Question

For \( \alpha, \beta, \gamma \neq 0 \). If \( \sin^{-1} \alpha + \sin^{-1} \beta + \sin^{-1} \gamma = \pi \) and \( (\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3 \alpha \beta \), then \( \gamma \) is equal to

• A. \( \frac{\sqrt{3}}{2} \)
• B. \( \frac{1}{\sqrt{2}} \)
• C. \( \frac{\sqrt{3} - 1}{2\sqrt{2}} \)
• D. \( \sqrt{3} \)

Answer 1

The Correct Option is A

To solve the problem, we need to find the value of \( \gamma \) given the equations:

1. \(\sin^{-1} \alpha + \sin^{-1} \beta + \sin^{-1} \gamma = \pi\)
2. \((\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3 \alpha \beta\)

Let's work through these step-by-step:

Step 1: Using the range of \(\sin^{-1}\)

The sum of the inverse sine functions equals \(\pi\). Since the range of \(\sin^{-1} x\) is \([-\frac{\pi}{2}, \frac{\pi}{2}]\), we deduce that it is possible if:

• \(\alpha, \beta, \gamma \in [1, -1]\).
• One of \(\alpha\), \(\beta\), or \(\gamma\) is either equal to 1 or -1 for the sum to equal \(\pi\).

Step 2: Using trigonometric identity for calculation

If \(\alpha = 1\), then \(\sin^{-1} \alpha = \frac{\pi}{2}\). Therefore:

• \(\sin^{-1} \beta + \sin^{-1} \gamma = \frac{\pi}{2}\)

Step 3: Verify the values using the given equation

Apply the identity where \(\sin^{-1} \beta + \sin^{-1} \gamma = \frac{\pi}{2}\) implies \(\beta = \cos \theta\) and \(\gamma = \cos \left(\frac{\pi}{2} - \theta\right) = \sin \theta\).

The equation given is:

• \((\alpha + \beta + \gamma)(\alpha - \gamma + \beta) = 3 \alpha \beta\)

Substitute values:

• \((1 + \cos \theta + \sin \theta)(1 - \sin \theta + \cos \theta) = 3 \cdot 1 \cdot \cos \theta\)

Expanding the left-hand side, we get:

• \((1 + \cos \theta + \sin \theta)(1 + \cos \theta - \sin \theta) = 1 + \cos^2 \theta - \sin^2 \theta + 2 \cos \theta - \sin \theta^2\)

Simplifying using trigonometrical identity \((\cos^2 \theta + \sin^2 \theta = 1)\), the expression becomes \(1 + 2 \cos \theta\).

Equate it to the right-hand side:

• \(1 + 2 \cos \theta = 3 \cos \theta\)

Simplifying gives:

• \(1 = \cos \theta\)

Thus after substituting back for \(\gamma\):\(\gamma = \sin \theta = \frac{\sqrt{3}}{2}\)

Conclusion

Thus, the value of \(\gamma\) is \(\frac{\sqrt{3}}{2}\).

Answer 2

Let $\sin^{-1} \alpha = A$, $\sin^{-1} \beta = B$, $\sin^{-1} \gamma = C$

$A + B + C = \pi$

$(\alpha + \beta)^2 - \gamma^2 = 3 \alpha \beta$

$\alpha^2 + \beta^2 - \gamma^2 = \alpha \beta$

$\frac{\alpha^2 + \beta^2 - \gamma^2}{2 \alpha \beta} = \frac{1}{2}$

$\Rightarrow \cos C = \frac{1}{2}$

$\sin C = \gamma$

$\cos C = \sqrt{1 - \gamma^2} = \frac{1}{2}$

$\gamma = \frac{\sqrt{3}}{2}$