If `sin^(-1)(a-a^2/3+a^3/9-...)+cos^(-1)(1+b+b^2+...)=pi/2` then find `a` and `b`

Options

A. `b = (2a -3)/(3a)`
B. `b = (3a -2)/(2a)`
C. `a = (3)/(2 - 3b)`
D. `a = (2)/(3 - 2b)`

Explanation

To solve the equation

\[ \sin^{-1}\left(a - \frac{a^2}{3} + \frac{a^3}{9} - \ldots\right) + \cos^{-1}(1 + b + b^2 + \ldots) = \frac{\pi}{2}, \]

we will break it down into two parts: the series for \(\sin^{-1}\) and the series for \(\cos^{-1}\).

### Step 1: Analyze the series for \(\sin^{-1}\)

The series for \(\sin^{-1}\) can be recognized as an infinite geometric series:

\[ S = a - \frac{a^2}{3} + \frac{a^3}{9} - \ldots \]

This series has a first term \(a\) and a common ratio \(r = -\frac{a}{3}\). The sum of an infinite geometric series can be calculated using the formula:

\[ S = \frac{a}{1 - r} \]

Substituting the values, we have:

\[ S = \frac{a}{1 - \left(-\frac{a}{3}\right)} = \frac{a}{1 + \frac{a}{3}} = \frac{3a}{3 + a} \]

### Step 2: Analyze the series for \(\cos^{-1}\)

The series for \(\cos^{-1}\) is:

\[ T = 1 + b + b^2 + \ldots \]

This is also an infinite geometric series with first term \(1\) and common ratio \(b\). The sum is given by:

\[ T = \frac{1}{1 - b} \]

### Step 3: Substitute the sums into the equation

Now we substitute the sums back into the original equation:

\[ \sin^{-1}\left(\frac{3a}{3 + a}\right) + \cos^{-1}\left(\frac{1}{1 - b}\right) = \frac{\pi}{2} \]

Using the identity \(\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}\), we can conclude that:

\[ \frac{3a}{3 + a} = \frac{1}{1 - b} \]

### Step 4: Set up the equation

From the above, we can equate:

\[ 3a(1 - b) = 3 + a \]

Expanding this gives:

\[ 3a - 3ab = 3 + a \]

Rearranging leads to:

\[ 2a - 3ab = 3 \]

### Step 5: Solve for \(b\)

Rearranging for \(b\):

\[ b = \frac{2a - 3}{3a} \]

### Step 6: Find values for \(a\) and \(b\)

To find specific values for \(a\) and \(b\), we can use a trial method. Let's try \(b = \frac{1}{2}\):

Substituting \(b = \frac{1}{2}\) into the equation for \(b\):

\[ \frac{1}{2} = \frac{2a - 3}{3a} \]

Cross-multiplying gives:

\[ 3a \cdot \frac{1}{2} = 2a - 3 \]

This simplifies to:

\[ \frac{3a}{2} = 2a - 3 \]

Multiplying through by \(2\) to eliminate the fraction:

\[ 3a = 4a - 6 \]

Rearranging gives:

\[ a = 6 \]

### Step 7: Find \(b\) using \(a\)

Now substituting \(a = 6\) back into the equation for \(b\):

\[ b = \frac{2(6) - 3}{3(6)} = \frac{12 - 3}{18} = \frac{9}{18} = \frac{1}{2} \]

### Final Result

Thus, the values of \(a\) and \(b\) are:

\[ \boxed{a = 6, b = \frac{1}{2}} \]