Question

$\frac{\cos {11}^o-\sin {11}^o}{\cos {11}^o+\sin {11}^o}=\cot {56}^o$

Answer 1

$\frac{\cos {11}^{\circ }-\sin {11}^{\circ }}{\cos {11}^{\circ }+\sin {11}^{\circ }}$

$=\frac{\cos {11}^{\circ }-\cos ({90}^{\circ }-{11}^{\circ })}{\cos {11}^{\circ }+\cos ({90}^{\circ }-{11}^{\circ })}$

$=\frac{\cos {11}^{\circ }-\cos {79}^{\circ }}{\cos {11}^{\circ }+\cos {79}^{\circ }}$

$=\frac{-2\sin \left(\frac{11+79}{2}\right)\sin \left(\frac{11-79}{2}\right)}{2\cos \left(\frac{11+79}{2}\right)\cos \left(\frac{11-79}{2}\right)}$

$=\frac{-\sin \left(\frac{90}{2}\right)\sin \left(\frac{-68}{2}\right)}{\cos \left(\frac{90}{2}\right)\cos \left(\frac{-68}{2}\right)}$

$=\frac{\sin {45}^{\circ }\sin {34}^{\circ }}{\cos {45}^{\circ }\cos {34}^{\circ }}$

$=\frac{\frac{1}{\sqrt{2}}\sin {34}^{\circ }}{\frac{1}{\sqrt{2}}\cos {34}^{\circ }}$

$=\frac{\sin {34}^{\circ }}{\cos {34}^{\circ }}$

$=\tan {34}^{\circ }$

$=\cot ({90}^{\circ }-{34}^{\circ })$ since $\cot (90-\theta )=\tan \theta$

$=\cot {56}^{\circ }$

Hence proved.