Q.Prove that $2\sin^{-1} \dfrac{3}{5} = \tan^{-1} \dfrac{24}{7}$.

Concept: Inverse Tangent Identity – We convert the left side into a tangent form using the double-angle formula for sine, then simplify to match the right side.

Step 1: Let $\theta = \sin^{-1} \frac{3}{5}$. Then $\sin \theta = \frac{3}{5}$, and $\cos \theta = \sqrt{1 - \frac{9}{25}} = \frac{4}{5}$ (positive since $\theta$ is acute).

Step 2: The left side is $2\theta$. Compute $\tan(2\theta)$ using the double-angle identity:

Here $\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{4/5} = \frac{3}{4}$.

Step 3: Substitute:

Since $2\theta$ lies in $(0, \pi)$ and $\tan(2\theta) = \frac{24}{7}$ with $2\theta$ acute, we have $2\theta = \tan^{-1}\frac{24}{7}$.

The core idea is that inverse trigonometric identities often become algebraic when you take a trigonometric function of both sides. Here, the left side is twice an inverse sine. If we let $\theta = \sin^{-1}\frac{3}{5}$, then $\sin\theta = \frac{3}{5}$ and we want to show $2\theta = \tan^{-1}\frac{24}{7}$. Taking the tangent of $2\theta$ and simplifying should give $\frac{24}{7}$, provided $2\theta$ lies in the principal range of $\tan^{-1}$.

Let’s walk through it.

1. Set up the substitution.

   Let $\theta = \sin^{-1}\frac{3}{5}$. Then $\sin\theta = \frac{3}{5}$ and, since $\sin^{-1}$ returns an angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\theta \in [0, \frac{\pi}{2}]$ (because $\frac{3}{5} > 0$). So $\theta$ is acute.

2. Find $\cos\theta$.

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

Since $\theta$ is acute, $\cos\theta > 0$, so $\cos\theta = \frac{4}{5}$.

3. Compute $\tan\theta$.

4. Apply the double-angle formula for tangent.

5. Check the range to confirm the equality. We have $\tan(2\theta) = \frac{24}{7}$. But $\tan^{-1}$ returns an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$. Is $2\theta$ in that interval? Since $\theta = \sin^{-1}\frac{3}{5} \approx 0.6435$ rad, $2\theta \approx 1.287$ rad, which is less than $\frac{\pi}{2} \approx 1.571$ rad. So $2\theta$ lies in $(0, \frac{\pi}{2})$, the principal range of $\tan^{-1}$. Therefore,

which is exactly $2\sin^{-1}\frac{3}{5} = \tan^{-1}\frac{24}{7}$.

Method: Proving an inverse-trig identity by taking a trig function of both sides

Use this general strategy to prove statements like $2\sin^{-1}a = \tan^{-1}b$.

Steps

Step 1: Let one side equal an angle.

Set $\theta$ equal to the inner inverse term, so a known ratio (here $\sin\theta$) is given. Deduce the other ratios from a right triangle or a Pythagorean identity, minding the sign from the principal range.

Step 2: Apply the trig function that matches the target side.

To reach a $\tan^{-1}$ target, compute $\tan$ of the left side using a double-angle formula, e.g.

Simplify to the number appearing on the right.

Step 3: Verify the angle lies in the target's principal range.

Equal tangents only give equal angles when both sit in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Estimate the angle numerically to confirm; only then conclude the two sides are equal.

Common Mistakes

Mistake 1: Concluding the identity from equal tangents alone.

Why it's wrong: $\tan(2\theta) = \frac{24}{7}$ does not by itself give $2\theta = \tan^{-1}\frac{24}{7}$ — that needs $2\theta$ inside $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Correct approach: verify $2\theta = 2\sin^{-1}\frac{3}{5} \approx 1.29$ rad is below $\frac{\pi}{2}$, then conclude.

Mistake 2: Taking $\cos\theta$ negative.

Why it's wrong: $\theta = \sin^{-1}\frac{3}{5}$ lies in $\left[0, \frac{\pi}{2}\right]$, where cosine is positive, so $\cos\theta = +\frac{4}{5}$. Correct approach: choose the positive root from the principal range, giving $\tan\theta = \frac{3}{4}$.