Question

Find the value of \[ \tan\!\left[\left(2\sin^{-1}\frac{2}{\sqrt{13}}\right)-2\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right)\right] \]

• A. \(\dfrac{31}{56}\)
• B. \(\dfrac{29}{56}\)
• C. \(\dfrac{33}{56}\)
• D. \(\dfrac{37}{56}\)

Hint:

When inverse trigonometric functions appear inside \(\tan(\cdot)\), first convert them into simple angles and use \(\tan(2\theta)\) identities before applying \(\tan(A-B)\).

Answer 1

The Correct Option is C

To find the value of \(\tan\!\left[\left(2\sin^{-1}\frac{2}{\sqrt{13}}\right)-2\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right)\right]\), we need to use the properties of inverse trigonometric functions and some trigonometric identities.

1. Define \(\sin^{-1}\frac{2}{\sqrt{13}} = \theta\). Then, \(\sin \theta = \frac{2}{\sqrt{13}}\).
2. Using the identity \(\cos \theta = \sqrt{1 - \sin^2 \theta}\), compute \(\cos \theta\): \(\cos \theta = \sqrt{1 - \left(\frac{2}{\sqrt{13}}\right)^2} = \sqrt{\frac{13}{13} - \frac{4}{13}} = \sqrt{\frac{9}{13}} = \frac{3}{\sqrt{13}}\).
3. Double angle identity for sine: \(\sin(2\theta) = 2\sin \theta \cos \theta\): \(\sin(2\theta) = 2 \times \frac{2}{\sqrt{13}} \times \frac{3}{\sqrt{13}} = \frac{12}{13}\).
4. Double angle identity for cosine: \(\cos(2\theta) = 2\cos^2 \theta - 1\): \(\cos(2\theta) = 2\left(\frac{3}{\sqrt{13}}\right)^2 - 1 = 2 \times \frac{9}{13} - 1 = \frac{18}{13} - \frac{13}{13} = \frac{5}{13}\).
5. Define \(\cos^{-1}\frac{3}{\sqrt{10}} = \phi\). Then, \(\cos \phi = \frac{3}{\sqrt{10}}\).
6. Using the identity \(\sin \phi = \sqrt{1 - \cos^2 \phi}\), compute \(\sin \phi\): \(\sin \phi = \sqrt{1 - \left(\frac{3}{\sqrt{10}}\right)^2} = \sqrt{\frac{10}{10} - \frac{9}{10}} = \frac{1}{\sqrt{10}}\).
7. Double angle identity for sine: \(\sin(2\phi) = 2\sin \phi \cos \phi\): \(\sin(2\phi) = 2 \times \frac{1}{\sqrt{10}} \times \frac{3}{\sqrt{10}} = \frac{6}{10} = \frac{3}{5}\).
8. Double angle identity for cosine: \(\cos(2\phi) = 2\cos^2 \phi - 1\): \(\cos(2\phi) = 2\left(\frac{3}{\sqrt{10}}\right)^2 - 1 = \frac{18}{10} - \frac{10}{10} = \frac{8}{10} = \frac{4}{5}\).
9. Now compute \(\tan(A - B)\) using the identity: \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\). Here, \(A = 2\theta\) and \(B = 2\phi\). We have:
   • \(\tan A = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{\frac{12}{13}}{\frac{5}{13}} = \frac{12}{5}\).
   • \(\tan B = \frac{\sin(2\phi)}{\cos(2\phi)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}\).
10. Substitute into the formula: \[ \begin{align*} \tan(A - B) & = \frac{\frac{12}{5} - \frac{3}{4}}{1 + \frac{12}{5} \cdot \frac{3}{4}} \\ & = \frac{\frac{48}{20} - \frac{15}{20}}{1 + \frac{36}{20}} \\ & = \frac{\frac{33}{20}}{\frac{56}{20}} \\ & = \frac{33}{56}. \end{align*} \]
11. Thus, the correct answer is \(\frac{33}{56}\).