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Question: The function \[f(x)={{\tan }^{-1}}(\sin x+\cos x)\] is increasing in 1\. \[\left( \dfrac{\pi }{4},...

The function f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x)={{\tan }^{-1}}(\sin x+\cos x) is increasing in
1. (π4,π2)\left( \dfrac{\pi }{4},\dfrac{\pi }{2} \right)
2. (−π2,π4)\left( -\dfrac{\pi }{2},\dfrac{\pi }{4} \right)
3. (0,π2)\left( 0,\dfrac{\pi }{2} \right)
4. (−π2,π2)\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)

Explanation

Solution

To find where the following function is increasing in, we need to first take differentiation of the following function. As we know already that whichever points the differentiation of the function is greater than 00 , the function is increasing and wherever it is less than 00 the function will be a decreasing function. Therefore first take the differentiation of the following function and then check at which point is the differentiation of the function greater than 00 .

Complete step by step answer:
Firstly we need to find where the function f(x)=tan⁡−1(sin⁡x+cos⁡x)f(x)={{\tan }^{-1}}(\sin x+\cos x) is increasing.
That’s why we need to start with taking the differentiation of this function. In taking the differentiation there are two functions in it that’s why we first start by taking the differentiation of inverse of tangent. And now since there is another function inside tan we after that take the function of it inside and differentiate it and then multiply it and that’s how we will get the differentiation of f(x)f(x) . which is:
f′(x)=cos⁡x−sin⁡x1+(sin⁡x+cos⁡x)2f'(x)=\dfrac{\cos x-\sin x}{1+{{(\sin x+\cos x)}^{2}}}
Now since we need to find where the function is increasing we check where the function is greater than 00 therefore.
cos⁡x−sin⁡x1+(sin⁡x+cos⁡x)2>0\dfrac{\cos x-\sin x}{1+{{(\sin x+\cos x)}^{2}}}>0
Now the denominator doesn’t affect if the function will be increasing or not that’s why we get,
cos⁡x−sin⁡x>0\cos x-\sin x>0
From this we get
cos⁡x>sin⁡x\cos x>\sin x
Now to find the answer that’s why we need to find the points where cos⁡x\cos x is always greater than sin⁡x\sin x . This happens at a certain interval that is from −π2-\dfrac{\pi }{2} to π4\dfrac{\pi }{4} . At π4\dfrac{\pi }{4} both sin and cos both of them have the same value and then sin is greater than cos.

So, the correct answer is “Option 2”.

Note: A function is increasing at an interval if its differentiate is greater than 00 and similarly it is decreasing if differentiate is less than 00 . To verify the answer, we can take the known values of trigonometric functions and check whether the answer is increasing or decreasing.