Question

Prove that $\frac{\cos 9^0+\sin 9^0}{\cos 9^0-\sin 9^0}=\cot {36}^0$

Answer 1

LHS $=\frac{(\cos 9^{\circ }+\sin 9^{\circ })}{(\cos 9^{\circ }-\sin 9^{\circ })}$

Divide by Nr and Dr by $\cos 9^{\circ }$

LHS $=\frac{1+\tan 9^{\circ }}{1-\tan 9^{\circ }}$

$=\frac{\tan {45}^{\circ }+\tan 9^{\circ }}{1-\tan {45}^{\circ }\tan 9^{\circ }}$ where $\tan {45}^{\circ }=1$

$=\tan (45+9)$ using compound angles formula

$=\tan {54}^{\circ }=\tan (90-36)=\cot {36}^{\circ }$

Hence Proved