$\tan^{-1} 3 + \tan^{-1}\lambda = \tan^{-1}\left(\dfrac{3+\lambda}{1-3\lambda}\right)$ is valid for what values of $\lambda$? (A) $\lambda \in \left(-\dfrac{1}{3},\ \dfrac{1}{3}\right)$ (B) $\lambda > \dfrac{1}{3}$ (C) $\lambda < \dfrac{1}{3}$ (D) All real values of $\lambda$

tan⁻¹ 3 + tan⁻¹λ = tan⁻¹((3+λ)/(1-3λ)) is valid for wha — CBSE Class 12 Mathematics 2020 PYQ

Question

\tan^{-1} 3 + \tan^{-1}\lambda = \tan^{-1}\left(\dfrac{3+\lambda}{1-3\lambda}\right)

CBSECBSE Class XII Board 2020MCQ· 1mImportance★★★★★

✓ Free question

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.

Tip

For $x>0$ : $\tan^{-1}\dfrac{1}{x} = \cot^{-1}x = \dfrac{\pi}{2}-\tan^{-1}x$ . For $x<0$ : $\tan^{-1}\dfrac{1}{x} = -\dfrac{\pi}{2}-\tan^{-1}x$ , but this is not the same as $\cot^{-1}x$ -- since $\cot^{-1}x$ always lies in $(0,\pi)$ (never negative), for $x<0$ it instead equals $\pi+\tan^{-1}\dfrac{1}{x}$ . Check $x=-1$ : $\cot^{-1}(-1)=\tfrac{3\pi}{4}$ , while $\tan^{-1}\dfrac{1}{-1}=-\tfrac{\pi}{4}$ -- these clearly are not equal, so never carry the $x>0$ shortcut over to negative $x$ .

The takeaway

Every inverse-tangent identity is the tangent addition formula read backwards. Learn the addition rule and its $xy$ conditions, then the subtraction, doubling, and complementary forms follow -- but always check the domain restriction on each alternate form before quoting it, since $\sin^{-1}$ , $\cos^{-1}$ and $\cot^{-1}$ each carry their own principal-range limits. That sign condition is where marks are won or lost.

The addition, subtraction, and doubling identities for tan⁻¹x are an important part of the CBSE Class 12 Inverse Trigonometric Functions chapter, and "tan inverse x plus tan inverse y formula with conditions" is a frequently searched topic because of the easy-to-miss xy conditions involved. These identities are tested regularly in both CBSE board exams and JEE Main inverse trigonometry problems.

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 $\frac{1}{3}$ .

Watch out

A common mistake is to also consider the denominator $1-3\lambda \neq 0$ , i.e. $\lambda \neq \frac{1}{3}$ . But that’s already covered: $3\lambda < 1$ excludes $\lambda = \frac{1}{3}$ automatically. The real pitfall is forgetting the product condition entirely and assuming the formula works for all $\lambda$ .

Matching with the options

(A) $\lambda \in \left(-\frac{1}{3}, \frac{1}{3}\right)$ — too restrictive; $\lambda$ can be much smaller than $-\frac{1}{3}$ and the formula still holds.
(B) $\lambda > \frac{1}{3}$ — exactly the opposite of the condition.
(C) $\lambda < \frac{1}{3}$ — correct.
(D) All real values — false, as shown.

Tip

Quick check: try $\lambda = 0$ . Then LHS = $\tan^{-1}3 + 0$ , RHS = $\tan^{-1}3$ — works. Try $\lambda = 1$ . Then LHS = $\tan^{-1}3 + \frac{\pi}{4} \approx 1.249 + 0.785 = 2.034$ rad, but RHS = $\tan^{-1}\frac{4}{-2} = \tan^{-1}(-2) \approx -1.107$ rad — clearly not equal. So the condition is real.

✓Final answer

\lambda < \frac{1}{3}

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 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 $\tan\theta = -\frac{1}{\sqrt{3}}$ .

Since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$ and $\tan$ is odd, $\tan\left(-\frac{\pi}{6}\right) = -\frac{1}{\sqrt{3}}$ .

And $-\frac{\pi}{6} \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ , so it is the principal value.

✓Final answer
$-\frac{\pi}{6}$
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 $x = \frac{1}{3}$ :

$\frac{2\cdot\frac{1}{3}}{1-\frac{1}{9}} = \frac{\frac{2}{3}}{\frac{8}{9}} = \frac{2}{3}\cdot\frac{9}{8} = \frac{18}{24} = \frac{3}{4}.$

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

✓Final answer
$\tan^{-1}\frac{3}{4}$
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

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

✓Final answer
$0$
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

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

✓Final answer
$\frac{3\pi}{4}$
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:

$1+ab = 1 + \frac{x}{y}\cdot\frac{x-y}{x+y} = \frac{y(x+y)+x(x-y)}{y(x+y)} = \frac{xy+y^2+x^2-xy}{y(x+y)} = \frac{x^2+y^2}{y(x+y)}.$

So the ratio is $1$ , giving

$\tan^{-1}(1) = \frac{\pi}{4}.$

✓Final answer
$\frac{\pi}{4}$
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

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

✓Final answer
$2\tan^{-1}x$
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 $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$ :

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

Since $\frac{5\pi}{14}$ lies in the principal range $(0,\pi)$ of $\cot^{-1}$ ,

$\cot^{-1}\left(\tan\frac{\pi}{7}\right)=\cot^{-1}\left(\cot\frac{5\pi}{14}\right)=\frac{5\pi}{14}.$

✓Final answer
$\frac{5\pi}{14}$
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 $\tan^{-1}(-x)=-\tan^{-1}x$ ,

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

This lies in the principal range, so it is the required value.

✓Final answer
$-\frac{\pi}{3}$
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 $\cot\frac{\pi}{6}=\sqrt{3}$ ,

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

Therefore

$\tan^{-1}(\sqrt{3})-\cot^{-1}(-\sqrt{3})=\frac{\pi}{3}-\frac{5\pi}{6}=\frac{2\pi-5\pi}{6}=-\frac{3\pi}{6}=-\frac{\pi}{2}.$

✓Final answer
$-\frac{\pi}{2}$
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

$\tan^{-1}\frac12+\tan^{-1}\frac13=\tan^{-1}\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}=\tan^{-1}\frac{\frac{5}{6}}{\frac{5}{6}}=\tan^{-1}1=\frac{\pi}{4}.$

✓Final answer
$\frac{\pi}{4}$
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:

$\frac{1}{2}\left(\tan^{-1}x + \tan^{-1}\frac{1}{x}\right) = \frac{\pi}{4}.$

Hence

$\tan\left(\frac{\pi}{4}\right) = 1.$

✓Final answer
$1$
CBSE 2025Set ANNUAL1 markMCQ
$x$
(i) $\dfrac{1}{\sqrt3}$
(ii) $\dfrac13$
(iii) $\dfrac14$
(iv) $\dfrac{1}{\sqrt5}$
›Reveal solutionSolution

Simplify the RHS first, then solve for $x$ .

$\sin^{-1}\left(\frac{\sqrt3}{2}\right) = \frac{\pi}{3}$

So the equation becomes:

$2\tan^{-1}x = \frac{\pi}{3} \implies \tan^{-1}x = \frac{\pi}{6}$

$x = \tan\frac{\pi}{6} = \frac{1}{\sqrt3}$

✓Final answer
$x=\dfrac{1}{\sqrt3}$

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

Applying it to this problem

Matching with the options

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

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