Question

Prove $\frac{\cos {20}^o+\sin {20}^o}{\cos {20}^o-\sin {20}^o}=\tan {65}^o$

Answer 1

L.H.S

$=\frac{\cos {20}^{\circ }+\sin {20}^{\circ }}{\cos {20}^{\circ }-\sin {20}^{\circ }}$

$=\frac{\sin ({90}^{\circ }-{20}^{\circ })+\sin {20}^{\circ }}{\sin ({90}^{\circ }-{20}^{\circ })-\sin {20}^{\circ }}$

$=\frac{\sin {70}^{\circ }+\sin {20}^{\circ }}{\sin {70}^{\circ }-\sin {20}^{\circ }}$

We know that

$\sin C+\sin D=2\sin \left(\frac{C+D}{2}\right)\cos \left(\frac{C-D}{2}\right)$

$\sin C-\sin D=2\cos \left(\frac{C+D}{2}\right)\sin \left(\frac{C-D}{2}\right)$

Therefore,

$=\frac{2\sin \left(\frac{{70}^{\circ }+{20}^{\circ }}{2}\right)\cos \left(\frac{{70}^{\circ }-{20}^{\circ }}{2}\right)}{2\cos \left(\frac{{70}^{\circ }+{20}^{\circ }}{2}\right)\sin \left(\frac{{70}^{\circ }-{20}^{\circ }}{2}\right)}$

$=\frac{\sin \left(\frac{{90}^{\circ }}{2}\right)\cos \left(\frac{{50}^{\circ }}{2}\right)}{\cos \left(\frac{{90}^{\circ }}{2}\right)\sin \left(\frac{{50}^{\circ }}{2}\right)}$

$=\frac{\sin {45}^{\circ }\cos {25}^{\circ }}{\cos {45}^{\circ }\sin {25}^{\circ }}$

$=\tan 45\cot \left({25}^{\circ }\right)$

$=1\times \tan ({90}^{\circ }-{65}^{\circ })$

$=\tan {65}^{\circ }$

R.H.S

Hence, proved.