Question

Write the function in the simplest form
${{\tan }^{-1}}\left( \dfrac{\cos x-\sin x}{\cos x+\sin x} \right)$

Answer 1

Hint: In this question, we are given a function which is the tangent inverse of another function involving sine and cosine. Therefore, we should try to convert the function inside the parenthesis into a function of tan so that the ${{\tan }^{-1}}$ gets cancelled with tan and we are left with a simple function.

Complete step-by-step answer:

Let us name the given function to be I. Then, from the question
$I={{\tan }^{-1}}\left( \dfrac{\cos x-\sin x}{\cos x+\sin x} \right)....................(1.1)$
We know from the trigonometric theory that
$\tan x=\dfrac{\sin x}{\cos x}......................(1.2)$
We can divide the numerator and denominator in the parenthesis of (1.1) by $\cos x$ to obtain
$I={{\tan }^{-1}}\left( \dfrac{\cos x-\sin x}{\cos x+\sin x} \right)={{\tan }^{-1}}\left( \dfrac{\dfrac{\cos x}{\cos x}-\dfrac{\sin x}{\cos x}}{\dfrac{\cos x}{\cos x}+\dfrac{\sin x}{\cos x}} \right)={{\tan }^{-1}}\left( \dfrac{1-\dfrac{\sin x}{\cos x}}{1+\dfrac{\sin x}{\cos x}} \right)$
Now, substituting the value of \[\dfrac{\sin x}{\cos x}\] from (1.2) into the above equation, we obtain
$I={{\tan }^{-1}}\left( \dfrac{1-\dfrac{\sin x}{\cos x}}{1+\dfrac{\sin x}{\cos x}} \right)={{\tan }^{-1}}\left( \dfrac{1-\tan x}{1+\tan x} \right)..............(1.3)$
Now, as $\tan \left( {{45}^{\circ }} \right)=1$, we can replace 1 in equation (1.3) and write it as
$I={{\tan }^{-1}}\left( \dfrac{1-\tan x}{1+\tan x} \right)={{\tan }^{-1}}\left( \dfrac{\tan \left( {{45}^{\circ }} \right)-\tan x}{1+\tan \left( {{45}^{\circ }} \right)\tan x} \right)..............(1.4)$
Also, we know that the formula for tangent of difference of two angles is given by
$\tan (a-b)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}$
Taking $a={{45}^{\circ }}$ and b=x in the above equation, we obtain
$\tan \left( {{45}^{\circ }}-x \right)=\dfrac{\tan \left( {{45}^{\circ }} \right)-\tan x}{1+\tan \left( {{45}^{\circ }} \right)\tan x}........(1.5)$
Therefore, using the value from equation (1.5) into the RHS of equation (1.4), we get
$I={{\tan }^{-1}}\left( \dfrac{\tan \left( {{45}^{\circ }} \right)-\tan x}{1+\tan \left( {{45}^{\circ }} \right)\tan x} \right)={{\tan }^{-1}}\left( \tan \left( {{45}^{\circ }}-x \right) \right)..............(1.6)$
Also, as ${{\tan }^{-1}}$ is the inverse function of tan, for any angle $\theta $, and the tangent function has a periodicity of ${{180}^{\circ }}$, we should have
${{\tan }^{-1}}\left( \tan \theta \right)=\theta +n\times {{180}^{\circ }}$ where n is any integer.
Therefore, using this expression with $\theta ={{45}^{\circ }}-x$ in equation (1.6), we obtain
$I={{\tan }^{-1}}\left( \tan \left( {{45}^{\circ }}-x \right) \right)={{45}^{\circ }}-x+n\times {{180}^{\circ }}\text{ , }n\in Z$
Thus, we have successfully simplified the expression given in the question to be
${{\tan }^{-1}}\left( \dfrac{\cos x-\sin x}{\cos x+\sin x} \right)={{45}^{\circ }}-x+n\times {{180}^{\circ }}\text{ , }n\in Z$
Which is the required answer to this question.

Note: We have expressed the angle in degrees while substituting 1 in equation (1.4). However, one can also write the expression as $\tan \left( \dfrac{\pi }{4} \right)=1$ and then obtain the final answer as ${{\tan }^{-1}}\left( \dfrac{\cos x-\sin x}{\cos x+\sin x} \right)=\dfrac{\pi }{4}-x+n\pi \text{, n}\in \text{Z}$. However, this answer is the same as obtained in the solution because we can express the angles in radian as $\pi ={{180}^{\circ }}$ and thus $\dfrac{\pi }{4}={{45}^{\circ }}$.