Q85 / 155 · WBJEE 2026 · Mathematics
If for two real numbers $a, b$ with $|a| \leq 1$ and $|b| \leq 1$, $\dfrac{1}{3} + \dfrac{\sin^{-1}a + \sin^{-1}b}{4} + \dfrac{(\sin^{-1}a + \sin^{-1}b)^2}{16} + \dfrac{(\sin^{-1}a + \sin^{-1}b)^3}{64} + \cdots = \dfrac{2(8 - 3\pi)}{3(16 + 3\pi)}$, then the value of $\sin^{-1}\left(a\sqrt{1 - b^2} + b\sqrt{1 - a^2}\right)$ is
- A$\dfrac{2(32 + 15\pi)}{3\pi - 8}$
- Correct answerB$-\dfrac{\pi}{4}$
- C$-\dfrac{3\pi}{4}$
- D$\dfrac{1}{3} + \dfrac{\pi}{4}$
Solution
Let $x = \dfrac{\sin^{-1}a + \sin^{-1}b}{4}$.
The given equation can be written as an infinite geometric progression for the terms after the first:
$\dfrac{1}{3} + x + x^2 + x^3 + \cdots = \dfrac{2(8 - 3\pi)}{3(16 + 3\pi)}$
Summing the infinite geometric series (where $|x| < 1$), we get:
$\dfrac{1}{3} + \dfrac{x}{1 - x} = \dfrac{16 - 6\pi}{3(16 + 3\pi)}$
$\dfrac{1 - x + 3x}{3(1 - x)} = \dfrac{16 - 6\pi}{3(16 + 3\pi)}$
$\dfrac{1 + 2x}{1 - x} = \dfrac{16 - 6\pi}{16 + 3\pi}$
Cross-multiplying to solve for $x$:
$(1 + 2x)(16 + 3\pi) = (1 - x)(16 - 6\pi)$
$16 + 3\pi + 32x + 6\pi x = 16 - 6\pi - 16x + 6\pi x$
$48x = -9\pi \Rightarrow x = -\dfrac{3\pi}{16}$
Since $x = \dfrac{\sin^{-1}a + \sin^{-1}b}{4}$, we have:
$\sin^{-1}a + \sin^{-1}b = 4 \left(-\dfrac{3\pi}{16}\right) = -\dfrac{3\pi}{4}$
We need to find the value of $\sin^{-1}\left(a\sqrt{1 - b^2} + b\sqrt{1 - a^2}\right)$.
Let $\alpha = \sin^{-1}a$ and $\beta = \sin^{-1}b$. Then $a = \sin\alpha$ and $b = \sin\beta$.
The expression becomes:
$\sin^{-1}(\sin\alpha \cos\beta + \cos\alpha \sin\beta) = \sin^{-1}(\sin(\alpha + \beta))$
Substitute $\alpha + \beta = -\dfrac{3\pi}{4}$:
$\sin^{-1}\left(\sin\left(-\dfrac{3\pi}{4}\right)\right) = \sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$
Answer: $-\dfrac{\pi}{4}$
The given equation can be written as an infinite geometric progression for the terms after the first:
$\dfrac{1}{3} + x + x^2 + x^3 + \cdots = \dfrac{2(8 - 3\pi)}{3(16 + 3\pi)}$
Summing the infinite geometric series (where $|x| < 1$), we get:
$\dfrac{1}{3} + \dfrac{x}{1 - x} = \dfrac{16 - 6\pi}{3(16 + 3\pi)}$
$\dfrac{1 - x + 3x}{3(1 - x)} = \dfrac{16 - 6\pi}{3(16 + 3\pi)}$
$\dfrac{1 + 2x}{1 - x} = \dfrac{16 - 6\pi}{16 + 3\pi}$
Cross-multiplying to solve for $x$:
$(1 + 2x)(16 + 3\pi) = (1 - x)(16 - 6\pi)$
$16 + 3\pi + 32x + 6\pi x = 16 - 6\pi - 16x + 6\pi x$
$48x = -9\pi \Rightarrow x = -\dfrac{3\pi}{16}$
Since $x = \dfrac{\sin^{-1}a + \sin^{-1}b}{4}$, we have:
$\sin^{-1}a + \sin^{-1}b = 4 \left(-\dfrac{3\pi}{16}\right) = -\dfrac{3\pi}{4}$
We need to find the value of $\sin^{-1}\left(a\sqrt{1 - b^2} + b\sqrt{1 - a^2}\right)$.
Let $\alpha = \sin^{-1}a$ and $\beta = \sin^{-1}b$. Then $a = \sin\alpha$ and $b = \sin\beta$.
The expression becomes:
$\sin^{-1}(\sin\alpha \cos\beta + \cos\alpha \sin\beta) = \sin^{-1}(\sin(\alpha + \beta))$
Substitute $\alpha + \beta = -\dfrac{3\pi}{4}$:
$\sin^{-1}\left(\sin\left(-\dfrac{3\pi}{4}\right)\right) = \sin^{-1}\left(-\dfrac{1}{\sqrt{2}}\right) = -\dfrac{\pi}{4}$
Answer: $-\dfrac{\pi}{4}$
Hence, the correct answer is option: 'B'.