A value of $$x$$ for which $$\sin\left(\cot^{-1}(1+x)\right) = \cos\left(\tan^{-1}x\right)$$, is :

The given equation is $$\sin\left(\cot^{-1}(1+x)\right) = \cos\left(\tan^{-1}x\right)$$. To solve for $$x$$, express both sides in terms of $$x$$. Consider the left side: $$\sin\left(\cot^{-1}(1+x)\right)$$. Let $$\alpha = \cot^{-1}(1+x)$$, so $$\cot \alpha = 1+x$$. In a right triangle, $$\cot \alpha = \frac{\text{adjacent}}{\text{opposite}}$$, so set adjacent = $$1+x$$ and opposite = 1. The hypotenuse is $$\sqrt{(1+x)^2 + 1^2} = \sqrt{1 + 2x + x^2 + 1} = \sqrt{x^2 + 2x + 2}$$. Thus, $$\sin \alpha = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 2x + 2}}$$. Now consider the right side: $$\cos\left(\tan^{-1}x\right)$$. Let $$\beta = \tan^{-1}x$$, so $$\tan \beta = x$$. In a right triangle, $$\tan \beta = \frac{\text{opposite}}{\text{adjacent}}$$, so set opposite = $$x$$ and adjacent = 1. The hypotenuse is $$\sqrt{x^2 + 1^2} = \sqrt{x^2 + 1}$$. Thus, $$\cos \beta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\sqrt{x^2 + 1}}$$. The equation becomes: $$$ \frac{1}{\sqrt{x^2 + 2x + 2}} = \frac{1}{\sqrt{x^2 + 1}} $$$ Since the denominators are positive, take reciprocals: $$$ \sqrt{x^2 + 2x + 2} = \sqrt{x^2 + 1} $$$ Square both sides to eliminate the square roots: $$$ (\sqrt{x^2 + 2x + 2})^2 = (\sqrt{x^2 + 1})^2 $$$ $$$ x^2 + 2x + 2 = x^2 + 1 $$$ Subtract $$x^2$$ from both sides: $$$ 2x + 2 = 1 $$$ Subtract 2 from both sides: $$$ 2x = -1 $$$ Divide by 2: $$$ x = -\frac{1}{2} $$$ Now, verify if $$x = -\frac{1}{2}$$ satisfies the original equation. Substitute $$x = -\frac{1}{2}$$: Left side: $$\sin\left(\cot^{-1}\left(1 + \left(-\frac{1}{2}\right)\right)\right) = \sin\left(\cot^{-1}\left(\frac{1}{2}\right)\right)$$. Let $$\theta = \cot^{-1}\left(\frac{1}{2}\right)$$, so $$\cot \theta = \frac{1}{2}$$. In a right triangle, adjacent = 1, opposite = 2, hypotenuse = $$\sqrt{1^2 + 2^2} = \sqrt{5}$$. Thus, $$\sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{2}{\sqrt{5}}$$. Right side: $$\cos\left(\tan^{-1}\left(-\frac{1}{2}\right)\right)$$. Let $$\phi = \tan^{-1}\left(-\frac{1}{2}\right)$$, so $$\tan \phi = -\frac{1}{2}$$. Since cosine is positive in the fourth quadrant, consider a triangle with opposite = $$\frac{1}{2}$$ (magnitude), adjacent = 1, hypotenuse = $$\sqrt{\left(\frac{1}{2}\right)^2 + 1^2} = \sqrt{\frac{1}{4} + 1} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$. Thus, $$\cos \phi = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$$. Both sides equal $$\frac{2}{\sqrt{5}}$$, so $$x = -\frac{1}{2}$$ satisfies the equation. Now, check the other options to confirm: - Option B: $$x = 1$$ Left side: $$\sin\left(\cot^{-1}(1+1)\right) = \sin\left(\cot^{-1}(2)\right)$$. Let $$\theta = \cot^{-1}(2)$$, so $$\cot \theta = 2$$. Adjacent = 2, opposite = 1, hypotenuse = $$\sqrt{2^2 + 1^2} = \sqrt{5}$$, so $$\sin \theta = \frac{1}{\sqrt{5}}$$. Right side: $$\cos\left(\tan^{-1}(1)\right) = \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$$. $$\frac{1}{\sqrt{5}} \neq \frac{1}{\sqrt{2}}$$, so not equal. - Option C: $$x = 0$$ Left side: $$\sin\left(\cot^{-1}(1+0)\right) = \sin\left(\cot^{-1}(1)\right) = \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$. Right side: $$\cos\left(\tan^{-1}(0)\right) = \cos(0) = 1$$. $$\frac{\sqrt{2}}{2} \neq 1$$, so not equal. - Option D: $$x = \frac{1}{2}$$ Left side: $$\sin\left(\cot^{-1}\left(1 + \frac{1}{2}\right)\right) = \sin\left(\cot^{-1}\left(\frac{3}{2}\right)\right)$$. Let $$\theta = \cot^{-1}\left(\frac{3}{2}\right)$$, so $$\cot \theta = \frac{3}{2}$$. Adjacent = $$\frac{3}{2}$$, opposite = 1, hypotenuse = $$\sqrt{\left(\frac{3}{2}\right)^2 + 1^2} = \sqrt{\frac{9}{4} + 1} = \sqrt{\frac{13}{4}} = \frac{\sqrt{13}}{2}$$, so $$\sin \theta = \frac{1}{\frac{\sqrt{13}}{2}} = \frac{2}{\sqrt{13}}$$. Right side: $$\cos\left(\tan^{-1}\left(\frac{1}{2}\right)\right)$$. Let $$\phi = \tan^{-1}\left(\frac{1}{2}\right)$$, so $$\tan \phi = \frac{1}{2}$$. Adjacent = 1, opposite = $$\frac{1}{2}$$, hypotenuse = $$\sqrt{1^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}$$, so $$\cos \phi = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}}$$. $$\frac{2}{\sqrt{13}} \neq \frac{2}{\sqrt{5}}$$, so not equal. Only $$x = -\frac{1}{2}$$ satisfies the equation, which corresponds to option A. Hence, the correct answer is Option A.