Question:

If $y = \tan^{-1}\left(\frac{\sin x + \cos x}{\cos x - \sin x}\right)$, then $\frac{dy}{dx} = $

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Simplifying expressions inside inverse trigonometric functions using trig identities almost always reduces the problem to a simple linear function!
Updated On: Jun 3, 2026
  • $1$
  • $-1$
  • $\frac{1}{2}$
  • $0$
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The Correct Option is A

Solution and Explanation

Step 1: Concept
Differentiate the function by first simplifying the trigonometric argument using standard inverse trigonometric identities.

Step 2: Meaning
We divide the numerator and the denominator inside the parentheses by $\cos x$ to express it in terms of $\tan x$.

Step 3: Analysis
\[ y = \tan^{-1}\left(\frac{\frac{\sin x}{\cos x} + 1}{1 - \frac{\sin x}{\cos x}}\right) = \tan^{-1}\left(\frac{1 + \tan x}{1 - \tan x}\right) \] Using the identity $\frac{1+\tan x}{1-\tan x} = \tan\left(\frac{\pi}{4} + x\right)$: \[ y = \tan^{-1}\left(\tan\left(\frac{\pi}{4} + x\right)\right) = \frac{\pi}{4} + x \] Now, differentiating both sides with respect to $x$: \[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{\pi}{4} + x\right) = 0 + 1 = 1 \]

Step 4: Conclusion
The derivative of the given function is $1$.

Final Answer: (A)
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