Differentiate w.r.t. $x$: $\tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right),\ -\dfrac{1}{\sqrt{3}} < \dfrac{x}{a} < \dfrac{1}{\sqrt{3}}$.

Differentiate w.r.t. x: tan⁻¹((3a² x - x³)/(a³ - 3 – Class 12 Mathematics

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Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The inverse tangent function obeys a family of addition and doubling identities that let you combine two arctangents into one. They come straight from the tangent addition formula, but they carry conditions you must respect.

The core addition identity

Start from $\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ . Put $A=\tan^{-1}x$ and $B=\tan^{-1}y$ , so $\tan A=x$ and $\tan B=y$ . Then

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\!\left(\frac{x+y}{1-xy}\right), \qquad xy<1.

The restriction $xy<1$ keeps the combined angle inside the principal range $(-\pi/2,\pi/2)$ .

Watch out

If $xy>1$ the raw formula lands in the wrong branch, so you must correct it:

$\tan^{-1}x+\tan^{-1}y = \pi + \tan^{-1}\!\left(\frac{x+y}{1-xy}\right) \ (x,y>0),$

and $-\pi+\tan^{-1}(\cdot)$ when $x,y<0$ . Ignoring this is the classic exam slip.

Subtraction

Replacing $y$ with $-y$ gives

\tan^{-1}x - \tan^{-1}y = \tan^{-1}\!\left(\frac{x-y}{1+xy}\right), \qquad xy>-1.

The doubling identity

Set $y=x$ in the addition formula:

2\tan^{-1}x = \tan^{-1}\!\left(\frac{2x}{1-x^2}\right), \qquad -1<x<1.

The same angle can also be rewritten through sine and cosine, which is handy in integration and in proofs -- but each alternate form only matches $2\tan^{-1}x$ on part of its domain, so the two forms carry different conditions:

2\tan^{-1}x = \sin^{-1}\!\left(\frac{2x}{1+x^2}\right), \qquad -1\le x\le 1,

2\tan^{-1}x = \cos^{-1}\!\left(\frac{1-x^2}{1+x^2}\right), \qquad x\ge 0.

Watch out

The $\cos^{-1}$ form needs $x\ge 0$ -- it fails for negative $x$ . Check $x=-1$ : $2\tan^{-1}(-1)=2\left(-\tfrac{\pi}{4}\right)=-\tfrac{\pi}{2}$ , but $\cos^{-1}\!\left(\tfrac{1-1}{1+1}\right)=\cos^{-1}(0)=\tfrac{\pi}{2}$ , the wrong sign entirely. The $\sin^{-1}$ form has no such restriction because $\sin^{-1}$ (unlike $\cos^{-1}$ ) can return a negative angle.

The complementary identity

For every real $x$ ,

\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.

This holds without restriction because $\tan^{-1}$ and $\cot^{-1}$ of the same value are complementary angles. …

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The given expression simplifies to $3\tan^{-1}\left(\frac{x}{a}\right)$ using the inverse tangent identity for triple angles, so its derivative is $\frac{3a}{a^2 + x^2}$ .

We start with the function

y = \tan^{-1}\left(\frac{3a^2 x - x^3}{a^3 - 3a x^2}\right)

and the condition $-\frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$ .

The key insight is that the fraction inside the inverse tangent resembles the formula for $\tan 3\theta$ in terms of $\tan \theta$ . Recall:

\tan 3\theta = \frac{3\tan\theta - \tan^3\theta}{1 - 3\tan^2\theta}

If we set $\tan\theta = \frac{x}{a}$ , then

\tan 3\theta = \frac{3\left(\frac{x}{a}\right) - \left(\frac{x}{a}\right)^3}{1 - 3\left(\frac{x}{a}\right)^2} = \frac{\frac{3x}{a} - \frac{x^3}{a^3}}{1 - \frac{3x^2}{a^2}} = \frac{\frac{3a^2 x - x^3}{a^3}}{\frac{a^2 - 3x^2}{a^2}} = \frac{3a^2 x - x^3}{a^3 - 3a x^2}

That’s exactly the argument of the inverse tangent. So

y = \tan^{-1}\big(\tan(3\theta)\big)

where $\theta = \tan^{-1}\left(\frac{x}{a}\right)$ .

Now, the identity $\tan^{-1}(\tan \alpha) = \alpha$ holds only when $\alpha$ lies in the principal branch $(-\pi/2, \pi/2)$ . Here $\alpha = 3\theta = 3\tan^{-1}(x/a)$ . The given condition $-\frac{1}{\sqrt{3}} < \frac{x}{a} < \frac{1}{\sqrt{3}}$ ensures that $\tan^{-1}(x/a)$ lies between $-\pi/6$ and $\pi/6$ , so $3\tan^{-1}(x/a)$ lies between $-\pi/2$ and $\pi/2$ . Perfect — we are safely inside the principal range.

Watch out

A common mistake is to forget the range condition. Without it, $\tan^{-1}(\tan 3\theta)$ might equal $3\theta - \pi$ or $3\theta + \pi$ , changing the derivative. Always check the interval.

Thus, …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 39 on this concept.

CBSE 2020Set 65/2/11 markMCQ
$\tan^{-1} 3 + \tan^{-1}\lambda = \tan^{-1}\left(\dfrac{3+\lambda}{1-3\lambda}\right)$
›Reveal solutionSolution
The inverse tangent addition formula $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$ holds only when $xy < 1$ . Here $x=3$ , $y=\lambda$ , so the condition is $3\lambda < 1$ , i.e. $\lambda < \frac{1}{3}$ . The correct option is (C).

The formula you’ve written —
$\tan^{-1} 3 + \tan^{-1}\lambda = \tan^{-1}\left(\frac{3+\lambda}{1-3\lambda}\right)$
— is the standard inverse tangent addition identity, but it comes with a hidden condition. Many students apply it blindly, and that’s where mistakes happen.

Let’s understand why the condition exists.

The core idea: the range of $\tan^{-1}$ and the product condition

Recall that $\tan^{-1} x$ (also written $\arctan x$ ) gives an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$ . So the sum of two such angles, $\tan^{-1}3 + \tan^{-1}\lambda$ , lies in $(-\pi, \pi)$ .

The formula
$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy}$
is derived from the tangent addition formula:
$\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$
If we set $A = \tan^{-1}x$ , $B = \tan^{-1}y$ , then $\tan(A+B) = \frac{x+y}{1-xy}$ .

But here’s the catch: $\tan^{-1}\frac{x+y}{1-xy}$ always gives an angle in $(-\frac{\pi}{2}, \frac{\pi}{2})$ . So the equality holds only when $A+B$ itself lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ .

When does $A+B$ stay inside that interval? It turns out the cleanest condition is $xy < 1$ .
$\tan^{-1}x + \tan^{-1}y = \tan^{-1}\frac{x+y}{1-xy} \quad \text{if and only if} \quad xy < 1$
If $xy = 1$ , the denominator is zero — the formula breaks. If $xy > 1$ , then $A+B$ falls outside $(-\frac{\pi}{2}, \frac{\pi}{2})$ , and the right-hand side would give a different principal value (you’d need to add or subtract $\pi$ ).

Applying it to this problem

Here $x = 3$ and $y = \lambda$ . So the condition for the formula to be valid is:

Write the product condition:

$xy < 1 \quad\Rightarrow\quad 3\lambda < 1$

Solve for $\lambda$ :

$\lambda < \frac{1}{3}$

That’s it. No further restrictions — $\lambda$ can be any real number less than $\frac{1}{3}$ . …
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CBSE 2026Set A1 markMCQ
$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) =$
(a) $\frac{\pi}{3}$
(b) $\frac{\pi}{6}$
(c) $-\frac{\pi}{3}$
(d) $-\frac{\pi}{6}$
›Reveal solutionSolution

$\tan^{-1}\left(-\frac{1}{\sqrt{3}}\right) = -\frac{\pi}{6}$ .

The principal value of $\tan^{-1}$ lies in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ .

We need the angle $\theta$ in this range with $\tan\theta = -\frac{1}{\sqrt{3}}$ .

…

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CBSE 2026Set A1 markMCQ
$2\tan^{-1}\frac{1}{3} =$
(a) $\tan^{-1}\frac{3}{2}$
(b) $\tan^{-1}\frac{3}{4}$
(c) $\tan^{-1}\frac{4}{3}$
(d) $\tan^{-1}\frac{2}{3}$
›Reveal solutionSolution

$2\tan^{-1}\frac{1}{3} = \tan^{-1}\frac{3}{4}$ .

Use the double-angle identity valid for $|x|<1$ :

$2\tan^{-1}x = \tan^{-1}\frac{2x}{1-x^2}.$

Here $x = \frac{1}{3}$ : …

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CBSE 2026Set A1 markMCQ
$x \in R,\ \cot(\tan^{-1}x + \cot^{-1}x) =$
(a) $1$
(b) $\frac{1}{2}$
(c) $0$
(d) $\frac{1}{3}$
›Reveal solutionSolution

$\cot\left(\tan^{-1}x + \cot^{-1}x\right) = \cot\frac{\pi}{2} = 0$ .

For all real $x$ , the complementary identity gives

$\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}.$

Therefore …

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CBSE 2026Set A1 markMCQ
$\tan^{-1}2 + \tan^{-1}3 =$
(a) $-\frac{\pi}{4}$
(b) $\frac{\pi}{4}$
(c) $\frac{3\pi}{4}$
(d) $\pi$
›Reveal solutionSolution

$\tan^{-1}2 + \tan^{-1}3 = \frac{3\pi}{4}$ .

Use the addition formula. With $a=2,\ b=3$ we have $ab=6>1$ , so

$\tan^{-1}a + \tan^{-1}b = \pi + \tan^{-1}\frac{a+b}{1-ab}.$

Compute:

$\frac{a+b}{1-ab} = \frac{5}{1-6} = \frac{5}{-5} = -1.$

So …

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CBSE 2026Set A1 markMCQ
$\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} =$
(a) $-\frac{3\pi}{4}$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{4}$
(d) $\frac{\pi}{3}$
›Reveal solutionSolution

$\tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} = \frac{\pi}{4}$ .

Use $\tan^{-1}a - \tan^{-1}b = \tan^{-1}\frac{a-b}{1+ab}$ with $a=\frac{x}{y}$ , $b=\frac{x-y}{x+y}$ .

Numerator:

$a-b = \frac{x}{y} - \frac{x-y}{x+y} = \frac{x(x+y) - y(x-y)}{y(x+y)} = \frac{x^2+xy-xy+y^2}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

Denominator: …

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CBSE 2026Set A1 markMCQ
$|x| \le 1,\ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) =$
(a) $2\cos^{-1}x$
(b) $2\sin^{-1}x$
(c) $2\tan^{-1}x$
(d) $\tan^{-1}2x$
›Reveal solutionSolution

$\cos^{-1}\frac{1-x^2}{1+x^2} = 2\tan^{-1}x$ (for $0\le x\le 1$ ).

Put $x = \tan\theta$ , so $\theta = \tan^{-1}x$ . Then

$\frac{1-x^2}{1+x^2} = \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta.$

Therefore …

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CBSE 2025Set E1 markMCQ
$\cot^{-1}\left(\tan\frac{\pi}{7}\right) =$
(a) $\frac{\pi}{7}$
(b) $\frac{5\pi}{14}$
(c) $\frac{9\pi}{14}$
(d) $\frac{3\pi}{14}$
›Reveal solutionSolution

Convert the tangent to a cotangent using complementary angles; the answer is $\frac{5\pi}{14}$ .

Use $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$ :

$\tan\frac{\pi}{7}=\cot\left(\frac{\pi}{2}-\frac{\pi}{7}\right)=\cot\frac{7\pi-2\pi}{14}=\cot\frac{5\pi}{14}.$

…

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CBSE 2025Set E1 markMCQ
$\tan^{-1}(-\sqrt{3}) =$
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(c) $\frac{2\pi}{3}$
(d) $-\frac{\pi}{3}$
›Reveal solutionSolution

$\tan^{-1}$ is an odd function with principal range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ ; the value is $-\frac{\pi}{3}$ .

Since $\tan\frac{\pi}{3}=\sqrt{3}$ and $\tan^{-1}(-x)=-\tan^{-1}x$ , …

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CBSE 2025Set E1 markMCQ
$\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3}) =$
(a) $0$
(b) $-\frac{\pi}{2}$
(c) $\pi$
(d) $\frac{\pi}{2}$
›Reveal solutionSolution

Evaluate each inverse function in its principal range and subtract; result $-\frac{\pi}{2}$ .

First, $\tan^{-1}(\sqrt{3})=\frac{\pi}{3}$ .

For $\cot^{-1}(-\sqrt{3})$ , the principal range of $\cot^{-1}$ is $(0,\pi)$ . We need $\cot\theta=-\sqrt{3}$ with $\theta\in(0,\pi)$ . Since $\cot\frac{\pi}{6}=\sqrt{3}$ ,

$\cot^{-1}(-\sqrt{3})=\pi-\frac{\pi}{6}=\frac{5\pi}{6}.$

…

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CBSE 2025Set E1 markMCQ
$\tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} =$
(a) $\pi$
(b) $\frac{\pi}{4}$
(c) $\frac{\pi}{2}$
(d) $\frac{\pi}{3}$
›Reveal solutionSolution

Use the sum formula for inverse tangents; the sum is $\frac{\pi}{4}$ .

When $xy<1$ , $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\dfrac{x+y}{1-xy}$ . Here $xy=\frac16<1$ , so …

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CBSE 2025Set E1 markMCQ
$\tan\left\{\frac{1}{2}\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right)\right\} =$
(a) $1$
(b) $\sqrt{3}$
(c) $0$
(d) $\infty$
›Reveal solutionSolution

$\tan^{-1}x + \tan^{-1}\frac1x = \frac{\pi}{2}$ ; half is $\frac{\pi}{4}$ ; $\tan\frac{\pi}{4} = 1$ .

For $x > 0$ there is a standard identity:

$\tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2}.$

Taking half: …

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Q.Differentiate w.r.t. $x$ : $\tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right),\ -\dfrac{1}{\sqrt{3}} < \dfrac{x}{a} < \dfrac{1}{\sqrt{3}}$ .

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Method: Trigonometric Substitution to Exploit the $\tan 3\theta$ Identity

Steps

Common Mistakes

The core idea: the range of $\tan^{-1}$ and the product condition

Applying it to this problem

Q.Differentiate w.r.t. xxx: tan⁡−1(3a2x−x3a3−3ax2), −13<xa<13\tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right),\ -\dfrac{1}{\sqrt{3}} < \dfrac{x}{a} < \dfrac{1}{\sqrt{3}}tan−1(a3−3ax23a2x−x3​), −3​1​<ax​<3​1​.

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Q.Differentiate w.r.t. $x$ : $\tan^{-1}\left(\dfrac{3a^2 x - x^3}{a^3 - 3ax^2}\right),\ -\dfrac{1}{\sqrt{3}} < \dfrac{x}{a} < \dfrac{1}{\sqrt{3}}$ .

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