Question

Evaluate $\cos 225°-\sin 225°+\tan 495°-cot495°$

Answer 1

Let us consider

$$\begin{gathered} \cos 225°=\cos \left(180°+45°\right)=-\cos 45°=-\frac{1}{\sqrt{2}} \\ \sin 225°=\sin \left(180°+45°\right)=-\sin 45°=-\frac{1}{\sqrt{2}} \\ \tan 495°=\tan \left(5\times 90°+45°\right)=-cot45°=-1 \\ cot495°=cot\left(5\times 90°+45°\right)=-\tan 45°=-1 \end{gathered}$$

Here $n=5$, which is an odd number.

$\therefore \tan$ changes to $cot$ and $cot$ changes to $\tan$ with negative sign

Now, consider $\cos 225°-\sin 225°+\tan 495°-cot495°$

$$\begin{gathered} -\frac{1}{\sqrt{2}}-\left(-\frac{1}{\sqrt{2}}\right)+\left(-1\right)-\left(-1\right) \\ -\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1+1 \\ 0 \end{gathered}$$

Hence, $\cos 225°-\sin 225°+\tan 495°-cot495°$ is $0$ .