Question

\(50 tan\) \(\left (3tan^{-1} (\frac{1}{2}) + 2 cos^{-1} (\frac{1}{√5})\right) + 4\sqrt2 tan \left (\frac{1}{2} tan^{-1} (2\sqrt2)\right)\) is equal to ____.

Answer 1

Correct Answer: 29

Let's solve the given expression: \(50 \tan(3\tan^{-1}(\frac{1}{2}) + 2\cos^{-1}(\frac{1}{\sqrt{5}})) + 4\sqrt{2} \tan(\frac{1}{2} \tan^{-1}(2\sqrt{2}))\).

Step 1: Simplify \(\tan^{-1}(\frac{1}{2})\) and \(\cos^{-1}(\frac{1}{\sqrt{5}})\)

\(\theta = \tan^{-1}(\frac{1}{2}) \Rightarrow \tan \theta = \frac{1}{2}\)

\(\phi = \cos^{-1}(\frac{1}{\sqrt{5}}) \Rightarrow \cos \phi = \frac{1}{\sqrt{5}}\), and \(\sin \phi = \sqrt{1 - \frac{1}{5}} = \frac{2}{\sqrt{5}}\)

Step 2: Use tangent and cosine addition formulas

For \(\tan (3\theta)\), use: \(\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}\)

Substituting \(\tan \theta = \frac{1}{2}\), find:

\(\tan 3\theta = \frac{3(\frac{1}{2}) - (\frac{1}{2})^3}{1 - 3(\frac{1}{2})^2} = \frac{\frac{3}{2} - \frac{1}{8}}{1 - \frac{3}{4}} = \frac{\frac{12 - 1}{8}}{\frac{1}{4}} = \frac{\frac{11}{8}}{\frac{1}{4}} = \frac{11}{2}\)

For \(\cos(2\phi)\), use: \(\cos 2\phi = 2\cos^2 \phi - 1\)

Substituting \(\cos \phi = \frac{1}{\sqrt{5}}\), we find:

\(\cos 2\phi = 2(\frac{1}{\sqrt{5}})^2 - 1 = 2(\frac{1}{5}) - 1 = \frac{2}{5} - 1 = -\frac{3}{5}\)

Now use the tangent addition formula:

\(\tan 3\theta + 2\phi = \frac{\tan 3\theta + \tan 2\phi}{1 - \tan 3\theta \tan 2\phi}\)

Substitute respective tangent values:

\(\tan 2\phi = \frac{2\ast\frac{1}{\sqrt{5}}\ast\frac{2}{\sqrt{5}}}{1-(\frac{2}{\sqrt{5}})^2} = \frac{\frac{4}{5}}{1-\frac{4}{5}} = 4\)

Now calculate:

\(\frac{\frac{11}{2} + 4}{1 - \frac{11}{2}\ast 4} = \frac{\frac{19}{2}}{1 - 22} = -\frac{19}{44}\)

Step 3: Calculate \(\tan(\frac{1}{2} \tan^{-1}(2\sqrt{2}))\)

\(\alpha = \tan^{-1}(2\sqrt{2}) \Rightarrow \tan \alpha = 2\sqrt{2}\)

Use \(\tan(\frac{\alpha}{2}) = \frac{\sin \alpha}{1 + \cos \alpha}\), with \(\sin \alpha = \frac{2\sqrt{2}}{\sqrt{9}} = \frac{2\sqrt{2}}{3}\) and \(\cos \alpha = \frac{1}{\sqrt{9}} = \frac{1}{3}\)

\(\tan(\frac{\alpha}{2}) = \frac{\frac{2\sqrt{2}}{3}}{1 + \frac{1}{3}} = \frac{\frac{2\sqrt{2}}{3}}{\frac{4}{3}} = \frac{2\sqrt{2}}{4} = \frac{\sqrt{2}}{2}\)

Step 4: Combine calculations

Calculate the total expression:

\(50(-\frac{19}{44}) + 4\sqrt{2}(\frac{\sqrt{2}}{2})\)

\(= 50 \ast -\frac{19}{44} + 4 \ast 1 = -\frac{950}{44} + 4 = -\frac{475}{22} + 4 = -21.59 + 4 = 29\)

Conclusion

The computed value is 29, which fits the specified range (29,29).