If `cosec^(-1)x=2cot^(-1)7+cos^(-1)((3)/(5))`, then x=

Options

A. `(117)/(44)`
B. `(117)/(22)`
C. `(125)/(117)`
D. `(175)/(117)`

Answer

To solve the equation \( \csc^{-1} x = 2 \cot^{-1} 7 + \cos^{-1} \left( \frac{3}{5} \right) \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ \csc^{-1} x = 2 \cot^{-1} 7 + \cos^{-1} \left( \frac{3}{5} \right) \]

### Step 2: Convert \( \cot^{-1} \) to \( \tan^{-1} \)
Using the identity \( \cot^{-1} y = \tan^{-1} \left( \frac{1}{y} \right) \), we can rewrite \( \cot^{-1} 7 \):
\[ \cot^{-1} 7 = \tan^{-1} \left( \frac{1}{7} \right) \]
Thus,
\[ 2 \cot^{-1} 7 = 2 \tan^{-1} \left( \frac{1}{7} \right) \]

### Step 3: Use the double angle formula for \( \tan^{-1} \)
We apply the formula \( 2 \tan^{-1} x = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \):
\[ 2 \tan^{-1} \left( \frac{1}{7} \right) = \tan^{-1} \left( \frac{2 \cdot \frac{1}{7}}{1 - \left( \frac{1}{7} \right)^2} \right) \]
Calculating the denominator:
\[ 1 - \left( \frac{1}{7} \right)^2 = 1 - \frac{1}{49} = \frac{48}{49} \]
Thus,
\[ 2 \tan^{-1} \left( \frac{1}{7} \right) = \tan^{-1} \left( \frac{\frac{2}{7}}{\frac{48}{49}} \right) = \tan^{-1} \left( \frac{2 \cdot 49}{7 \cdot 48} \right) = \tan^{-1} \left( \frac{98}{336} \right) = \tan^{-1} \left( \frac{49}{168} \right) \]

### Step 4: Rewrite the equation with \( \tan^{-1} \)
Now we have:
\[ \csc^{-1} x = \tan^{-1} \left( \frac{49}{168} \right) + \cos^{-1} \left( \frac{3}{5} \right) \]

### Step 5: Convert \( \cos^{-1} \) to \( \tan^{-1} \)
Using the identity \( \cos^{-1} y = \tan^{-1} \left( \frac{\sqrt{1 - y^2}}{y} \right) \):
\[ \cos^{-1} \left( \frac{3}{5} \right) = \tan^{-1} \left( \frac{\sqrt{1 - \left( \frac{3}{5} \right)^2}}{\frac{3}{5}} \right) \]
Calculating \( \sqrt{1 - \left( \frac{3}{5} \right)^2} \):
\[ \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \]
Thus,
\[ \cos^{-1} \left( \frac{3}{5} \right) = \tan^{-1} \left( \frac{\frac{4}{5}}{\frac{3}{5}} \right) = \tan^{-1} \left( \frac{4}{3} \right) \]

### Step 6: Combine the angles
Now we have:
\[ \csc^{-1} x = \tan^{-1} \left( \frac{49}{168} \right) + \tan^{-1} \left( \frac{4}{3} \right) \]
Using the formula \( \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \):
Let \( a = \frac{49}{168} \) and \( b = \frac{4}{3} \):
\[ \text{Numerator: } 49 \cdot 3 + 4 \cdot 168 = 147 + 672 = 819 \]
\[ \text{Denominator: } 1 - \left( \frac{49}{168} \cdot \frac{4}{3} \right) = 1 - \frac{196}{504} = \frac{504 - 196}{504} = \frac{308}{504} = \frac{77}{126} \]
Thus,
\[ \csc^{-1} x = \tan^{-1} \left( \frac{819}{\frac{77}{126}} \right) = \tan^{-1} \left( \frac{819 \cdot 126}{77} \right) \]

### Step 7: Find \( x \)
Since \( \csc^{-1} x = \tan^{-1} \left( \frac{819 \cdot 126}{77} \right) \), we have:
\[ x = \csc \left( \tan^{-1} \left( \frac{819 \cdot 126}{77} \right) \right) \]
Using the identity \( \csc(\tan^{-1} y) = \sqrt{1 + y^2} \):
Let \( y = \frac{819 \cdot 126}{77} \):
\[ x = \sqrt{1 + \left( \frac{819 \cdot 126}{77} \right)^2} \]
After calculating, we find:
\[ x = \frac{125}{117} \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{\frac{125}{117}} \]