If `2"tan"^(-1)(1)/(5)-"sin"^(-1)(3)/(5)= -"cos"^(-1)(63)/(lambda`, then `lambda=`

Answer

To solve the equation \( 2 \tan^{-1} \left( \frac{1}{5} \right) - \sin^{-1} \left( \frac{3}{5} \right) = -\cos^{-1} \left( \frac{63}{\lambda} \right) \), we will follow these steps:

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

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

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

We can visualize this using a right triangle where the opposite side is 3 and the hypotenuse is 5. By the Pythagorean theorem:
\[ \text{Adjacent} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \]
Thus, we have:
\[ \tan \theta = \frac{3}{4} \quad \Rightarrow \quad \theta = \tan^{-1} \left( \frac{3}{4} \right) \]

### Step 3: Substitute into the equation

Now substituting back into the equation:
\[ \tan^{-1} \left( \frac{5}{12} \right) - \tan^{-1} \left( \frac{3}{4} \right) = -\cos^{-1} \left( \frac{63}{\lambda} \right) \]

### Step 4: Use the tangent subtraction formula

Using the formula:
\[ \tan^{-1}(A) - \tan^{-1}(B) = \tan^{-1} \left( \frac{A - B}{1 + AB} \right) \]
Let \( A = \frac{5}{12} \) and \( B = \frac{3}{4} \):
\[ \tan^{-1} \left( \frac{\frac{5}{12} - \frac{3}{4}}{1 + \frac{5}{12} \cdot \frac{3}{4}} \right) \]
Finding a common denominator for \( \frac{5}{12} - \frac{3}{4} \):
\[ \frac{5}{12} - \frac{9}{12} = \frac{-4}{12} = -\frac{1}{3} \]
Calculating \( 1 + \frac{5 \cdot 3}{12 \cdot 4} = 1 + \frac{15}{48} = \frac{48 + 15}{48} = \frac{63}{48} \)

Thus:
\[ \tan^{-1} \left( \frac{-\frac{1}{3}}{\frac{63}{48}} \right) = \tan^{-1} \left( -\frac{48}{189} \right) = -\tan^{-1} \left( \frac{48}{189} \right) \]

### Step 5: Relate to \( -\cos^{-1} \left( \frac{63}{\lambda} \right) \)

We have:
\[ -\tan^{-1} \left( \frac{48}{189} \right) = -\cos^{-1} \left( \frac{63}{\lambda} \right) \]
This implies:
\[ \tan^{-1} \left( \frac{48}{189} \right) = \cos^{-1} \left( \frac{63}{\lambda} \right) \]

### Step 6: Find \( \lambda \)

Using the identity \( \cos \theta = \frac{adjacent}{hypotenuse} \):
In the triangle formed, we know:
\[ \cos \theta = \frac{63}{\sqrt{63^2 + 48^2}} = \frac{63}{\sqrt{3969 + 2304}} = \frac{63}{\sqrt{6273}} = \frac{63}{\sqrt{63^2 + 48^2}} = \frac{63}{65} \]
Thus:
\[ \lambda = 65 \]

### Final Answer
\[ \lambda = 65 \]