Question

If $f(x)=\tan^{-1}\left(\frac{3\cos x-5\sin x}{5\cos x+3\sin x}\right)$ , then the value $f^{\prime}(1)$ is

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. -1
• E. $-\frac{1}{4}$

Hint:

Logic Tip: The derivative of $y = \tan^{-1}\left(\frac{a\cos x - b\sin x}{b\cos x + a\sin x}\right)$ is always exactly $-1$, regardless of what constants $a$ and $b$ are, because it always simplifies to a constant minus $x$.

Answer 1

The Correct Option is D

Concept:
Expressions inside inverse trigonometric functions can often be simplified using compound angle identities. The goal is to transform the inner expression into the form $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$, allowing the $\tan^{-1}$ and $\tan$ functions to neutralize each other.
Step 1: Divide numerator and denominator by the leading cosine term.
To create the "1" in the denominator required by the tangent compound formula, divide every term inside the parentheses by $5\cos x$: $$f(x) = \tan^{-1}\left( \frac{\frac{3\cos x}{5\cos x} - \frac{5\sin x}{5\cos x}}{\frac{5\cos x}{5\cos x} + \frac{3\sin x}{5\cos x}} \right)$$
Step 2: Simplify into terms of tangent.
$$f(x) = \tan^{-1}\left( \frac{\frac{3}{5} - \tan x}{1 + \frac{3}{5}\tan x} \right)$$
Step 3: Apply the tangent difference identity.
Let $\tan \alpha = \frac{3}{5}$. The expression perfectly matches the expansion of $\tan(\alpha - x)$: $$f(x) = \tan^{-1}\left( \frac{\tan \alpha - \tan x}{1 + \tan \alpha \tan x} \right)$$ $$f(x) = \tan^{-1}( \tan(\alpha - x) )$$ Assuming principal values, the inverse tangent cancels the tangent: $$f(x) = \alpha - x$$
Step 4: Differentiate the simplified function.
Since $\alpha = \tan^{-1}\left(\frac{3}{5}\right)$, it is a constant. Its derivative is 0. $$f'(x) = \frac{d}{dx}(\alpha - x)$$ $$f'(x) = 0 - 1 = -1$$
Step 5: Evaluate at the specific point.
The derivative $f'(x) = -1$ is a constant. Therefore, the value is the same for any $x$: $$f'(1) = -1$$