The result $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ is true when value of $xy$ is __________.

The result tan⁻¹x-tan⁻¹y=tan⁻¹((x-y)/(1+xy)) is true when value of xy is _____.

NCERT Exemplar · Q47

\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)

Rajasthan RbseShort· 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 formula $\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)$ holds only when $xy > -1$ . This condition ensures the sum/difference of angles stays within the principal value range of $\tan^{-1}$ .

Why this condition matters

The inverse tangent function $\tan^{-1}$ (also written as $\arctan$ ) returns an angle in the principal branch: $(-\frac{\pi}{2}, \frac{\pi}{2})$ . When we combine two such angles by addition or subtraction, the result must also lie in this interval for the formula to be valid directly.

The standard identity for the tangent of a difference is:

\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}

If we set $x = \tan A$ and $y = \tan B$ , then $A = \tan^{-1}x$ and $B = \tan^{-1}y$ (taking principal values). So:

\tan(\tan^{-1}x - \tan^{-1}y) = \frac{x - y}{1 + xy}

Taking $\tan^{-1}$ on both sides gives:

\tan^{-1}x - \tan^{-1}y = \tan^{-1}\left(\frac{x-y}{1+xy}\right)

but only if $\tan^{-1}x - \tan^{-1}y$ itself lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ . Otherwise, we need to add or subtract $\pi$ to bring it into the principal range.

Step-by-step reasoning

Understand the domain of $\tan^{-1}$

The principal value of $\tan^{-1}$ is always in $(-\frac{\pi}{2}, \frac{\pi}{2})$ . So both $\tan^{-1}x$ and $\tan^{-1}y$ are angles strictly between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$ .
When does the difference stay in $(-\frac{\pi}{2}, \frac{\pi}{2})$ ?

The difference of two numbers each in $(-\frac{\pi}{2}, \frac{\pi}{2})$ can be as large as nearly $\pi$ (if one is near $\frac{\pi}{2}$ and the other near $-\frac{\pi}{2}$ ) or as small as $-\pi$ . So it can fall outside the principal range.
Relate the condition to $xy$

Consider the sign of $1+xy$ . The denominator $1+xy$ in the formula determines whether the angle sum/difference crosses $\frac{\pi}{2}$ or $-\frac{\pi}{2}$ .
- If $xy > -1$ , then $1+xy > 0$ . The angle $\tan^{-1}x - \tan^{-1}y$ lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ , so the formula holds as written. …

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Q.The result $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ is true when value of $xy$ is __________.

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Why this condition matters

Step-by-step reasoning

Method: State the domain condition for an inverse-tangent difference

Steps

Common Mistakes

Q.The result tan⁡−1x−tan⁡−1y=tan⁡−1(x−y1+xy)\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)tan−1x−tan−1y=tan−1(1+xyx−y​) is true when value of xyxyxy is __________.

Why this condition matters

Step-by-step reasoning

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Q.The result $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left(\frac{x-y}{1+xy}\right)$ is true when value of $xy$ is __________.

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