Prove that \( \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right) = \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right) \)
Solution:
Let
\[ \theta = \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) \]
Using identity:
\[ \tan(A+B) = \frac{\tan A + \tan B}{1 – \tan A \tan B} \]
\[ \tan \theta = \frac{\frac{1}{4} + \frac{2}{9}}{1 – \frac{1}{4}\cdot\frac{2}{9}} \]
\[ = \frac{\frac{9 + 8}{36}}{1 – \frac{2}{36}} \]
\[ = \frac{17/36}{34/36} \]
\[ = \frac{1}{2} \]
Thus,
\[ \theta = \tan^{-1}\left(\frac{1}{2}\right) \]
Now let
\[ \alpha = \tan^{-1}\left(\frac{1}{2}\right) \]
Construct a triangle:
- Opposite = 1
- Adjacent = 2
- Hypotenuse = √5
So,
\[ \sin \alpha = \frac{1}{\sqrt{5}}, \quad \cos \alpha = \frac{2}{\sqrt{5}} \]
Using double angle identity:
\[ \cos(2\alpha) = 2\cos^2\alpha – 1 \]
\[ = 2\left(\frac{2}{\sqrt{5}}\right)^2 – 1 \]
\[ = 2 \cdot \frac{4}{5} – 1 = \frac{8}{5} – 1 = \frac{3}{5} \]
Hence,
\[ 2\alpha = \cos^{-1}\left(\frac{3}{5}\right) \Rightarrow \alpha = \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right) \]
Also,
\[ \sin(2\alpha) = 2\sin\alpha \cos\alpha \]
\[ = 2 \cdot \frac{1}{\sqrt{5}} \cdot \frac{2}{\sqrt{5}} = \frac{4}{5} \]
\[ 2\alpha = \sin^{-1}\left(\frac{4}{5}\right) \Rightarrow \alpha = \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right) \]
Therefore,
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right) = \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right) \]
Final Answer:
\[ \tan^{-1}\left(\frac{1}{4}\right) + \tan^{-1}\left(\frac{2}{9}\right) = \frac{1}{2}\cos^{-1}\left(\frac{3}{5}\right) = \frac{1}{2}\sin^{-1}\left(\frac{4}{5}\right) \]