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

The core idea here is the complementary relationship between inverse sine and inverse cosine. For any number $t$ in $[-1,1]$, we have

This is because $\sin \theta = t$ and $\cos(\frac{\pi}{2} - \theta) = t$ — they are complementary angles.

The problem gives us one sum involving these functions with different arguments ($x$ and $y$), and asks for another sum with the arguments swapped. The natural move is to add the two expressions together and see what cancels.

Let’s denote:

and

We want $S_2$.

1. Add $S_1$ and $S_2$:

Rearranging:

2. Apply the complementary identity to each pair:

So:

3. Substitute the known value of $S_1$:

Therefore:

The key here is to simplify the messy fraction inside the inverse cotangent by rewriting $\sqrt{1 \pm \sin x}$ in a more manageable form. When you see $1 \pm \sin x$, think of the identity $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$ and the perfect square $(\sin\frac{x}{2} \pm \cos\frac{x}{2})^2 = 1 \pm \sin x$. That’s the whole trick — once you recognise the squares, the square roots become absolute values, and the range of $x$ tells you which sign to take.

Let’s walk through it.

1. Rewrite the square roots. Since $\sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$, we have

Therefore

2. Determine the signs using the given interval.

   For $x \in (0, \pi/4)$, we have $\frac{x}{2} \in (0, \pi/8)$. In this range, $\sin\frac{x}{2} > 0$ and $\cos\frac{x}{2} > 0$, and importantly $\cos\frac{x}{2} > \sin\frac{x}{2}$ (since $\tan\frac{x}{2} < 1$).

   Hence:

   • $\sin\frac{x}{2} + \cos\frac{x}{2} > 0$, so $\left|\sin\frac{x}{2} + \cos\frac{x}{2}\right| = \sin\frac{x}{2} + \cos\frac{x}{2}$.
   • $\sin\frac{x}{2} - \cos\frac{x}{2} < 0$, so $\left|\sin\frac{x}{2} - \cos\frac{x}{2}\right| = \cos\frac{x}{2} - \sin\frac{x}{2}$.

3. Substitute into the fraction.

   The numerator becomes

The denominator becomes

So the fraction inside the inverse cotangent is

4. Now the problem reduces to

Recall the identity: $\cot^{-1}(-t) = \pi - \cot^{-1}(t)$ for $t > 0$ (this holds because the principal value branch of $\cot^{-1}$ is $(0,\pi)$).

Here $\cot\frac{x}{2} > 0$ since $\frac{x}{2} \in (0,\pi/8)$. Therefore

5. Simplify further. Since $\frac{x}{2} \in (0,\pi/8) \subset (0,\pi)$, the principal value of $\cot^{-1}(\cot\theta)$ is exactly $\theta$ for $\theta \in (0,\pi)$. Hence

So the whole expression equals

The identity comes from the double-angle relation $\dfrac{2\tan\theta}{1+\tan^2\theta}=\sin 2\theta$, but $\sin^{-1}$ only returns its principal value in $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$, so the equality holds only on a restricted range.

1. Substitute $x=\tan\theta$, with $\theta=\tan^{-1}x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

So the left side is $\sin^{-1}(\sin 2\theta)$ and the equation reads $\sin^{-1}(\sin 2\theta)=2\theta$.

2. Apply the range condition. $\sin^{-1}(\sin y)=y$ holds exactly when $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Here $y=2\theta=2\tan^{-1}x$, so we need

3. Solve for $x$. Since $\tan^{-1}$ is increasing,

4. Check the boundary and exterior. At $x=1$: LHS $=\sin^{-1}(1)=\tfrac{\pi}{2}$, RHS $=2\tan^{-1}1=\tfrac{\pi}{2}$. At $x=-1$: both sides $=-\tfrac{\pi}{2}$. For $x=2>1$: LHS $=\sin^{-1}(0.8)\approx0.927$ while RHS $=2\tan^{-1}2\approx2.214$ are unequal, so the identity fails outside $[-1,1]$.

Step 1 — The identity that unlocks it.

For all $x \in \mathbb{R}$ (principal branches):

Step 2 — Split the expression to expose that identity.

This is the whole trick: pair the inverse functions off so only one of them survives, and its range is then easy to state.

Step 3 — Use the principal range of $\cot^{-1}$.

For every real $x$,

Step 4 — Bound the expression.

Adding $\pi$ throughout,

Step 5 — Match with the given inequality.

We are told

Comparing the two lower bounds and the two upper bounds:

Step 6 — Spot-check. At $x = 0$: $2\tan^{-1}0 + 3\cot^{-1}0 = 0 + 3\cdot\frac{\pi}{2} = \frac{3\pi}{2}$, which indeed lies in $(\pi, 2\pi)$ ✓. As $x \to \infty$ the value $\to \pi$, and as $x \to -\infty$ it $\to 2\pi$ — the bounds are approached but never reached ✓.

ANSWER: B

Concept & Intuition

When you see an inverse tangent of a rational function like $\frac{3-2x}{1+6x}$, your first instinct might be to use the quotient rule inside the chain rule — messy and error-prone. Instead, notice the structure: it looks exactly like the formula for $\tan^{-1} A - \tan^{-1} B$ because

Here, if we set $A = 3$ and $B = 2x$, then

which matches perfectly. This trick turns a complicated derivative into a trivial one.

Step-by-step solution

1. Recognize the pattern We have $y = \tan^{-1}\left(\frac{3 - 2x}{1 + 6x}\right)$. Compare with the identity:

Choose $a = 3$ and $b = 2x$. Then

So we can write:

2. Differentiate term by term The derivative of $\tan^{-1} 3$ is zero because it’s a constant. The derivative of $\tan^{-1}(2x)$ is given by the chain rule:

Therefore,

3. Match with the options The result $-\frac{2}{1+4x^2}$ corresponds exactly to option (C).

Step 1 — Rewrite the general term

For the $k$-th term of the series,

This matches the subtraction formula

with $A = k+1$ and $B = k$. Hence

Step 2 — Telescope the sum

Summing from $k=1$ to $k=n$:

since every intermediate term cancels, leaving only the last and first pieces.

Step 3 — Combine into a single inverse tangent

Using the subtraction formula again with $A = n+1$, $B=1$: