$f(x)=tan^{-1}(sin~x+cos~x)$ is increasing in:

Question

Accepted Answer

To determine where the function $ f(x) = \tan^{-1}(\sin x + \cos x) $ is increasing, we need to analyze the derivative of the function. The function $ f(x) $ is increasing where its derivative $ f'(x) $ is positive.

First, let's find the derivative of $ f(x) $:

$$ f'(x) = \frac{d}{dx} \left( \tan^{-1}(\sin x + \cos x) \right) $$

Using the chain rule, the derivative is:

$$ f'(x) = \frac{1}{1 + (\sin x + \cos x)^2} \cdot (\cos x - \sin x) $$

For $ f(x) $ to be increasing, we need:

$$ \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} > 0 $$

Since the denominator $ 1 + (\sin x + \cos x)^2 $ is always positive, we focus on the numerator:

$$ \cos x - \sin x > 0 $$

This inequality simplifies to:

$$ \cos x > \sin x $$

The inequality $ \cos x > \sin x $ holds true in the interval $ (-\frac{\pi}{2}, \frac{\pi}{4}) $. This is because:

- At $ x = -\frac{\pi}{2} $, $\cos x = 0$ and $\sin x = -1$, so $\cos x > \sin x$.
- At $ x = \frac{\pi}{4} $, $\cos x = \sin x = \frac{\sqrt{2}}{2}$, so the inequality is not strictly satisfied at this endpoint, but it holds just before reaching $\frac{\pi}{4}$.

Now, let's examine why the other options are incorrect:

1. **Option 1: $(\frac{\pi}{4}, \frac{\pi}{2})$**

In this interval, $\sin x$ becomes greater than $\cos x$ as $x$ approaches $\frac{\pi}{2}$. Therefore, $\cos x - \sin x$ becomes negative, making $f'(x)$ negative, which means $f(x)$ is not increasing.

2. **Option 3: $(0, \frac{\pi}{2})$**

In this interval, $\sin x$ is generally greater than $\cos x$ as $x$ increases from 0 to $\frac{\pi}{2}$. Thus, $\cos x - \sin x$ becomes negative, making $f'(x)$ negative, indicating that $f(x)$ is not increasing.

3. **Option 4: $(- \frac{\pi}{2}, \frac{\pi}{2})$**

While $\cos x > \sin x$ holds in part of this interval, it does not hold throughout the entire interval. Specifically, as $x$ approaches $\frac{\pi}{2}$, $\sin x$ becomes greater than $\cos x$, making $\cos x - \sin x$ negative, and thus $f(x)$ is not increasing throughout the entire interval.

Therefore, the correct answer is Option 2: $(- \frac{\pi}{2}, \frac{\pi}{4})$, where the function $ f(x) $ is indeed increasing.