The value of $x$ for which $\sin(\cot^{-1}(1+x)) = \cos(\tan^{-1}x)$ is
Step-by-Step Solution
Key Concept: Use the definitions of inverse trigonometric functions to convert the equation into algebraic form
From $\cot^{-1}(x+1) = \theta$, we have $\cot\theta = x+1$. From $\tan^{-1}x = \phi$, we have $\tan\phi = x$. Given $\sin\theta = \cos\phi$, we can express $\sin\theta = \frac{1}{\sqrt{1+(x+1)^2}}$ and $\cos\phi = \frac{1}{\sqrt{1+x^2}}$. Setting them equal: $(x+1)^2 + 1 = x^2 + 1$, which simplifies to $x^2 + 2x + 1 = x^2$, giving $x = -\frac{1}{2}$.
Correct Answer: -1/2