Question

Simplify:
$\frac{\cos {27}^0+\sin {27}^0}{\cos {27}^0-\sin {27}^0}=\tan {72}^0$

Answer 1

L.H.S

$\frac{\cos {27}^0+\sin {27}^0}{\cos {27}^0-\sin {27}^0}$

$\Rightarrow \frac{\cos ({90}^0-{63}^0)+\sin {27}^0}{\cos ({90}^0-{63}^0)-\sin {27}^0}$

$\Rightarrow \frac{\sin {63}^0+\sin {27}^0}{\sin {63}^0-\sin {27}^0}$

We know that

$\sin C+\sin D=2\sin \frac{C+D}{2}.\cos \frac{C-D}{2}$

$\sin C-\sin D=2\cos \frac{C+D}{2}.\sin \frac{C-D}{2}$

Therefore,

$\Rightarrow \frac{2\sin \frac{63+27}{2}.\cos \frac{63-27}{2}}{2\cos \frac{63+27}{2}.\sin \frac{63-27}{2}}$

$\Rightarrow \frac{2\sin \frac{{90}^0}{2}.\cos \frac{{36}^0}{2}}{2\cos \frac{{90}^0}{2}.\sin \frac{{36}^0}{2}}$

$\Rightarrow \frac{2\sin {45}^0.\cos {18}^0}{2\cos {45}^0.\sin {18}^0}$

$\Rightarrow \tan {45}^0.\cot {18}^0$

$\Rightarrow 1\times \cot ({90}^0-{72}^0)$

$\Rightarrow \tan {72}^0$

R.H.S

Hence, proved.