Prove that 2sin⁻¹(3/5) = tan⁻¹(24/7)

Prove that \(2\sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right)\)

Solution:

Let

\[ \theta = \sin^{-1}\left(\frac{3}{5}\right) \]

Then,

\[ \sin \theta = \frac{3}{5} \]

Consider a right triangle:

So,

\[ \cos \theta = \frac{4}{5} \]

Now, we use the identity:

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

First find \(\tan \theta\):

\[ \tan \theta = \frac{3}{4} \]

Now,

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

\[ = \frac{\frac{6}{4}}{1 – \frac{9}{16}} \]

\[ = \frac{\frac{6}{4}}{\frac{7}{16}} \]

\[ = \frac{6}{4} \times \frac{16}{7} \]

\[ = \frac{24}{7} \]

Thus,

\[ \tan(2\theta) = \frac{24}{7} \]

Taking inverse tangent:

\[ 2\theta = \tan^{-1}\left(\frac{24}{7}\right) \]

But \(\theta = \sin^{-1}(3/5)\), hence:

\[ 2\sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right) \]

\[ 2\sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{24}{7}\right) \]

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