If $\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}$, then $\cot^{-1}x+\cot^{-1}y$ equals (A) $\frac{\pi}{5}$ (B) $\frac{2\pi}{5}$ (C) $\frac{3\pi}{5}$ (D) $\pi$

> [!TLDR] > The problem uses the inverse-tangent sum identity and the complementary relationship between $\tan^{-1}$ and $\cot^{-1}$. Given $\tan^{-1}x + \tan^{-1}y = \frac{4\pi}{5}$, we find $\cot^{-1}x + \cot^{-1}y = \frac{\pi}{5}$. The key insight is that for any real $x$, $\tan^{-1}x$ and $\cot^{-1}x$ are complementary: they add up to $\frac{\pi}{2}$. This is because $\cot \theta = \tan\left(\frac{\pi}{2} - \theta\right)$, so the inverse functions obey the same shift. That single relationship turns the problem into a simple subtraction. Let’s walk through it. 1. **Recall the complementary identity** For any real $x$, $$ \tan^{-1}x + \cot^{-1}x = \frac{\pi}{2} $$ This holds for all $x$ (the principal-value branches are chosen so that the sum is constant). The same is true for $y$: $$ \tan^{-1}y + \cot^{-1}y = \frac{\pi}{2} $$ 2. **Add the two complementary equations** $$ (\tan^{-1}x + \cot^{-1}x) + (\tan^{-1}y + \cot^{-1}y) = \frac{\pi}{2} + \frac{\pi}{2} = \pi $$ Rearranging: $$ (\tan^{-1}x + \tan^{-1}y) + (\cot^{-1}x + \cot^{-1}y) = \pi $$ 3. **Substitute the given sum** We know $\tan^{-1}x + \tan^{-1}y = \frac{4\pi}{5}$. So: $$ \frac{4\pi}{5} + (\cot^{-1}x + \cot^{-1}y) = \pi $$ 4. **Solve for the required sum** $$ \cot^{-1}x + \cot^{-1}y = \pi - \frac{4\pi}{5} = \frac{\pi}{5} $$ …

If tan⁻¹x+tan⁻¹y=(4π)/(5), then cot⁻¹x+cot⁻¹y equals (A) (π)/(5) (B) (2π)/(5) (C) (3π)/(5) (D) π

NCERT Exemplar · Q30

\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}

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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 problem uses the inverse-tangent sum identity and the complementary relationship between $\tan^{-1}$ and $\cot^{-1}$ . Given $\tan^{-1}x + \tan^{-1}y = \frac{4\pi}{5}$ , we find $\cot^{-1}x + \cot^{-1}y = \frac{\pi}{5}$ .

The key insight is that for any real $x$ , $\tan^{-1}x$ and $\cot^{-1}x$ are complementary: they add up to $\frac{\pi}{2}$ . This is because $\cot \theta = \tan\left(\frac{\pi}{2} - \theta\right)$ , so the inverse functions obey the same shift. That single relationship turns the problem into a simple subtraction.

Let’s walk through it.

Recall the complementary identity For any real $x$ ,

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

This holds for all $x$ (the principal-value branches are chosen so that the sum is constant). The same is true for $y$ :

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

Add the two complementary equations

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

Rearranging:

(\tan^{-1}x + \tan^{-1}y) + (\cot^{-1}x + \cot^{-1}y) = \pi

Substitute the given sum We know $\tan^{-1}x + \tan^{-1}y = \frac{4\pi}{5}$ . So:

\frac{4\pi}{5} + (\cot^{-1}x + \cot^{-1}y) = \pi

Solve for the required sum $\cot^{-1}x + \cot^{-1}y = \pi - \frac{4\pi}{5} = \frac{\pi}{5}$ …

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Q.If $\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}$ , then $\cot^{-1}x+\cot^{-1}y$ equals
(A) $\frac{\pi}{5}$
(B) $\frac{2\pi}{5}$
(C) $\frac{3\pi}{5}$
(D) $\pi$

Concept understanding — Inverse Tangent Identity

Inverse Tangent Identities

The core addition identity

Subtraction

The doubling identity

The complementary identity

Method: Relating $\sum\cot^{-1}$ to $\sum\tan^{-1}$ via the complementary identity

Steps

Common Mistakes

Q.If tan⁡−1x+tan⁡−1y=4π5\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}tan−1x+tan−1y=54π​, then cot⁡−1x+cot⁡−1y\cot^{-1}x+\cot^{-1}ycot−1x+cot−1y equals (A) π5\frac{\pi}{5}5π​ (B) 2π5\frac{2\pi}{5}52π​ (C) 3π5\frac{3\pi}{5}53π​ (D) π\piπ

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Q.If $\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}$ , then $\cot^{-1}x+\cot^{-1}y$ equals
(A) $\frac{\pi}{5}$
(B) $\frac{2\pi}{5}$
(C) $\frac{3\pi}{5}$
(D) $\pi$

Unlock everything free for 14 days