If $3\tan^{-1}x+\cot^{-1}x=\pi$, then $x$ equals (A) $0$ (B) $1$ (C) $-1$ (D) $\frac{1}{2}$

If 3tan⁻¹x+cot⁻¹x=π, then x equals (A) 0 (B) 1 (C) -1 (D) (1)/(2)

NCERT Exemplar · Q22

3\tan^{-1}x+\cot^{-1}x=\pi

Odisha ChseMCQ· 1mImportance★★★★★

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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 key idea is to use the identity $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ to simplify the given equation. Substituting this reduces the problem to $2\tan^{-1}x = \frac{\pi}{2}$ , giving $x = 1$ .

We start with the equation:

3\tan^{-1}x + \cot^{-1}x = \pi

The core insight here is the inverse tangent identity: for any real $x$ , $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$ . This holds because $\tan^{-1}x$ and $\cot^{-1}x$ are complementary angles — their sum is always a right angle. This identity is the backbone of the solution.

Let’s work through it step by step.

Apply the identity Replace $\cot^{-1}x$ with $\frac{\pi}{2} - \tan^{-1}x$ :

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

Simplify the left-hand side Combine the $\tan^{-1}x$ terms:

(3\tan^{-1}x - \tan^{-1}x) + \frac{\pi}{2} = \pi

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

Isolate the inverse tangent term Subtract $\frac{\pi}{2}$ from both sides:

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

Solve for $\tan^{-1}x$ Divide by 2:

\tan^{-1}x = \frac{\pi}{4}

Take the tangent of both sides Since $\tan(\tan^{-1}x) = x$ and $\tan\frac{\pi}{4} = 1$ , we get: …

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Q.If $3\tan^{-1}x+\cot^{-1}x=\pi$ , then $x$ equals
(A) $0$
(B) $1$
(C) $-1$
(D) $\frac{1}{2}$

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Method: Solving equations that mix $\tan^{-1}$ and $\cot^{-1}$

Steps

Common Mistakes

Q.If 3tan⁡−1x+cot⁡−1x=π3\tan^{-1}x+\cot^{-1}x=\pi3tan−1x+cot−1x=π, then xxx equals (A) 000 (B) 111 (C) −1-1−1 (D) 12\frac{1}{2}21​

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Q.If $3\tan^{-1}x+\cot^{-1}x=\pi$ , then $x$ equals
(A) $0$
(B) $1$
(C) $-1$
(D) $\frac{1}{2}$

Unlock everything free for 14 days