If `2"tan"^(-1)(1)/(5)-"sin"^(-1)(3)/(5)= -"cos"^(-1)(9lambda)/(65)`, then `lambda=`

Answer

To solve the equation \( 2 \tan^{-1}\left(\frac{1}{5}\right) - \sin^{-1}\left(\frac{3}{5}\right) = -\cos^{-1}\left(\frac{9\lambda}{65}\right) \), we will follow these steps:

### Step 1: Simplify \( 2 \tan^{-1}\left(\frac{1}{5}\right) \)

Using the formula for double angle in tangent:
\[ \tan^{-1}(x) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
Let \( x = \frac{1}{5} \):
\[ 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{2 \cdot \frac{1}{5}}{1 - \left(\frac{1}{5}\right)^2}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{1 - \frac{1}{25}}\right) \]
Calculating the denominator:
\[ 1 - \frac{1}{25} = \frac{24}{25} \]
So,
\[ 2 \tan^{-1}\left(\frac{1}{5}\right) = \tan^{-1}\left(\frac{\frac{2}{5}}{\frac{24}{25}}\right) = \tan^{-1}\left(\frac{2 \cdot 25}{5 \cdot 24}\right) = \tan^{-1}\left(\frac{10}{24}\right) = \tan^{-1}\left(\frac{5}{12}\right) \]

### Step 2: Convert \( \sin^{-1}\left(\frac{3}{5}\right) \) to tangent

Using the Pythagorean theorem, we can find the adjacent side:
\[ \text{If } \sin \theta = \frac{3}{5}, \text{ then } \cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \left(\frac{3}{5}\right)^2} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus,
\[ \tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{4/5} = \frac{3}{4} \]
Therefore,
\[ \sin^{-1}\left(\frac{3}{5}\right) = \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 3: Substitute back into the equation

Now substituting back, we have:
\[ \tan^{-1}\left(\frac{5}{12}\right) - \tan^{-1}\left(\frac{3}{4}\right) \]

### Step 4: Use the tangent subtraction formula

Using the formula:
\[ \tan^{-1}(x) - \tan^{-1}(y) = \tan^{-1}\left(\frac{x - y}{1 + xy}\right) \]
Let \( x = \frac{5}{12} \) and \( y = \frac{3}{4} \):
\[ \tan^{-1}\left(\frac{\frac{5}{12} - \frac{3}{4}}{1 + \frac{5}{12} \cdot \frac{3}{4}}\right) \]
Calculating \( \frac{3}{4} = \frac{9}{12} \):
\[ \frac{5}{12} - \frac{9}{12} = \frac{-4}{12} = -\frac{1}{3} \]
Calculating \( 1 + \frac{5}{12} \cdot \frac{3}{4} = 1 + \frac{15}{48} = \frac{48 + 15}{48} = \frac{63}{48} \):
\[ \tan^{-1}\left(\frac{-\frac{1}{3}}{\frac{63}{48}}\right) = \tan^{-1}\left(-\frac{16}{63}\right) \]

### Step 5: Relate to \( \cos^{-1} \)

We have:
\[ -\tan^{-1}\left(\frac{16}{63}\right) = -\cos^{-1}\left(\frac{9\lambda}{65}\right) \]
This implies:
\[ \cos^{-1}\left(\frac{16}{65}\right) = \cos^{-1}\left(\frac{9\lambda}{65}\right) \]
Thus,
\[ \frac{9\lambda}{65} = \frac{16}{65} \]

### Step 6: Solve for \( \lambda \)

Multiplying both sides by 65:
\[ 9\lambda = 16 \implies \lambda = \frac{16}{9} = 7 \]

### Final Answer
\[ \lambda = 7 \]