Question

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

Answer 1

Hint: 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,
\[\cos (180+45)^\circ-\sin (180+45)^\circ+\tan 495^\circ-\cot 495^\circ \\
= -\cos 45^\circ+\sin 45^\circ+\tan 495^\circ-\cot 495^\circ\]
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.