(a) Write $\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$ in the simplest form. OR (b) Find the domain of $\sin^{-1}\sqrt{x-1}$.

(a) Write sin⁻¹(fracx√(1+x²)) in the simplest form. OR (b) Find the domain of sin⁻¹√(x-1).

Question

\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)

(OR)

(b) Find the domain of

\sin^{-1}\sqrt{x-1}

CBSECBSE Class XII Board 2025Subjective· 2mImportance★★★★★

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Part (a)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. …

Part (b)Concept understanding — Domain Of Composite Function

Domain of a Composite Function

Picture a building with two doors: the first opens with a blue pass, the second with a red pass. Walking through $f(g(x))$ means passing the inner door $g$ first, then the outer door $f$ . The domain of the composite is simply: which inputs make it through both doors?

The intuition

Take $f(x) = \sqrt{x}$ and $g(x) = x - 5$ , so $f(g(x)) = \sqrt{x-5}$ .

The inner $g(x) = x - 5$ accepts every real number.
The outer $f$ accepts only non-negative inputs.

So the real question is: which $x$ make $g(x)$ land inside the domain of $f$ ? The domain of the composite is not just the domain of $g$ , nor just the domain of $f$ — it is the overlap seen through $g$ .

The precise statement

\text{Domain}(f \circ g) = \{\, x \in \text{Domain}(g) \;\mid\; g(x) \in \text{Domain}(f) \,\}.

Two steps, in order:

Keep only the $x$ that $g$ can handle.
Among those, keep only the $x$ for which $g(x)$ is something $f$ can handle.

Watch out

A frequent mistake is to restrict $x$ using the domain of $f$ directly. The restriction comes from $g(x)$ lying in $\text{Domain}(f)$ , not from $x$ itself.

A worked check

For $f(x) = \sqrt{x}$ , $g(x) = \dfrac{1}{x-1}$ :

Domain of $g$ : $x \ne 1$ .
Outer condition: $g(x) \ge 0 \Rightarrow \dfrac{1}{x-1} \ge 0 \Rightarrow x - 1 > 0 \Rightarrow x > 1$ .

So $\text{Domain}(f\circ g) = (1, \infty)$ — the condition from $f$ already excludes $x = 1$ .

Order matters …

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$\sin^{-1}\!\frac{x}{\sqrt{1+x^2}}=\tan^{-1}x$ .
Domain $=[1,2]$ .

Part (a)

Substitute $x=\tan\theta$ , where $\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ . Then $\sqrt{1+x^2}=\sqrt{1+\tan^2\theta}=\sec\theta$ , and

\frac{x}{\sqrt{1+x^2}}=\frac{\tan\theta}{\sec\theta}=\sin\theta.

Hence …

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

…

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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 $x = \frac{1}{3}$ : …

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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 …

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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 …

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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: …

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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 …

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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 $\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right)$ :

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

…

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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 $\tan^{-1}(-x)=-\tan^{-1}x$ , …

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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 $\cot\frac{\pi}{6}=\sqrt{3}$ ,

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

…

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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 …

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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: …

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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: …

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Q.(a) Write sin⁡−1(x1+x2)\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)sin−1(1+x2​x​) in the simplest form.

Part (a)

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Q.(a) Write $\sin^{-1}\left(\frac{x}{\sqrt{1+x^2}}\right)$ in the simplest form.

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