Question

Evaluate:$\cos {225}^{\circ }-\sin {225}^{\circ }+\tan {495}^{\circ }-\cot {495}^{\circ }$

Answer 1

Consider,$\cos {225}^{\circ }-\sin {225}^{\circ }$

$=\cos ({180}^{\circ }+{45}^{\circ })-\sin ({180}^{\circ }+{45}^{\circ })$

$=-\cos {45}^{\circ }+\sin {45}^{\circ }$

$=-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}$

$=0$

$=\frac{-2}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{-2\sqrt{2}}{2}=-\sqrt{2}$

Consider $\tan {495}^{\circ }-\cot {495}^{\circ }$

$=\tan {495}^{\circ }-\cot {495}^{\circ }$

$=\tan (5\times {90}^{\circ }+{45}^{\circ })-\cot (5\times {90}^{\circ }+{45}^{\circ })$

Here $n=5$ is odd , so $\tan$ changes to $\cot$ and $\cot$ changes to $\tan$

$=-\cot {45}^{\circ }-(-\tan {45}^{\circ })$ since ${495}^{\circ }=5\times {90}^{\circ }+{45}^{\circ }$ is in second quadrant,hence $\tan {495}^{\circ }$ is negative.

$=-1+1=0$

$\therefore \cos {225}^{\circ }-\sin {225}^{\circ }+\tan {495}^{\circ }-\cot {495}^{\circ }=0-\sqrt{2}=-\sqrt{2}$