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Chapter 5: Problem 50

Verify the identity. $$\tan \left(\cos ^{-1} \frac{x+1}{2}\right)=\frac{\sqrt{4-(x+1)^{2}}}{x+1}$$

Short Answer

Expert verified
The identity \( \tan \left(\cos ^{-1} \frac{x+1}{2}\right) = \frac{\sqrt{4-(x+1)^{2}}}{x+1} \) is verified.

Step by step solution

01

Apply Pythagorean identity

The first aim is to replace the inverse cosine function with a single variable. Suppose \( \cos^{-1} \frac{x + 1}{2} = 胃 \). Then, \( \cos 胃 = \frac{x + 1}{2} \) and we can draw a right-angled triangle using this information. The length of the adjacent side is \(x + 1\), the hypotenuse is 2 and by the Pythagorean theorem, the opposite side is \( \sqrt{4 - (x + 1)^2} \).
02

Find the tangent of 胃

Using the triangle from Step 1, the tangent of 胃 is found as the ratio of the opposite side to the adjacent side. So, \( \tan 胃 = \frac{\sqrt{4 - (x + 1)^2}}{x + 1} \).
03

Substitute 胃 back into equation

Remember from step 1, we set \( 胃 = \cos^{-1} \frac{x + 1}{2} \). Replacing in the equation from step2, we get \( \tan \left(\cos ^{-1} \frac{x+1}{2}\right) = \frac{\sqrt{4-(x+1)^{2}}}{x+1} \), thus verifying the original identity.

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