Q.The value of $\sin(2\tan^{-1}(0.75))$ is equal to
(A) $0.75$
(B) $1.5$
(C) $0.96$
(D) $\sin 1.5$

Concept: Inverse Tangent Identity — use the double-angle formula for sine in terms of $\tan \theta$, then substitute $\theta = \tan^{-1}(0.75)$.

Let $\theta = \tan^{-1}(0.75)$, so $\tan \theta = 0.75 = \frac{3}{4}$.

We need $\sin(2\theta)$. Recall:

Substitute $\tan\theta = \frac{3}{4}$: …

When you see an expression like $\sin(2\tan^{-1}x)$, the instinct should be to reach for a formula that connects sine of a double angle directly to the tangent of the original angle. Why? Because the inverse tangent gives you an angle whose tangent you know exactly — here, $\tan\theta = 0.75$. But sine of twice that angle isn't immediately obvious from just the tangent value. You could draw a right triangle, find the hypotenuse, then compute $\sin\theta$ and $\cos\theta$, and finally use $\sin 2\theta = 2\sin\theta\cos\theta$. That works, but the identity below is faster and avoids square roots entirely.

This identity comes from writing $\sin 2\theta = 2\sin\theta\cos\theta$ and then dividing numerator and denominator by $\cos^2\theta$:

It's valid whenever $\cos\theta \neq 0$, which is true here since $\tan\theta = 0.75$ is finite.

Now let's apply it step by step.

1. Set the angle.

   Let $\theta = \tan^{-1}(0.75)$. By definition, $\tan\theta = 0.75$ and $\theta$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ — specifically in the first quadrant since $0.75 > 0$.

2. Plug into the double-angle identity.

   We want $\sin(2\theta)$. Using the formula:

3. Compute the numerator and denominator.

   Numerator: $2 \times 0.75 = 1.5$.

   Denominator: $1 + (0.75)^2 = 1 + 0.5625 = 1.5625$.

4. Divide.

You can verify: $1.5625 \times 0.96 = 1.5$, so it's exact. …

Method: Evaluating $\sin$ (or $\cos$) of a multiple of an inverse-tangent angle

For $\sin(2\tan^{-1}t)$ and similar, use a double-angle identity written directly in terms of $\tan\theta$, so you never need square roots or a triangle.

Steps

Step 1: Name the angle.

Let $\theta=\tan^{-1}t$, so $\tan\theta=t$ and $\theta\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. The target is $\sin 2\theta$.

Step 2: Use the tangent-form double-angle identity.

Instead of $\sin2\theta=2\sin\theta\cos\theta$, use the version that depends only on $\tan\theta$:

This follows from dividing $2\sin\theta\cos\theta$ by $\sin^2\theta+\cos^2\theta=1$ and then by $\cos^2\theta$. …

Common Mistakes

Mistake 1: Treating $\sin(2\tan^{-1}0.75)$ as $2\sin(\tan^{-1}0.75)$.

Why it's wrong: the $2$ multiplies the angle, not the sine; sine is not linear. Correct approach: use the double-angle identity $\sin 2\theta=\tfrac{2\tan\theta}{1+\tan^2\theta}$.

Mistake 2: Answering $1.5$ (just $2\times 0.75$) or $0.75$ (the tangent itself). …