Question

Considering the principal values of inverse trigonometric functions, the value of \[ \tan\!\left(2\sin^{-1}\!\frac{2}{\sqrt{13}}-2\cos^{-1}\!\frac{3}{\sqrt{10}}\right) \] is equal to:

• A. \(\dfrac{33}{56}\)
• B. \(-\dfrac{33}{56}\)
• C. \(\dfrac{16}{63}\)
• D. \(-\dfrac{16}{63}\)

Hint:

Always convert inverse trigonometric expressions into basic ratios before applying multiple-angle identities.

Answer 1

The Correct Option is A

To solve the given problem, we need to find the value of \( \tan\!\left(2\sin^{-1}\!\frac{2}{\sqrt{13}}-2\cos^{-1}\!\frac{3}{\sqrt{10}}\right) \) using inverse trigonometric functions and their properties.

1. First, let's simplify the terms inside the tangent function, starting with \( 2\sin^{-1}\!\left(\frac{2}{\sqrt{13}}\right) \).
   • Let \( \theta = \sin^{-1}\!\left(\frac{2}{\sqrt{13}}\right) \), which means \(\sin\theta = \frac{2}{\sqrt{13}}\).
   • Using the identity \(\sin^2\theta + \cos^2\theta = 1\), we find \(\cos\theta = \sqrt{1 - \left(\frac{2}{\sqrt{13}}\right)^2} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}\).
   The double angle identity for sine gives us: \[ \sin(2\theta) = 2 \sin\theta \cos\theta = 2 \times \frac{2}{\sqrt{13}} \times \frac{3}{\sqrt{13}} = \frac{12}{13} \]
2. Next, calculate \( 2\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right) \).
   • Let \( \phi = \cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right) \), which means \(\cos\phi = \frac{3}{\sqrt{10}}\).
   • Again, using the identity \(\sin^2\phi + \cos^2\phi = 1\), we find \(\sin\phi = \sqrt{1 - \left(\frac{3}{\sqrt{10}}\right)^2} = \sqrt{\frac{1}{10}} = \frac{1}{\sqrt{10}}\).
   The double angle identity for cosine gives us: \[ \cos(2\phi) = 2\cos^2\phi - 1 = 2\left(\frac{3}{\sqrt{10}}\right)^2 - 1 = \frac{18}{10} - 1 = \frac{8}{10} = \frac{4}{5} \]
3. Now, use the difference formula for tangent: \[ \tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b} \] Here, \(a = 2\sin^{-1}\!\left(\frac{2}{\sqrt{13}}\right)\) and \(b = 2\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right)\). We have: \[ \tan a = \frac{\sin 2\theta}{\cos 2\theta} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5} \] \(\tan b\) can be given by: \[ \tan b = \frac{\sin 2\phi}{\cos 2\phi} = \frac{\sqrt{1 - \left(\frac{4}{5}\right)^2}}{\frac{4}{5}} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4} \] Finally, we substitute these values into the tangent difference formula: \[ \tan\left(a - b\right) = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \times \frac{3}{4}} = \frac{\frac{48}{20} - \frac{15}{20}}{1 + \frac{36}{20}} = \frac{\frac{33}{20}}{\frac{56}{20}} = \frac{33}{56} \]

Hence, the value of \( \tan\left(2\sin^{-1}\frac{2}{\sqrt{13}} - 2\cos^{-1}\frac{3}{\sqrt{10}}\right) \) is \(\frac{33}{56}\). This matches the given correct answer.