos {15}^{\circ }$ when $A={45}^{\circ },B={30}^{\circ }$

• A. $\left(i\right)\frac{\sqrt{3}+1}{\sqrt{3}-1}\left(ii\right)\frac{\sqrt{3}+1}{2\sqrt{2}}$
• B. $\left(i\right)\frac{\sqrt{2}+1}{\sqrt{3}-1}\left(ii\right)\frac{\sqrt{2}+1}{2\sqrt{2}}$
• C. $\left(i\right)\frac{\sqrt{3}-1}{\sqrt{3}+1}\left(ii\right)\frac{\sqrt{3}+1}{2\sqrt{2}}$
• D. $\left(i\right)\frac{\sqrt{3}+1}{\sqrt{3}-1}\left(ii\right)\frac{\sqrt{3}+1}{\sqrt{2}}$

Answer 1

The correct option is A $\left(i\right)\frac{\sqrt{3}+1}{\sqrt{3}-1}\left(ii\right)\frac{\sqrt{3}+1}{2\sqrt{2}}$

$tan{75}^0$
$=tan({45}^0+{30}^0)$
$=\frac{tan{45}^0+tan{30}^0}{1-tan{45}^{0}tan{30}^0}$
$=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$
$=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Now
$cos\left({15}^0\right)$
$=cos({45}^0-{30}^0)$
$=cos{45}^{0}sin{30}^0+sin{45}^{0}cos{30}^0$
$=\frac{1}{2\sqrt{2}}+\frac{\sqrt{3}}{2\sqrt{2}}$
$=\frac{\sqrt{3}+1}{2\sqrt{2}}$