Question

These are numeric entry type questions.
If $\frac{\cos 18^{\circ }+\sin 18^{\circ }}{\cos 18^{\circ }-\sin 18^{\circ }}=\tan A$, then $A=$____

Answer 1

Answer: 63

$\frac{\cos 18^{\circ }+\sin 18^{\circ }}{\cos 18^{\circ }-\sin 18^{\circ }}=\tan A$
$\frac{\cos (63-45)+\sin (63-45)}{\cos (63-45)-\sin (63-45)}=\tan A$
$=\frac{\cos 63\cos 45+\sin 63\sin 45+\sin 63\cos 45-\cos 63\sin 45}{\cos 63\cos 45+\sin 63\sin 45-\sin 63\cos 45+\cos 63\sin 45}$
$=\frac{\frac{\cos 63}{\sqrt{2}}+\frac{\sin 63}{\sqrt{2}}+\frac{\sin 63}{\sqrt{2}}-\frac{\cos 63}{\sqrt{2}}}{\sqrt{2}}+\frac{\sin 63}{\sqrt{2}}-\frac{\sin 63}{\sqrt{2}}+\frac{\cos 63}{\sqrt{2}}$
$=\frac{2/}{\sqrt{2}}\sin 63$
$\frac{2}{\sqrt{2}}\cos 63$
$\Rightarrow A=63^{\circ }$