lf `asin^-1 x -bcos^-1 x=c,`then `asin^-1 x +bcos^-1` equal to

Options

A. 0
B. `(pi ab + c (b -a))/(a + b)`
C. `(pi)/(2)`
D. `(pi ab + c (a -b))/(a + b)`

Explanation

To solve the equation \( a \sin^{-1} x - b \cos^{-1} x = c \) and find \( a \sin^{-1} x + b \cos^{-1} x \), we can follow these steps:

### Step 1: Write down the given equation
We start with the equation:
\[ a \sin^{-1} x - b \cos^{-1} x = c \]

### Step 2: Use the identity for inverse trigonometric functions
We know that:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
From this, we can express \( \cos^{-1} x \) in terms of \( \sin^{-1} x \):
\[ \cos^{-1} x = \frac{\pi}{2} - \sin^{-1} x \]

### Step 3: Substitute \( \cos^{-1} x \) in the original equation
Substituting this into the original equation gives:
\[ a \sin^{-1} x - b \left( \frac{\pi}{2} - \sin^{-1} x \right) = c \]
Expanding this, we have:
\[ a \sin^{-1} x - \frac{b\pi}{2} + b \sin^{-1} x = c \]

### Step 4: Combine like terms
Combining the \( \sin^{-1} x \) terms:
\[ (a + b) \sin^{-1} x - \frac{b\pi}{2} = c \]

### Step 5: Solve for \( \sin^{-1} x \)
Rearranging gives:
\[ (a + b) \sin^{-1} x = c + \frac{b\pi}{2} \]
Thus, we can express \( \sin^{-1} x \) as:
\[ \sin^{-1} x = \frac{c + \frac{b\pi}{2}}{a + b} \]

### Step 6: Find \( a \sin^{-1} x + b \cos^{-1} x \)
Now, we can find \( a \sin^{-1} x + b \cos^{-1} x \):
Using the identity for \( \cos^{-1} x \):
\[ b \cos^{-1} x = b \left( \frac{\pi}{2} - \sin^{-1} x \right) \]
So,
\[ a \sin^{-1} x + b \cos^{-1} x = a \sin^{-1} x + b \left( \frac{\pi}{2} - \sin^{-1} x \right) \]
This simplifies to:
\[ = a \sin^{-1} x + \frac{b\pi}{2} - b \sin^{-1} x \]
\[ = (a - b) \sin^{-1} x + \frac{b\pi}{2} \]

### Step 7: Substitute \( \sin^{-1} x \) back into the equation
Substituting \( \sin^{-1} x \) from Step 5:
\[ = (a - b) \left( \frac{c + \frac{b\pi}{2}}{a + b} \right) + \frac{b\pi}{2} \]
This leads to:
\[ = \frac{(a - b)(c + \frac{b\pi}{2})}{a + b} + \frac{b\pi}{2} \]

### Step 8: Combine the terms
To combine the terms, we can express \( \frac{b\pi}{2} \) in terms of a common denominator:
\[ = \frac{(a - b)(c + \frac{b\pi}{2}) + \frac{b\pi}{2}(a + b)}{a + b} \]
This gives us the final expression for \( a \sin^{-1} x + b \cos^{-1} x \).

### Final Result
Thus, the value of \( a \sin^{-1} x + b \cos^{-1} x \) is:
\[ \frac{(a - b)c + \frac{(a + b)b\pi}{2}}{a + b} \]