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Chapter 5: Problem 30

Show that $$ \cos \left(\tan ^{-1} t\right)=\frac{1}{\sqrt{1+t^{2}}} $$ for every number \(t\).

Short Answer

Expert verified
To show that \( \cos(\tan^{-1}(t)) = \frac{1}{\sqrt{1+t^2}} \) for every number t, we first let 胃 be an angle such that \( \tan^{-1}(t) = \theta \), and construct a right-angled triangle with angle 胃, opposite side t, and adjacent side 1. Using the Pythagorean theorem, we find the length of the hypotenuse as \( h = \sqrt{t^2 + 1} \). Then, we calculate the cosine of 胃 using the triangle definition, \( \cos(\theta) = \frac{1}{\sqrt{t^2 + 1}} \). Finally, we substitute 胃 by its definition, and obtain the identity: \( \cos(\tan^{-1}(t)) = \frac{1}{\sqrt{1+t^2}} \).

Step by step solution

01

Express the inverse tangent as an angle

Let 胃 be an angle such that \( \tan^{-1}(t) = \theta \). This means that the tangent of 胃 is equal to t, or \( \tan(\theta) = t \).
02

Use the right-angled triangle definition of tangent

Let's construct a right-angled triangle with angle 胃, such that the tangent of this angle is t. To do this, we can set the length of the opposite side as t, and the length of the adjacent side as 1. This will give us a right triangle, with the opposite side as t, adjacent side as 1, and angle 胃. In this triangle, notice that \( \tan(\theta) = \frac{opposite}{adjacent} = \frac{t}{1} = t \), which corresponds to the definition we established in Step 1.
03

Find the length of the hypotenuse

We can now use the Pythagorean theorem to find the length of the hypotenuse of our right-angled triangle: \( hypotenuse^2 = opposite^2 + adjacent^2 \) Let's denote the length of the hypotenuse as "h". Then, \( h^2 = t^2 + 1^2 = t^2 + 1 \) Now, take the square root of both sides, we get: \( h = \sqrt{t^2 + 1} \)
04

Calculate the cosine of 胃 using the right-angled triangle

From the right-angled triangle definition of cosine, we have: \( \cos(\theta) = \frac{adjacent}{hypotenuse} \) Substitute the values from our triangle and the expression of the hypotenuse obtained above: \( \cos(\theta) = \frac{1}{\sqrt{t^2 + 1}} \)
05

Replace 胃 by its original definition

Recall that 胃 was the angle such that \( \tan^{-1}(t) = \theta \). Now substitute 胃 by its definition: \( \cos(\tan^{-1}(t)) = \frac{1}{\sqrt{t^2 + 1}} \) Therefore, for every number t, we have proven the given identity: $$ \cos\left(\tan^{-1}(t) \right) = \frac{1}{\sqrt{1+t^{2}}} $$

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