Prove sin⁻¹(4/5) + 2tan⁻¹(1/3) = π/2

Prove that \( \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} \)

Solution:

Let

\[ \theta = \tan^{-1}\left(\frac{1}{3}\right) \]

Then,

\[ \tan \theta = \frac{1}{3} \]

Using double angle identity:

\[ \tan(2\theta) = \frac{2\tan\theta}{1 – \tan^2\theta} \]

\[ = \frac{2 \cdot \frac{1}{3}}{1 – \frac{1}{9}} = \frac{2/3}{8/9} = \frac{3}{4} \]

\[ \Rightarrow 2\theta = \tan^{-1}\left(\frac{3}{4}\right) \]

Now consider:

\[ \sin^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{3}{4}\right) \]

Let

\[ \alpha = \sin^{-1}\left(\frac{4}{5}\right) \Rightarrow \sin \alpha = \frac{4}{5} \]

Then,

\[ \cos \alpha = \frac{3}{5} \Rightarrow \tan \alpha = \frac{4}{3} \]

So,

\[ \alpha = \tan^{-1}\left(\frac{4}{3}\right) \]

Thus,

\[ \tan^{-1}\left(\frac{4}{3}\right) + \tan^{-1}\left(\frac{3}{4}\right) \]

Using identity:

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

Here,

\[ \frac{4}{3} \cdot \frac{3}{4} = 1 \]

Hence,

\[ \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} \]

\[ \sin^{-1}\left(\frac{4}{5}\right) + 2\tan^{-1}\left(\frac{1}{3}\right) = \frac{\pi}{2} \]

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