For what value of x, 2tan⁻¹x=sin⁻¹((√3)/(2))? (i) – Class 12 Mathematics | ConceptFirst

Question of 108

x

(i)

\dfrac{1}{\sqrt3}

(ii)

\dfrac13

(iii)

\dfrac14

(iv)

\dfrac{1}{\sqrt5}

Odisha ChseOdisha CHSE +2 Science Board Exam 2025MCQ· 1mImportance★★★★★

0% · 0/108 Questions

🔒 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 →

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 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 →

←Previous chapterCh 1Relations and Functions