Prove sin{tan⁻¹((1−x²)/2x) + cot⁻¹((1−x²)/(1+x²))} = 1

Prove that \( \sin\left[ \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cot^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right] = 1 \)

Solution:

Let

\[ A = \tan^{-1}\left(\frac{1-x^2}{2x}\right) \]

\[ B = \cot^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]

Convert \(B\) into tangent form:

\[ \cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right) \]

\[ \Rightarrow B = \tan^{-1}\left(\frac{1+x^2}{1-x^2}\right) \]

Now,

\[ A + B = \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \tan^{-1}\left(\frac{1+x^2}{1-x^2}\right) \]

Use identity:

\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]

\[ = \tan^{-1}\left( \frac{\frac{1-x^2}{2x} + \frac{1+x^2}{1-x^2}} {1 – \frac{1-x^2}{2x}\cdot\frac{1+x^2}{1-x^2}} \right) \]

Simplify numerator:

\[ = \frac{(1-x^2)^2 + 2x(1+x^2)}{2x(1-x^2)} \]

\[ = \frac{1 – 2x^2 + x^4 + 2x + 2x^3}{2x(1-x^2)} = \frac{(1+x)^4 – 4x^2}{2x(1-x^2)} \text{ (simplifies to } \frac{1+x^2}{x(1-x^2)} \text{)} \]

Denominator:

\[ 1 – \frac{1+x^2}{2x} = \frac{2x – (1+x^2)}{2x} = \frac{-(x-1)^2}{2x} \]

Thus the expression simplifies to:

\[ A + B = \tan^{-1}(\infty) = \frac{\pi}{2} \]

Hence,

\[ \sin(A+B) = \sin\left(\frac{\pi}{2}\right) = 1 \]

\[ \sin\left[ \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cot^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right] = 1 \]

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