Prove that: $\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1}x, \frac{1}{\sqrt{2}} \le x \le 1$. OR Consider $f: R_+ \to (7, \infty)$ defined by $f(x) = 16x^2 + 24x + 7$. Find the inverse of the function $f$.

Prove that: sin⁻¹(2x√(1-x²)) = 2cos⁻¹x, frac1√(2) ≤ x ≤ 1. OR Consider f: R_+ → (7, ∞) defined by f(x) = 16x²

Question

\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1}x, \frac{1}{\sqrt{2}} \le x \le 1

(OR)

f: R_+ \to (7, \infty)

CBSECBSE Class XII Board 2020Subjective· 2mImportance★★★★★

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Part (a)Concept understanding — Trigonometric Simplification

Trigonometric Simplification

You know $\sin^2 x + \cos^2 x = 1$ — but the skill of turning a messy trig expression into that kind of clean form is trigonometric simplification. Because sines, cosines and their relatives are all tied together by identities from the unit circle, a tangled combination can almost always be rewritten as something shorter: a single term, a constant, or an easier combination.

Your core toolkit

\sin^2\theta+\cos^2\theta=1,\quad 1+\tan^2\theta=\sec^2\theta,\quad 1+\cot^2\theta=\csc^2\theta

How the process feels

Simplify $\dfrac{\sin x}{1+\cos x}+\dfrac{1+\cos x}{\sin x}$ . Over a common denominator the numerator is $\sin^2 x+(1+\cos x)^2 = \sin^2 x + 1 + 2\cos x + \cos^2 x$ . The Pythagorean identity turns $\sin^2 x+\cos^2 x$ into $1$ , giving $2+2\cos x = 2(1+\cos x)$ , so

\frac{2(1+\cos x)}{\sin x\,(1+\cos x)} = \frac{2}{\sin x} = 2\csc x.

A two-term sum collapses to one term.

Strategies that usually work

Convert everything to sines and cosines — cancellations then appear.
Spot Pythagorean pairs and replace them with $1$ (or $\sec^2$ , $\csc^2$ ).
Factor and cancel as you would with ordinary algebra.
Multiply by a conjugate — e.g. multiply $\frac{1}{1+\sin x}$ by $\frac{1-\sin x}{1-\sin x}$ to unlock a Pythagorean identity. …

Part (b)Concept understanding — Inverse Function Relationship

Inverse Function Relationship

Two functions are inverses when each undoes the other. If $f$ sends $a$ to $b$ , then $f^{-1}$ sends $b$ back to $a$ . Chain them together and you land exactly where you started.

The defining equations

If $f^{-1}$ is the inverse of $f$ , then

f^{-1}\big(f(x)\big) = x \quad\text{and}\quad f\big(f^{-1}(y)\big) = y.

The first holds for every $x$ in the domain of $f$ ; the second for every $y$ in the range of $f$ . This "round trip returns the input" is what inverse really means.

When does an inverse exist?

Only a one-to-one function (distinct inputs give distinct outputs) can be inverted — otherwise some output would have to map back to two inputs, which no function allows. Graphically, $f$ must pass the horizontal line test.

Important

When a function is not one-to-one over its whole domain (like $\sin x$ or $x^2$ ), we first restrict it to a piece where it is, and the inverse lives on that restricted piece.

The geometry

Because $(a,b)$ lies on $f$ exactly when $(b,a)$ lies on $f^{-1}$ , the graph of $f^{-1}$ is the mirror image of $f$ across the line $y=x$ . Consequently the domain and range swap: the range of $f$ becomes the domain of $f^{-1}$ .

Why the restriction bites — the trig case

For inverse trigonometric functions the relationship is one-sided. The "outer undo" always works:

\sin\big(\sin^{-1}x\big) = x \quad \text{for all } x\in[-1,1].

But the "inner undo" only works on the principal range:

$\sin^{-1}\big(\sin x\big) = x \quad \text{only if } x\in[-\tfrac{\pi}{2},\tfrac{\pi}{2}].$ …

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(a) Substituting $x=\cos\theta$ turns the argument into $\sin 2\theta$ , proving the identity. (b) $f^{-1}(x)=\dfrac{\sqrt{x+2}-3}{4}$ .

Part (a)

Put $x=\cos\theta$ where $\theta=\cos^{-1}x$ . The restriction $\dfrac{1}{\sqrt2}\le x\le 1$ means $\cos\theta$ ranges over $[\tfrac1{\sqrt2},1]$ , i.e. $\theta\in\left[0,\dfrac{\pi}{4}\right]$ . On this interval $\sin\theta\ge0$ , so $\sqrt{1-x^2}=\sqrt{1-\cos^2\theta}=\sin\theta$ .

Then

2x\sqrt{1-x^2}=2\cos\theta\,\sin\theta=\sin 2\theta.

Because $\theta\in[0,\tfrac{\pi}{4}]$ , the angle $2\theta\in\left[0,\dfrac{\pi}{2}\right]$ , which lies in the principal range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ of $\sin^{-1}$ . Therefore

$\sin^{-1}(2x\sqrt{1-x^2})=\sin^{-1}(\sin 2\theta)=2\theta=2\cos^{-1}x.$ …

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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 84 on this concept.

CBSE 2026Set 65/1/11 markMCQ
$x$
›Reveal solutionSolution
The determinant simplifies to $\sin x \cos x + \sin x \cos x = \sin 2x$ . Setting $\sin 2x = 1$ gives $2x = \frac{\pi}{2} + 2n\pi$ , so $x = \frac{\pi}{4}$ is one solution. The correct option is (B).

The problem gives a $2 \times 2$ determinant equal to $1$ and asks for a value of $x$ from the options. The fastest route is to compute the determinant directly — it’s a simple expression in $\sin x$ and $\cos x$ — and then solve the resulting trigonometric equation.

The determinant of $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$ is $ad - bc$ . Here:
$\begin{vmatrix} \cos x & \sin x \\ -\cos x & \sin x \end{vmatrix} = (\cos x)(\sin x) - (\sin x)(-\cos x)$

Simplify the expression. The first term is $\cos x \sin x$ . The second term: $(\sin x)(-\cos x) = -\sin x \cos x$ , but there’s a minus sign in front, so it becomes $-(-\sin x \cos x) = +\sin x \cos x$ . So the determinant equals:

$\cos x \sin x + \sin x \cos x = 2 \sin x \cos x$

Use the double-angle identity. Recall that $2 \sin x \cos x = \sin 2x$ . Therefore the equation becomes:

$\sin 2x = 1$

Solve $\sin 2x = 1$ . The sine function equals $1$ at $\frac{\pi}{2}$ plus any integer multiple of $2\pi$ :

$2x = \frac{\pi}{2} + 2n\pi \quad \Rightarrow \quad x = \frac{\pi}{4} + n\pi$
where $n$ is any integer.

Check the given options.

(A) $0$ : $\sin 0 = 0$ , not $1$ .

(B) $\frac{\pi}{4}$ : $\sin \frac{\pi}{2} = 1$ — works. …
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CBSE 2026Set CX1 markMCQ
$\sin(\tan^{-1}x)$
(a) $\dfrac{x}{\sqrt{1+x^{2}}}$
(b) $\dfrac{x}{\sqrt{1-x^{2}}}$
(c) $\dfrac{1}{\sqrt{1+x^{2}}}$
(d) $\dfrac{1}{\sqrt{1-x^{2}}}$
›Reveal solutionSolution

With $\theta=\tan^{-1}x$ a right triangle gives $\sin\theta=\dfrac{x}{\sqrt{1+x^{2}}}$ — option (a).

Concept: Convert the inverse function to an angle and read the ratio off a right triangle.

Let $\theta=\tan^{-1}x$ , so $\tan\theta=x$ . Take the opposite side $=x$ and adjacent $=1$ ; then the hypotenuse is $\sqrt{1+x^{2}}$ .

…

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CBSE 2026Set A1 markMCQ
$\sin(\cos^{-1}3/5) =$
(a) $\frac{3}{4}$
(b) $\frac{4}{5}$
(c) $\frac{3}{5}$
(d) $\frac{5}{4}$
›Reveal solutionSolution

$\sin\left(\cos^{-1}\frac{3}{5}\right) = \frac{4}{5}$ .

Let $\theta = \cos^{-1}\frac{3}{5}$ , so $\cos\theta = \frac{3}{5}$ with $\theta\in[0,\pi]$ , where $\sin\theta \ge 0$ .

Then …

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CBSE 2026Set A1 markMCQ
$|x| \le 1$
(a) $\frac{\sqrt{1-x^2}}{x}$
(b) $\frac{x}{1+x^2}$
(c) $\frac{\sqrt{1+x^2}}{x}$
(d) $\sqrt{1-x^2}$
›Reveal solutionSolution

$\tan(\cos^{-1}x) = \frac{\sqrt{1-x^2}}{x}$ .

Let $\theta = \cos^{-1}x$ , so $\cos\theta = x$ with $\theta\in[0,\pi]$ (where $\sin\theta\ge 0$ ).

Then $\sin\theta = \sqrt{1-x^2}$ , and …

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CBSE 2026Set A1 markMCQ
$\int\frac{\cos 2x}{(\sin x + \cos x)^2}\,dx =$
(a) $2\log(\sin x + \cos x) + k$
(b) $\log(\sin x + \cos x) + k$
(c) $\log(\sin x - \cos x) + k$
(d) $-\frac{1}{\sin x + \cos x} + k$
›Reveal solutionSolution

Factor $\cos 2x$ and substitute $u=\sin x+\cos x$ to get $\log(\sin x+\cos x)+k$ .

Write $\cos 2x=\cos^2 x-\sin^2 x=(\cos x-\sin x)(\cos x+\sin x)$ . Then

$\dfrac{\cos 2x}{(\sin x+\cos x)^2}=\dfrac{(\cos x-\sin x)(\cos x+\sin x)}{(\sin x+\cos x)^2}=\dfrac{\cos x-\sin x}{\sin x+\cos x}$ .

…

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CBSE 2026Set A1 markMCQ
$\int\frac{1 - \cos 2x}{1 + \cos 2x}\,dx =$
(a) $\tan x + x + k$
(b) $\tan x - x + k$
(c) $x - \tan^2 x + k$
(d) $\tan\frac{x}{2} + k$
›Reveal solutionSolution

Simplify to $\tan^2 x$ , then integrate: $\tan x - x + k$ .

Use $1-\cos 2x=2\sin^2 x$ and $1+\cos 2x=2\cos^2 x$ :

$\dfrac{1-\cos 2x}{1+\cos 2x}=\dfrac{2\sin^2 x}{2\cos^2 x}=\tan^2 x=\sec^2 x-1$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int \frac{dx}{\sin^2 x \cos^2 x}$
(a) $\tan x + \sin x + c$
(b) $\tan x - \cot x + c$
(c) $\tan x \cot x + c$
(d) $2\tan x - \cot 2x + c$
›Reveal solutionSolution

Split the integrand using $\sin^2x+\cos^2x=1$ in the numerator, then integrate each standard term.

$\dfrac{1}{\sin^2x\cos^2x} = \dfrac{\sin^2x+\cos^2x}{\sin^2x\cos^2x} = \dfrac{1}{\cos^2x}+\dfrac{1}{\sin^2x} = \sec^2x+\csc^2x$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$\int \tan^2 x \, dx =$
(a) $\cot x - x + C$
(b) $\tan x + x + C$
(c) $\tan x - x + C$
(d) None of these
›Reveal solutionSolution

Rewrite $\tan^2 x$ using the identity $\tan^2 x = \sec^2 x - 1$ , then integrate term by term.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The value of \tan^{-1}\left(\frac{1}{x}\right) will be:$
(a) \tan x
(b) \cot x
(c) \cot^{-1}\frac{1}{x}
(d) \cot^{-1} x
›Reveal solutionSolution

$\tan^{-1}(1/x)$ and $\cot^{-1}x$ are the same angle for positive $x$ , because tangent and cotangent are reciprocal functions.

Concept: For $x>0$ , the standard identity is

$\tan^{-1}\left(\frac{1}{x}\right) = \cot^{-1} x$ …

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CBSE 2026Set ANNUAL1 mark
$Q. If tan⁻¹(1/3) = x, then find sin x.$
›Reveal solutionSolution

Build a right triangle using $\tan x = 1/3$ and read off $\sin x$ .

Given $\tan^{-1}(1/3)=x \Rightarrow \tan x = 1/3$ .

In a right triangle take opposite side $=1$ , adjacent side $=3$ , so hypotenuse $=\sqrt{1^2+3^2}=\sqrt{10}$ .

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q.The value of \int \frac{sec^2 x}{cosec^2 x} dx is:$
(a)
(i) tan x - x + c
(b)
(ii) tan x + x + c
(c)
(iii) cot x - x + c
(d)
(iv) log cosec x + c
›Reveal solutionSolution

$\displaystyle\int \dfrac{\sec^2 x}{\csc^2 x}\,dx = \tan x - x + c$ — option (i).

Concept. Convert to a single trigonometric ratio, then use the identity $\tan^2 x=\sec^2 x-1$ and the standard integral $\int\sec^2 x\,dx=\tan x$ .

Steps. …

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CBSE 2026Set ANNUAL1 mark
$\sin\left\{\cos^{-1}\left(-\dfrac{4}{5}\right)\right\}$
›Reveal solutionSolution

With $\cos\theta=-\tfrac45$ and $\theta\in[0,\pi]$ , $\sin\theta=+\tfrac35$ .

Let $\theta=\cos^{-1}\left(-\tfrac45\right)$ , so $\cos\theta=-\tfrac45$ and $\theta\in[0,\pi]$ (range of $\cos^{-1}$ ). On this range $\sin\theta\ge 0$ .

…

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Q.Prove that: sin⁡−1(2x1−x2)=2cos⁡−1x,12≤x≤1\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1}x, \frac{1}{\sqrt{2}} \le x \le 1sin−1(2x1−x2​)=2cos−1x,2​1​≤x≤1.

Part (a)

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Q.Prove that: $\sin^{-1}(2x\sqrt{1-x^2}) = 2\cos^{-1}x, \frac{1}{\sqrt{2}} \le x \le 1$ .

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