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Question: Prove the given inverse trigonometric expression, \[{{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \r...

Prove the given inverse trigonometric expression, tan⁡−1(cos⁡x1+sin⁡x)=π4−x2,x∈(−π2,π2){{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2},x\in \left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right) .

Explanation

Solution

Hint: In this question, we have the inverse tan function, but we don’t have a tan function inside that inverse function. So, first, we need to convert that part in terms of tan. Using the formula cos⁡x=cos⁡2x2−sin⁡2x2\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2} , sin⁡x=2.sin⁡x2.cos⁡x2\sin x=2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2} and the identity 1=sin⁡2x2+cos⁡2x21={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}. Put these all, and solve further.

Complete step-by-step solution -
According to the question, we have
tan⁡−1(cos⁡x1+sin⁡x){{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)……………(1)
In equation (1), we have the tan inverse of cos⁡x1+sin⁡x\dfrac{\cos x}{1+\sin x}.
Here we have to remove this inverse of tan. For removing we have to make the term cos⁡x1+sin⁡x\dfrac{\cos x}{1+\sin x} in the form of tan.
Let’s proceed with the numerator.
We know that,
cos⁡x=cos⁡2x2−sin⁡2x2\cos x={{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}………….(2)
We also know that,
sin⁡x=2.sin⁡x2.cos⁡x2\sin x=2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2}……………..(3)
1=sin⁡2x2+cos⁡2x21={{\sin }^{2}}\dfrac{x}{2}+{{\cos }^{2}}\dfrac{x}{2}……………..(4)
Adding equation (3) and equation (4), we get
1+sin⁡x=cos⁡2x2+sin⁡2x2+2.sin⁡x2.cos⁡x21+\sin x={{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2} ………….(5)
From equation (1), we have tan⁡−1(cos⁡x1+sin⁡x){{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right) .
Putting equation (2) and equation (5) in equation (1), we get
tan⁡−1(cos⁡2x2−sin⁡2x2cos⁡2x2+sin⁡2x2+2cos⁡x2.sin⁡x2){{\tan }^{-1}}\left( \dfrac{{{\cos }^{2}}\dfrac{x}{2}-{{\sin }^{2}}\dfrac{x}{2}}{{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2\cos \dfrac{x}{2}.\sin \dfrac{x}{2}} \right) ……………(6)
We also know that,
A2−B2=(A+B). (A−B){{A}^{2}}-{{B}^{2}}=\left( A+B \right).\text{ }\left( A-B \right) .
Similarily equation (2) can be written as,
(cos⁡x2−sin⁡x2).(cos⁡x2+sin⁡x2)\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right).\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right) …………..(7)
And also,
(A+B)2=A2+B2+2.A.B{{\left( A+B \right)}^{2}}={{A}^{2}}+{{B}^{2}}+2.A.B .
Similarily equation (5) can be written as,
cos⁡2x2+sin⁡2x2+2.sin⁡x2.cos⁡x2=(cos⁡x2+sin⁡x2)2{{\cos }^{2}}\dfrac{x}{2}+{{\sin }^{2}}\dfrac{x}{2}+2.\sin \dfrac{x}{2}.\cos \dfrac{x}{2}={{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}…………….(8)
Using equation (7) and equation (8), we can write equation (6) as,
tan⁡−1((cos⁡x2−sin⁡x2)(cos⁡x2+sin⁡x2)(cos⁡x2+sin⁡x2)2){{\tan }^{-1}}\left( \dfrac{\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right)\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}{{{\left( \cos \dfrac{x}{2}+\sin \dfrac{x}{2} \right)}^{2}}} \right)
(cos⁡x2−sin⁡x2)\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right) is common in the numerator as well as the denominator. So, we can cancel one (cos⁡x2−sin⁡x2)\left( \cos \dfrac{x}{2}-\sin \dfrac{x}{2} \right) in both numerator and denominator. Our equation will look like, tan⁡−1(cos⁡x2−sin⁡x2cos⁡x2+sin⁡x2){{\tan }^{-1}}\left( \dfrac{\cos \dfrac{x}{2}-\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}+\sin \dfrac{x}{2}} \right) .
Dividing by cos⁡x2\cos \dfrac{x}{2} in numerator and denominator, we get
tan⁡−1(1−sin⁡x2cos⁡x21+sin⁡x2cos⁡x2){{\tan }^{-1}}\left( \dfrac{1-\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}}{1+\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}} \right)………………(9)
We know that, sin⁡x2cos⁡x2=tan⁡x2\dfrac{\sin \dfrac{x}{2}}{\cos \dfrac{x}{2}}=\tan \dfrac{x}{2} ……………..(10)
Using equation (10), equation (9) can be written as
tan⁡−1(1−tan⁡x21+tan⁡x2){{\tan }^{-1}}\left( \dfrac{1-\tan \dfrac{x}{2}}{1+\tan \dfrac{x}{2}} \right)
tan⁡−1(1−tan⁡x21+1.tan⁡x2){{\tan }^{-1}}\left( \dfrac{1-\tan \dfrac{x}{2}}{1+1.\tan \dfrac{x}{2}} \right)……………(11)
We also know, tan⁡π4=1\tan \dfrac{\pi }{4}=1 ……………(12)
Using equation (12), equation (11) can be written as
tan⁡−1(tan⁡π4−tan⁡x21+tan⁡π4.tan⁡x2){{\tan }^{-1}}\left( \dfrac{\tan \dfrac{\pi }{4}-\tan \dfrac{x}{2}}{1+\tan \dfrac{\pi }{4}.\tan \dfrac{x}{2}} \right)
=tan⁡−1(tan⁡(π4−x2))={{\tan }^{-1}}\left( \tan \left( \dfrac{\pi }{4}-\dfrac{x}{2} \right) \right)
=π4−x2=\dfrac{\pi }{4}-\dfrac{x}{2}
So, tan⁡−1(cos⁡x1+sin⁡x)=π4−x2{{\tan }^{-1}}\left( \dfrac{\cos x}{1+\sin x} \right)=\dfrac{\pi }{4}-\dfrac{x}{2} .
Therefore, LHS=RHS.
Hence, proved.

Note: In this question, we have to be careful during the simplification of the inverse tan function. As tan⁡π4=1\tan \dfrac{\pi }{4}=1 and also tan⁡(5π4)=1\tan \left( \dfrac{5\pi }{4} \right)=1 . If we take tan⁡(5π4)=1\tan \left( \dfrac{5\pi }{4} \right)=1 , then it will be wrong. As in RHS, we have π4\dfrac{\pi }{4} , so we have to continue with tan⁡(π4)=1\tan \left( \dfrac{\pi }{4} \right)=1 .