Question

If for two real numbers \(a,b\) with \(|a|\le1\) and \(|b|\le1\), \[ \frac{1}{3}+\frac{\sin^{-1}a+\sin^{-1}b}{4}+\frac{(\sin^{-1}a+\sin^{-1}b)^{2}}{16}+\frac{(\sin^{-1}a+\sin^{-1}b)^{3}}{64}+\dots=\frac{2(8-3\pi)}{3(16+3\pi)}, \] then the value of \[ \sin^{-1}(a\sqrt{1-b^{2}}+b\sqrt{1-a^{2}}) \] is:

• A. $\frac{2(32+15\pi)}{3\pi-8}$
• B. $-\frac{\pi}{4}$
• C. $-\frac{3\pi}{4}$
• D. $\frac{1}{3}+\frac{\pi}{4}$

Hint:

Always look for patterns that let you substitute a single variable for a recurring expression. Grouping the term $(\sin^{-1}a + \sin^{-1}b)$ into a single variable $S$ upfront simplifies the algebra down to a basic linear equation.

Answer 1

The Correct Option is C

Concept: The left-hand side of the given equation can be modeled as an infinite geometric series. Let us substitute a single placeholder variable $S = \sin^{-1}a + \sin^{-1}b$, and apply the standard sum formula for an infinite geometric progression: $$S_{\infty} = \frac{\text{First term } (A)}{1 - \text{Common ratio } (R)}$$ Step 1: Evaluate the infinite geometric series expression.
Let us write out the terms of the series from the left side of our equation: $$\text{Series} = \frac{1}{3} + \left[ \frac{S}{4} + \frac{S^2}{16} + \frac{S^3}{64} + \dots \right]$$ Notice that starting from the second term, the expression forms an infinite geometric progression with a first term of $A = \frac{S}{4}$ and a common ratio of $R = \frac{S}{4}$. Applying our sum formula: $$\text{Series} = \frac{1}{3} + \frac{\frac{S}{4}}{1 - \frac{S}{4}} = \frac{1}{3} + \frac{S}{4 - S}$$ Combine these terms over a single common denominator: $$\text{Series} = \frac{(4 - S) + 3S}{3(4 - S)} = \frac{4 + 2S}{3(4 - S)} \quad \cdots (1)$$

Step 2: Equate the series expression to the given constant value.
Set our simplified expression from equation (1) equal to the fraction given on the right side of the problem: $$\frac{2(2 + S)}{3(4 - S)} = \frac{2(8 - 3\pi)}{3(16 + 3\pi)}$$ Cancel out the common fraction multiplier of $\frac{2}{3}$ from both sides of the equation: $$\frac{2 + S}{4 - S} = \frac{8 - 3\pi}{16 + 3\pi}$$

Step 3: Solve the equation for the variable $S$.
Cross-multiply the denominators to solve for our sum variable $S$: $$(2 + S)(16 + 3\pi) = (4 - S)(8 - 3\pi)$$ $$32 + 6\pi + S(16 + 3\pi) = 32 - 12\pi - S(8 - 3\pi)$$ Cancel out the constant 32 from both sides and collect all terms containing $S$ on the left: $$S(16 + 3\pi) + S(8 - 3\pi) = -12\pi - 6\pi$$ $$S(16 + 3\pi + 8 - 3\pi) = -18\pi \quad \Rightarrow \quad S(24) = -18\pi$$ $$S = -\frac{18\pi}{24} = -\frac{3\pi}{4}$$ This gives us the value of our sum variable: $\sin^{-1}a + \sin^{-1}b = -\frac{3\pi}{4}$.

Step 4: Apply the inverse sine addition identity.
Recall the standard addition identity for inverse sine functions: $$\sin^{-1}a + \sin^{-1}b = \sin^{-1}\left(a\sqrt{1-b^2} + b\sqrt{1-a^2}\right)$$ Notice that this matches the target expression requested by the problem statement exactly. Therefore, the value of the target expression is simply equal to our solved value for $S$: $$\text{Target Value} = -\frac{3\pi}{4}$$ This matches option (C) perfectly.