Question

Considering only the principal values of the inverse trigonometric functions, the value of \(\tan(\sin^{-1}(\frac{3}{5})-2\cos^{-1}(\frac{2}{\sqrt5}))\) is

• A. \(\frac{7}{24}\)
• B. \(\frac{-7}{24}\)
• C. \(\frac{-5}{24}\)
• D. \(\frac{5}{24}\)

Answer 1

The Correct Option is B

Let \[ 2 \cos^{-1} \left( \frac{2}{\sqrt{5}} \right) = \theta. \]

Step 1: Expressing in Terms of Half-Angle 

\[ \frac{2}{\sqrt{5}} = \cos \frac{\theta}{2} \]

Therefore, \[ \tan \frac{\theta}{2} = \frac{1}{2}. \]

Step 2: Calculating Individual Trigonometric Functions

\[ \tan \sin^{-1} \left( \frac{3}{5} \right) = \frac{3}{4}, \quad \tan \cos^{-1} \left( \frac{2}{\sqrt{5}} \right) = \frac{4}{3}. \]

Step 3: Applying the Tangent Subtraction Formula

\[ \tan \left( \sin^{-1} \frac{3}{5} - \cos^{-1} \frac{2}{\sqrt{5}} \right) = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \frac{3}{4} \cdot \frac{4}{3}} \]

Final Calculation

\[ = \frac{\frac{9}{12} - \frac{16}{12}}{1 + \frac{12}{12}} \] \[ = \frac{-7}{24} \]

Answer 2

To solve the problem, we evaluate the expression involving inverse trigonometric functions and find the value of:

\[ \tan \left( \sin^{-1}\left(\frac{3}{5}\right) - 2 \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \right) \]

1. Let:
\[ A = \sin^{-1}\left(\frac{3}{5}\right) \] \[ B = \cos^{-1}\left(\frac{2}{\sqrt{5}}\right) \]

2. Find \(\sin A\) and \(\cos A\):
\[ \sin A = \frac{3}{5} \] \[ \cos A = \sqrt{1 - \sin^2 A} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]

3. Find \(\cos B\) and \(\sin B\):
\[ \cos B = \frac{2}{\sqrt{5}} \] \[ \sin B = \sqrt{1 - \cos^2 B} = \sqrt{1 - \left(\frac{2}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{4}{5}} = \sqrt{\frac{1}{5}} = \frac{1}{\sqrt{5}} \]

4. Calculate \(\tan(2B)\):
Using double angle formula:
\[ \tan(2B) = \frac{2 \tan B}{1 - \tan^2 B} \] First find \(\tan B\):
\[ \tan B = \frac{\sin B}{\cos B} = \frac{\frac{1}{\sqrt{5}}}{\frac{2}{\sqrt{5}}} = \frac{1}{2} \] Then, \[ \tan(2B) = \frac{2 \times \frac{1}{2}}{1 - \left(\frac{1}{2}\right)^2} = \frac{1}{1 - \frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3} \]

5. Calculate \(\tan(A - 2B)\):
Using subtraction formula:
\[ \tan (A - 2B) = \frac{\tan A - \tan(2B)}{1 + \tan A \tan(2B)} \] Calculate \(\tan A\):
\[ \tan A = \frac{\sin A}{\cos A} = \frac{3/5}{4/5} = \frac{3}{4} \] Now, \[ \tan (A - 2B) = \frac{\frac{3}{4} - \frac{4}{3}}{1 + \frac{3}{4} \times \frac{4}{3}} = \frac{\frac{9}{12} - \frac{16}{12}}{1 + 1} = \frac{-\frac{7}{12}}{2} = -\frac{7}{24} \]

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