Question

The function $ f(x) = \tan^{-1}(\sin x + \cos x) $ is an increasing function in.

• A. \( \left( \frac{\pi}{4}, \frac{\pi}{2} \right) \)
• B. \( (0, \frac{\pi}{2}) \)
• C. \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \)
• D. \( \left( -\frac{\pi}{2}, \pi \right) \)

Hint:

When dealing with trigonometric functions and their inverses, remember to check the sign of the derivative to determine intervals of increase or decrease.

Answer 1

The Correct Option is C

To find the interval where \( f(x) = \tan^{-1}(\sin x + \cos x) \) is increasing, we need to find where the derivative \( f'(x) \) is positive. Let us first differentiate \( f(x) \). The derivative of \( \tan^{-1}(y) \) with respect to \( x \) is: \[ f'(x) = \frac{1}{1 + y^2} \cdot \frac{dy}{dx} \] where \( y = \sin x + \cos x \). Now, differentiate \( y = \sin x + \cos x \): \[ \frac{dy}{dx} = \cos x - \sin x \] Thus, the derivative of \( f(x) \) is: \[ f'(x) = \frac{\cos x - \sin x}{1 + (\sin x + \cos x)^2} \] For \( f(x) \) to be increasing, we need \( f'(x)>0 \). This requires: \[ \cos x - \sin x>0 \] which simplifies to: \[ \cos x>\sin x \] This inequality holds for: \[ x \in \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \] Thus, the function is increasing in the interval \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \), which is the correct answer. 
Thus, the correct answer is \( \left( -\frac{\pi}{2}, \frac{\pi}{4} \right) \).