Find the value of $\sin\left(\cos^{-1} \dfrac{4}{5} + \tan^{-1} \dfrac{2}{3}\right)$.

Find the value of sin(cos⁻¹ (4)/(5) + tan⁻¹ (2)/(3)).

Question

\sin\left(\cos^{-1} \dfrac{4}{5} + \tan^{-1} \dfrac{2}{3}\right)

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

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Concept understanding — Inverse Cosine Addition

Inverse Cosine Addition – From Intuition to Formula

Suppose you know $\cos A = x$ and $\cos B = y$ and want the angle $A+B$ — that is, $\cos^{-1} x + \cos^{-1} y$ in terms of $x$ and $y$ .

The answer is not simply $\cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ — that's the cosine of the sum, not the sum itself. The real formula is subtler, because inverse cosine returns an angle in a fixed range.

The Intuition

$\cos^{-1} x$ is "the angle whose cosine is $x$ ", and by definition it lies in $[0, \pi]$ . So $\cos^{-1} x + \cos^{-1} y$ is a sum of two angles each in $[0, \pi]$ — anywhere from $0$ to $2\pi$ .

Inverse cosine is not linear, so take the cosine of the sum using the addition formula:

\cos(\cos^{-1} x + \cos^{-1} y) = \cos(\cos^{-1} x)\cos(\cos^{-1} y) - \sin(\cos^{-1} x)\sin(\cos^{-1} y)

Since $\cos(\cos^{-1} x) = x$ and $\sin(\cos^{-1} x) = \sqrt{1-x^2}$ (positive because $\cos^{-1} x \in [0,\pi]$ ):

\cos(\cos^{-1} x + \cos^{-1} y) = xy - \sqrt{1-x^2}\sqrt{1-y^2}

The sum itself is the inverse cosine of that expression — only if the sum lies in $[0,\pi]$ , the range of $\cos^{-1}$ .

The Precise Statement

\cos^{-1} x + \cos^{-1} y = \begin{cases} \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}\right), & \text{if } x+y \ge 0 \\[6pt] 2\pi - \cos^{-1}\left(xy - \sqrt{1-x^2}\sqrt{1-y^2}\right), & \text{if } x+y < 0 \end{cases}

Why the case split? $\cos^{-1}$ always returns an angle in $[0,\pi]$ . When $x+y \ge 0$ the sum lies in $[0,\pi]$ , so it equals the inverse cosine directly. When $x+y < 0$ the sum lies in $(\pi, 2\pi)$ , so we use $\cos^{-1}(-t) = \pi - \cos^{-1} t$ to bring it back into range.

Watch out

A common mistake is writing $\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2})$ without checking $x+y \ge 0$ . This is false when $x+y < 0$ — you then need $2\pi$ minus that inverse cosine.

A Quick Example

Let $x = y = -\frac{1}{2}$ . Then $\cos^{-1}(-\frac12) = \frac{2\pi}{3}$ , so the true sum is $\frac{4\pi}{3}$ . …

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Express each inverse trigonometric function as an angle in a right triangle, find the sine and cosine of each angle using Pythagoras, then apply the sine addition formula. The value is $\boxed{\frac{17\sqrt{13}}{65}}$ .

The key insight is to interpret inverse trigonometric functions geometrically. When you see $\cos^{-1} \frac{4}{5}$ , think of an angle $\alpha$ in a right triangle where the adjacent side is 4 and the hypotenuse is 5. Similarly, $\tan^{-1} \frac{2}{3}$ represents an angle $\beta$ where opposite is 2 and adjacent is 3. Once we have these triangles, we can find all the trigonometric ratios we need and apply the sine addition formula.

Let me denote $\alpha = \cos^{-1} \frac{4}{5}$ and $\beta = \tan^{-1} \frac{2}{3}$ . We need $\sin(\alpha + \beta)$ .

\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta

Step 1: Find $\sin\alpha$ and $\cos\alpha$ from the first angle

Since $\alpha = \cos^{-1} \frac{4}{5}$ , we know $\cos\alpha = \frac{4}{5}$ .

In a right triangle with adjacent = 4 and hypotenuse = 5, the opposite side is:

\text{opposite} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3

Therefore, $\sin\alpha = \frac{3}{5}$ .

Step 2: Find $\sin\beta$ and $\cos\beta$ from the second angle

Since $\beta = \tan^{-1} \frac{2}{3}$ , we know $\tan\beta = \frac{2}{3}$ .

In a right triangle with opposite = 2 and adjacent = 3, the hypotenuse is: …

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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 14 on this concept.

CBSE 2026Set A1 markMCQ
$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) =$
(a) $\frac{\pi}{6}$
(b) $\frac{\pi}{3}$
(c) $\frac{5\pi}{6}$
(d) $\frac{7\pi}{6}$
›Reveal solutionSolution

$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) = \frac{5\pi}{6}$ .

The range of $\cos^{-1}$ is $[0,\pi]$ , but $\frac{7\pi}{6}\notin[0,\pi]$ , so we cannot just cancel.

First find the actual cosine value:

$\cos\frac{7\pi}{6} = -\cos\frac{\pi}{6} = -\frac{\sqrt{3}}{2}.$

…

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CBSE 2025Set E1 markMCQ
$\cos^{-1}\left(\cos\frac{8\pi}{5}\right) =$
(a) $\frac{8\pi}{5}$
(b) $\frac{2\pi}{5}$
(c) $\frac{\pi}{5}$
(d) $\frac{3\pi}{5}$
›Reveal solutionSolution

Reduce the angle to the principal range $[0,\pi]$ of $\cos^{-1}$ ; the answer is $\frac{2\pi}{5}$ .

The angle $\frac{8\pi}{5}$ is not in $[0,\pi]$ , so we cannot just cancel. Use $\cos(2\pi-\theta)=\cos\theta$ :

$\cos\frac{8\pi}{5}=\cos\left(2\pi-\frac{8\pi}{5}\right)=\cos\frac{2\pi}{5}.$

…

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CBSE 2025Set E1 markMCQ
$\sin\left\{\sin^{-1}\frac{1}{5} + \cos^{-1}x\right\} = 1,\ \Rightarrow x =$
(a) $1$
(b) $0$
(c) $\frac{4}{5}$
(d) $\frac{1}{5}$
›Reveal solutionSolution

Set the bracket equal to $\frac{\pi}{2}$ and solve; $x=\frac15$ .

Since $\sin\theta=1$ only when $\theta=\frac{\pi}{2}$ ,

$\sin^{-1}\frac15+\cos^{-1}x=\frac{\pi}{2}.$

…

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CBSE 2025Set E1 markMCQ
$\cos^{-1}x + \sec^{-1}\frac{1}{x} =$
(a) $\frac{\pi}{2}$
(b) $\cos^{-1}(2x^2 - 1)$
(c) $\cos^{-1}(1 - 2x^2)$
(d) $\cos^{-1}2x$
›Reveal solutionSolution

$\sec^{-1}\frac1x = \cos^{-1}x$ , so the sum is $2\cos^{-1}x$ , which equals $\cos^{-1}(2x^2-1)$ .

Use $\sec^{-1}(y) = \cos^{-1}\frac{1}{y}$ , so

$\sec^{-1}\frac{1}{x} = \cos^{-1}x.$

Therefore

$\cos^{-1}x + \sec^{-1}\frac{1}{x} = 2\cos^{-1}x.$ …

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CBSE 2024Set D1 markMCQ
$\cos^{-1}\left(-\frac{1}{2}\right) =$
(a) $\frac{2\pi}{3}$
(b) $\frac{\pi}{3}$
(c) $\frac{\pi}{6}$
(d) $\frac{\pi}{2}$
›Reveal solutionSolution

The principal value lies in $[0,\pi]$ : $\cos^{-1}(-\tfrac12)=\tfrac{2\pi}{3}$ .

The range of $\cos^{-1}$ is $[0,\pi]$ . Since $\cos\dfrac{2\pi}{3} = -\dfrac{1}{2}$ and $\dfrac{2\pi}{3}\in[0,\pi]$ :

…

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CBSE 2024Set D1 markMCQ
$x \in [-1, 1],\ \cos^{-1} x =$
(a) $\frac{\pi}{2} - \cot^{-1} x$
(b) $\frac{\pi}{2} - \sin^{-1} x$
(c) $\frac{\pi}{2} - \tan^{-1} x$
(d) $\frac{\pi}{2} - \sec^{-1} x$
›Reveal solutionSolution

Complementary identity: $\cos^{-1}x=\tfrac\pi2-\sin^{-1}x$ for $x\in[-1,1]$ .

For $x\in[-1,1]$ we have $\sin^{-1}x + \cos^{-1}x = \dfrac{\pi}{2}$ .

…

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CBSE 2024Set D1 markMCQ
$\cos^{-1}\left(\cos\frac{7\pi}{6}\right) =$
(a) $\frac{7\pi}{6}$
(b) $\frac{5\pi}{6}$
(c) $\frac{\pi}{3}$
(d) $\frac{\pi}{6}$
›Reveal solutionSolution

$\cos^{-1}$ must return a value in $[0,\pi]$ , so the answer is $\tfrac{5\pi}{6}$ , not $\tfrac{7\pi}{6}$ .

Since $\tfrac{7\pi}{6}\notin[0,\pi]$ , we cannot simply cancel. Compute $\cos\dfrac{7\pi}{6} = -\dfrac{\sqrt3}{2}$ .

…

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CBSE 2023Set E1 markMCQ
$x\in[-1,1],\ \sin\left[2(\sin^{-1}x+\cos^{-1}x)\right] =$
(a) $0$
(b) $1$
(c) $-1$
(d) $\dfrac{1}{2}$
›Reveal solutionSolution

$\sin\left[2(\sin^{-1}x+\cos^{-1}x)\right]=0$ .

For all $x\in[-1,1]$ , $\sin^{-1}x+\cos^{-1}x=\dfrac{\pi}{2}$ . Hence

…

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CBSE 2023Set A1 mark
$\cos(\sec^{-1}x + \text{cosec}^{-1}x)$
›Reveal solutionSolution

$\sec^{-1}x+\text{cosec}^{-1}x=\dfrac{\pi}{2}$ for $|x|\ge1$ , so $\cos\left(\sec^{-1}x+\text{cosec}^{-1}x\right)=\cos\dfrac{\pi}{2}=0$ .

…

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CBSE 2022Set HE2191 mark
$\sin^{-1} x + \cos^{-1} x =$
›Reveal solutionSolution

This is a standard inverse trigonometric identity valid for $x\in[-1,1]$ .

Let $\sin^{-1}x = \theta$ , so $x = \sin\theta = \cos\left(\dfrac{\pi}{2}-\theta\right)$ , giving $\cos^{-1}x = \dfrac{\pi}{2}-\theta$ . Adding, …

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CBSE 2021Set I1 markMCQ
$\cos^{-1}(2x) + \sin^{-1}(2x) = \ldots\ldots , \ 2x \in [-1,1]$
(a) $\dfrac{\pi}{2}$
(b) $\dfrac{\pi}{4}$
(c) $\pi$
(d) $0$
›Reveal solutionSolution

For all valid arguments, $\sin^{-1}t+\cos^{-1}t=\tfrac{\pi}{2}$ .

There is a standard identity: for every $t\in[-1,1]$ ,

$\sin^{-1}t+\cos^{-1}t=\dfrac{\pi}{2}$ .

…

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CBSE 2019Set 65/3/11 mark
$y = \sin^{-1} x + \cos^{-1} x$
›Reveal solutionSolution
The sum $\sin^{-1} x + \cos^{-1} x$ is constant ( $\pi/2$ ) for all $x$ in $[-1, 1]$ , so its derivative is zero: $\frac{dy}{dx} = 0$ .

Concept and Intuition

The problem asks for the derivative of $y = \sin^{-1} x + \cos^{-1} x$ . A brute-force approach would differentiate each inverse trig function separately using known formulas, then add. But that misses the deeper point.

There is a beautiful identity: for any $x$ in $[-1, 1]$ ,
$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}.$
Why? Think geometrically. If $\theta = \sin^{-1} x$ , then $\sin \theta = x$ and $\theta \in [-\pi/2, \pi/2]$ . The complementary angle $\pi/2 - \theta$ has cosine equal to $x$ , so $\cos^{-1} x = \pi/2 - \theta$ . Adding them gives $\pi/2$ .

Since $y$ is constant, its derivative is zero — no calculation needed. This is the elegant, concept-first way.

Watch out

A common mistake is to differentiate each term separately and get $\frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0$ , which is correct but misses why the sum is constant. The identity is the real insight.

Step-by-Step Solution

Recall the fundamental identity For any $x \in [-1, 1]$ ,

$\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}.$
This holds because if $\sin^{-1} x = \theta$ , then $\cos(\pi/2 - \theta) = \sin \theta = x$ , and $\pi/2 - \theta$ lies in $[0, \pi]$ , the principal range of $\cos^{-1}$ . …
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Q.Find the value of sin⁡(cos⁡−145+tan⁡−123)\sin\left(\cos^{-1} \dfrac{4}{5} + \tan^{-1} \dfrac{2}{3}\right)sin(cos−154​+tan−132​).

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Q.Find the value of $\sin\left(\cos^{-1} \dfrac{4}{5} + \tan^{-1} \dfrac{2}{3}\right)$ .

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