Question

Find the value of $\cos 225^\circ-\sin 225^\circ+\tan 495^\circ-\cot 495^\circ$.

Answer 1

For angle between 0 to $2\pi$, all trigonometric angles can be rotated by $\dfrac{\pi}{2}$, $\pi$ and $2\pi$ following the sign convention of the angles in the four quadrants.
In the first quadrant, all trigonometric angles are positive. In the second quadrant, only $sin$ and $cosec$ are positive. In the third quadrant, only $tan$ and $cot$ are positive. And in the fourth quadrant, $cos$ and $sec$ is positive.

Complete step-by-step answer:
Given expression is $\cos 225^\circ-\sin 225^\circ+\tan 495^\circ-\cot 495^\circ$.
Using the transformation $\cos (\pi+x) = -\cos x$ and $\sin (\pi+x) = -\sin x$, simplify the expression as,

= -\cos 45^\circ+\sin 45^\circ+\tan 495^\circ-\cot 495^\circ$$ Using the transformation $$\tan (2\pi+x) = \tan x$$ and $$\cot (2\pi+x) = \cot x$$, simplify the expression as, $$-\cos 45^\circ+\sin 45^\circ+\tan (360+135)^\circ-\cot (360+135)^\circ\\\ = -\cos 45^\circ+\sin 45^\circ+\tan 135^\circ-\cot 135^\circ$$ Using the transformation $$\tan \left(\dfrac{\pi}{2}+x\right) = -\tan x$$ and $$\cot \left(\dfrac{\pi}{2}+x\right) = -\cot x$$, simplify the remaining expression as, $$-\cos 45^\circ+\sin 45^\circ+\tan 135^\circ-\cot 135^\circ\\\ = -\cos 45^\circ+\sin 45^\circ-\tan 45^\circ+\cot 45^\circ$$ Using the value $$\cos 45^\circ = \sin 45^\circ = \dfrac{1}{\sqrt{2}}$$ and $$\tan 45^\circ = \cot 45^\circ = 1$$, the value of the expression is, $$\begin{align*}-\cos 45^\circ+\sin 45^\circ-\tan 45^\circ+\cot 45^\circ &= -\dfrac{1}{\sqrt{2}}+-\dfrac{1}{\sqrt{2}}-1+1\\\ &= 0\end{align*}$$ **Hence the value of the expression is 0.** **Note:** You can also simplify the expression using the trigonometric formula for the sum of two angles. For example, use the formula for $$\cos (A \pm B)$$ and so on.