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

Step 1: Evaluate the first inverse trigonometric term Let \[ A=\sin^{-1}\!\left(\frac{2}{\sqrt{13}}\right) \] Then, \[ \sin A=\frac{2}{\sqrt{13}},\quad \cos A=\frac{3}{\sqrt{13}},\quad \tan A=\frac{2}{3} \] Using the identity: \[ \tan 2A=\frac{2\tan A}{1-\tan^2 A} \] \[ \tan 2A=\frac{2\cdot\frac23}{1-\left(\frac23\right)^2} =\frac{\frac{4}{3}}{1-\frac49} =\frac{\frac{4}{3}}{\frac59} =\frac{12}{5} \]
Step 2: Evaluate the second inverse trigonometric term Let \[ B=\cos^{-1}\!\left(\frac{3}{\sqrt{10}}\right) \] Then, \[ \cos B=\frac{3}{\sqrt{10}},\quad \sin B=\frac{1}{\sqrt{10}},\quad \tan B=\frac{1}{3} \] \[ \tan 2B=\frac{2\tan B}{1-\tan^2 B} =\frac{\frac{2}{3}}{1-\frac19} =\frac{\frac{2}{3}}{\frac89} =\frac{3}{4} \]
Step 3: Use the identity for \(\tan(\alpha-\beta)\) \[ \tan(2A-2B)=\frac{\tan2A-\tan2B}{1+\tan2A\tan2B} \] \[ =\frac{\frac{12}{5}-\frac{3}{4}}{1+\frac{12}{5}\cdot\frac{3}{4}} =\frac{\frac{48-15}{20}}{\frac{20+36}{20}} =\frac{33}{56} \] Final Answer: \[ \boxed{\dfrac{33}{56}} \]