If $\tan^{-1} x - \cot^{-1} x = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$, $x > 0$, then find the value of $x$ and hence find the value of $\sec^{-1}\left(\dfrac{2}{x}\right)$.

> [!TLDR] > The key idea is to rewrite $\cot^{-1}x$ as $\frac{\pi}{2} - \tan^{-1}x$, then solve the resulting linear equation for $\tan^{-1}x$. This gives $x = \sqrt{3}$, and then $\sec^{-1}\left(\frac{2}{x}\right) = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$. The problem gives an equation involving inverse trigonometric functions. The first step is always to simplify the left-hand side using a standard identity. For $x > 0$, we know that $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$. This is a direct consequence of the complementary angle relationship in a right triangle: if an angle has tangent $x$, its complement has cotangent $x$. Let’s work through it. 1. **Rewrite the left-hand side** Using $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$, we get: \[ \tan^{-1}x - \left(\frac{\pi}{2} - \tan^{-1}x\right) = 2\tan^{-1}x - \frac{\pi}{2}. \] 2. **Set equal to the right-hand side** The given equation becomes: \[ 2\tan^{-1}x - \frac{\pi}{2} = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right). \] We know $\tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6}$, since $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}}$. 3. **Solve for $\tan^{-1}x$** \[ 2\tan^{-1}x - \frac{\pi}{2} = \frac{\pi}{6} \] \[ 2\tan^{-1}x = \frac{\pi}{2} + \frac{\pi}{6} = \frac{2\pi}{3} \] \[ \tan^{-1}x = \frac{\pi}{3}. \] 4. **Find $x$** Since $\tan\frac{\pi}{3} = \sqrt{3}$, we have $x = \sqrt{3}$. > [!WARNING] > A common mistake is to forget the condition $x > 0$ and try to use $\cot^{-1}x = \pi - \tan^{-1}x$ (which holds for $x 0$ ensures the simpler identity works. …

If tan⁻¹ x - cot⁻¹ x = tan⁻¹(dfrac1√(3)), x > 0, then f — CBSE Class 12 Mathematics 2019 PYQ

Question

\tan^{-1} x - \cot^{-1} x = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)

CBSECBSE Class XII Board 2019Subjective· 4mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

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. …

Start your 14-day free trial to unlock the full solution →

The key idea is to rewrite $\cot^{-1}x$ as $\frac{\pi}{2} - \tan^{-1}x$ , then solve the resulting linear equation for $\tan^{-1}x$ . This gives $x = \sqrt{3}$ , and then $\sec^{-1}\left(\frac{2}{x}\right) = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}$ .

The problem gives an equation involving inverse trigonometric functions. The first step is always to simplify the left-hand side using a standard identity. For $x > 0$ , we know that $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$ . This is a direct consequence of the complementary angle relationship in a right triangle: if an angle has tangent $x$ , its complement has cotangent $x$ .

Let’s work through it.

Rewrite the left-hand side Using $\cot^{-1}x = \frac{\pi}{2} - \tan^{-1}x$ , we get:

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

Set equal to the right-hand side The given equation becomes:

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

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

Solve for $\tan^{-1}x$

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

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

\tan^{-1}x = \frac{\pi}{3}.

Find $x$ Since $\tan\frac{\pi}{3} = \sqrt{3}$ , we have $x = \sqrt{3}$ .

Watch out

A common mistake is to forget the condition $x > 0$ and try to use $\cot^{-1}x = \pi - \tan^{-1}x$ (which holds for $x < 0$ ). Here, $x > 0$ ensures the simpler identity works. …

Start your 14-day free trial to unlock the full solution →

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

…

Start your 14-day free trial to see the full solution →
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 $x = \frac{1}{3}$ : …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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 $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$ :

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

…

Start your 14-day free trial to see the full solution →
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 $\tan^{-1}(-x)=-\tan^{-1}x$ , …

Start your 14-day free trial to see the full solution →
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 $\cot\frac{\pi}{6}=\sqrt{3}$ ,

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

…

Start your 14-day free trial to see the full solution →
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 …

Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →
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: …

Start your 14-day free trial to see the full solution →

Q.If tan⁡−1x−cot⁡−1x=tan⁡−1(13)\tan^{-1} x - \cot^{-1} x = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)tan−1x−cot−1x=tan−1(3​1​), x>0x > 0x>0, then find the value of xxx and hence find the value of sec⁡−1(2x)\sec^{-1}\left(\dfrac{2}{x}\right)sec−1(x2​).

🎓Unlock everything free for 14 days

Q.If $\tan^{-1} x - \cot^{-1} x = \tan^{-1}\left(\dfrac{1}{\sqrt{3}}\right)$ , $x > 0$ , then find the value of $x$ and hence find the value of $\sec^{-1}\left(\dfrac{2}{x}\right)$ .

Unlock everything free for 14 days