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Question: Write the function in the simplest form \({{\tan }^{-1}}\left( \dfrac{\cos x-\sin x}{\cos x+\sin x...

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

Explanation

Solution

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{{\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(cos⁡x−sin⁡xcos⁡x+sin⁡x)....................(1.1)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=sin⁡xcos⁡x......................(1.2)\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\cos x to obtain
I=tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)=tan⁡−1(cos⁡xcos⁡x−sin⁡xcos⁡xcos⁡xcos⁡x+sin⁡xcos⁡x)=tan⁡−1(1−sin⁡xcos⁡x1+sin⁡xcos⁡x)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 sin⁡xcos⁡x\dfrac{\sin x}{\cos x} from (1.2) into the above equation, we obtain
I=tan⁡−1(1−sin⁡xcos⁡x1+sin⁡xcos⁡x)=tan⁡−1(1−tan⁡x1+tan⁡x)..............(1.3)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⁡(45∘)=1\tan \left( {{45}^{\circ }} \right)=1, we can replace 1 in equation (1.3) and write it as
I=tan⁡−1(1−tan⁡x1+tan⁡x)=tan⁡−1(tan⁡(45∘)−tan⁡x1+tan⁡(45∘)tan⁡x)..............(1.4)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)=tan⁡a−tan⁡b1+tan⁡atan⁡b\tan (a-b)=\dfrac{\tan a-\tan b}{1+\tan a\tan b}
Taking a=45∘a={{45}^{\circ }} and b=x in the above equation, we obtain
tan⁡(45∘−x)=tan⁡(45∘)−tan⁡x1+tan⁡(45∘)tan⁡x........(1.5)\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(tan⁡(45∘)−tan⁡x1+tan⁡(45∘)tan⁡x)=tan⁡−1(tan⁡(45∘−x))..............(1.6)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{{\tan }^{-1}} is the inverse function of tan, for any angle θ\theta , and the tangent function has a periodicity of 180∘{{180}^{\circ }}, we should have
tan⁡−1(tan⁡θ)=θ+n×180∘{{\tan }^{-1}}\left( \tan \theta \right)=\theta +n\times {{180}^{\circ }} where n is any integer.
Therefore, using this expression with θ=45∘−x\theta ={{45}^{\circ }}-x in equation (1.6), we obtain
I=tan⁡−1(tan⁡(45∘−x))=45∘−x+n×180∘ , n∈ZI={{\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(cos⁡x−sin⁡xcos⁡x+sin⁡x)=45∘−x+n×180∘ , n∈Z{{\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⁡(π4)=1\tan \left( \dfrac{\pi }{4} \right)=1 and then obtain the final answer as tan⁡−1(cos⁡x−sin⁡xcos⁡x+sin⁡x)=π4−x+nπ, n∈Z{{\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 π=180∘\pi ={{180}^{\circ }} and thus π4=45∘\dfrac{\pi }{4}={{45}^{\circ }}.