Prove tan⁻¹((1−x²)/2x) + cot⁻¹((1−x²)/2x) = π/2

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

Solution:

Let

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

Then,

\[ \tan \theta = \frac{1-x^2}{2x} \]

Now recall the identity:

\[ \cot^{-1}(t) = \tan^{-1}\left(\frac{1}{t}\right) \quad \text{(for } t > 0\text{)} \]

So,

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

Thus the given expression becomes:

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

Using identity:

\[ \tan^{-1}a + \tan^{-1}b = \frac{\pi}{2}, \quad \text{if } ab = 1 \text{ and } a>0 \]

Here,

\[ \frac{1-x^2}{2x} \cdot \frac{2x}{1-x^2} = 1 \]

Hence,

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

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

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