Evaluate or simplify: $\cos \left(\tan ^{-1} u\right)$

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04-11-2026
Mathematics

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Evaluate or simplify: $\cos \left(\tan ^{-1} u\right)$

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-04T01:57:12-05:00

Let $\theta = \tan^{-1} u$, implying $\tan \theta = u$.
Utilize the identity $\sec^2 \theta = 1 + \tan^2 \theta$ to relate cosine and tangent.
Express $\cos \theta$ as $\frac{1}{\sqrt{1 + \tan^2 \theta}}$ given that $\cos \theta > 0$ in the range of the inverse tangent function.
The simplified expression is $\boxed{\frac{1}{\sqrt{1 + u^2}}}$.

Explanation

Defining the Variables and Goal

We want to simplify the expression $\cos(\tan^{-1} u)$ where $u$ is any real number. Let $\theta = \tan^{-1} u$. By the definition of the inverse tangent function, this implies that $\tan \theta = u$ and $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Within this interval, the cosine function is always positive. Our goal is to express $\cos \theta$ purely in terms of $u$.

Relating Cosine and Tangent Identities

To relate $\cos \theta$ to $\tan \theta$, we use the trigonometric identity $\sec^2 \theta = 1 + \tan^2 \theta$. Since $\sec \theta = \frac{1}{\cos \theta}$, we can rewrite this identity as $\frac{1}{\cos^2 \theta} = 1 + \tan^2 \theta$. Substituting $\tan \theta = u$ into this equation, we get $\frac{1}{\cos^2 \theta} = 1 + u^2$.

Solving for the Final Expression

From the equation $\frac{1}{\cos^2 \theta} = 1 + u^2$, we can solve for $\cos^2 \theta$ by taking the reciprocal: $\cos^2 \theta = \frac{1}{1 + u^2}$. Since the range of the inverse tangent function is $(-\frac{\pi}{2}, \frac{\pi}{2})$, the cosine of this angle is always positive. Therefore, taking the positive square root of both sides gives us $\cos \theta = \frac{1}{\sqrt{1 + u^2}}$. Thus, $\cos(\tan^{-1} u) = \frac{1}{\sqrt{1 + u^2}}$.

Examples

In computer graphics, when calculating the orientation of a surface or a camera, we often deal with angles and their trigonometric ratios. For example, if you know the slope $u = \tan \theta$ of a line segment, finding the cosine of the angle $\theta$ tells you the horizontal component of a unit vector along that line. The identity $\cos(\tan^{-1} u) = \frac{1}{\sqrt{1+u^2}}$ allows programmers to calculate this projection directly from the slope without needing to compute the inverse tangent function explicitly, which saves computational resources and improves efficiency in rendering engines.