Q.Show that $\tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right)=\frac{4-\sqrt7}{3}$ and justify why the other value $\frac{4+\sqrt7}{3}$ is ignored.

Concept: Half-angle formula for tangent in terms of sine, and range analysis for inverse trigonometric functions.

Step 1: Let $\theta = \sin^{-1}\frac{3}{4}$, so $\sin\theta = \frac{3}{4}$ and $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Since $\frac{3}{4}>0$, $\theta$ is in the first quadrant, so $\cos\theta = \sqrt{1-\sin^2\theta} = \sqrt{1-\frac{9}{16}} = \frac{\sqrt7}{4}$.

Step 2: Use the half-angle formula for tangent:

$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} = \frac{1-\frac{\sqrt7}{4}}{\frac{3}{4}} = \frac{4-\sqrt7}{3}.$ …

Concept and Intuition

When you see an expression like $\tan\left(\frac{1}{2}\sin^{-1}\frac{3}{4}\right)$, the natural instinct is to work from the inside out. Let the inverse sine produce an angle — call it $\theta$ — so that $\sin\theta = \frac{3}{4}$. Then the problem reduces to finding $\tan(\theta/2)$.

The half-angle formula for tangent is your best friend here. There are several forms, but the one that avoids square roots in the denominator is:

This formula is derived from $\tan(\theta/2) = \sin\theta/(1+\cos\theta)$ and its conjugate, and it's particularly clean when you already know $\sin\theta$.

The twist: when you solve, you'll get two possible numeric values because the algebra involves a square root. But only one of them corresponds to the actual angle. The angle $\frac{1}{2}\sin^{-1}\frac{3}{4}$ is half of an acute angle (since $\sin^{-1}(3/4)$ is acute), so it must also be acute — hence its tangent is positive. That's why we discard the larger, positive-but-invalid value.

Step-by-Step Solution

1. Set up the substitution.

   Let $\theta = \sin^{-1}\frac{3}{4}$. Then $\sin\theta = \frac{3}{4}$, and by definition $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Since $\frac{3}{4} > 0$, $\theta$ is in the first quadrant: $0 < \theta < \frac{\pi}{2}$.

2. Find $\cos\theta$.

   Using $\sin^2\theta + \cos^2\theta = 1$:

Since $\theta$ is acute, $\cos\theta > 0$, so:

3. Apply the half-angle formula for tangent. Use the form $\tan\frac{\theta}{2} = \frac{1 - \cos\theta}{\sin\theta}$. Substitute the known values:

This gives the required result directly.

4. Why is the other value $\frac{4+\sqrt{7}}{3}$ ignored? The alternative half-angle formula $\tan\frac{\theta}{2} = \frac{\sin\theta}{1 + \cos\theta}$ would give:

Rationalising: $\frac{3}{4+\sqrt{7}} \cdot \frac{4-\sqrt{7}}{4-\sqrt{7}} = \frac{3(4-\sqrt{7})}{16-7} = \frac{4-\sqrt{7}}{3}$, same result.

But where does $\frac{4+\sqrt{7}}{3}$ come from? If you had used the formula $\tan\frac{\theta}{2} = \pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}$, the square root would produce both signs:

Rationalising the inside: $\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}} = \frac{4-\sqrt{7}}{\sqrt{16-7}} = \frac{4-\sqrt{7}}{3}$. So the positive root gives $\frac{4-\sqrt{7}}{3}$, and the negative root gives $-\frac{4-\sqrt{7}}{3}$, not $\frac{4+\sqrt{7}}{3}$. …

Method: Tangent of half an inverse-sine angle

This method handles any expression of the form $\tan\left(\tfrac{1}{2}\sin^{-1}k\right)$ (or $\tfrac12\cos^{-1}k$): let the inverse function define a single angle, find its cosine, then apply a half-angle formula while letting the quadrant fix the sign.

Steps

Step 1: Name the inner angle and pin down its quadrant.

Set $\theta = \sin^{-1}k$, so $\sin\theta = k$ and, by definition of the principal branch, $\theta \in \left[-\tfrac{\pi}{2}, \tfrac{\pi}{2}\right]$. The sign of $k$ tells you the quadrant. This step is what removes all ambiguity later — the inverse function has already chosen one specific angle for you.

Step 2: Get $\cos\theta$ with the correct sign.

Use the Pythagorean identity, and pick the sign from the quadrant found in Step 1:

Because $\sin^{-1}$ returns an angle in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, $\cos\theta$ is always $\ge 0$ here — take the positive root.

Step 3: Apply a sign-safe half-angle formula.

Prefer the form that avoids a $\pm$ ambiguity:

$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}.$ …

Common Mistakes

Mistake 1: Taking $\cos\theta=\pm\tfrac{\sqrt7}{4}$ and keeping the negative root.

Why it's wrong: $\theta=\sin^{-1}\tfrac34$ is in $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$, where cosine is never negative. Correct approach: for a positive sine argument, $\theta$ is acute, so $\cos\theta=+\tfrac{\sqrt7}{4}$; the negative choice is what wrongly produces $\tfrac{4+\sqrt7}{3}$.

Mistake 2: Using the $\pm$ square-root half-angle form and not resolving the sign. …