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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
Yes, the trigonometric identity \( \tan \left(\cos^{-1} \frac{x+1}{2}\right) = \frac{\sqrt{4-(x+1)^{2}}}{x+1} \) holds true.

Step by step solution

01

Expressing \( \tan \) as ratio of sine and cosine and determine \( \cos \cos^{-1} \)

To begin with, one need to remember that \( \tan \theta = \frac{\sin \theta}{\cos \theta} \). So, \( \tan \left(\cos^{-1} \frac{x+1}{2}\right) = \frac{\sin{\left(\cos^{-1} \frac{x+1}{2}\right)}}{\cos{\left(\cos^{-1} \frac{x+1}{2}\right)}} \). Now, use the property of inverse cosine function, that \( \cos \left(\cos^{-1} \alpha\right) = \alpha \), thus the denominator of the fraction is \( \frac{x+1}{2} \).
02

Calculating the value of \( \sin \cos^{-1} \)

To get the numerator of the fraction, one needs to find \( \sin \theta \) when \( \cos \theta = \frac{x+1}{2} \). From the Pythagorean theorem, we have \( \sin^2 \theta + \cos^2 \theta = 1 \). So, after rearranging, \( \sin \theta = \sqrt{1 - \cos^2 \theta} = \sqrt{1 - \left(\frac{x+1}{2}\right)^2} = \sqrt{1 - \frac{(x+1)^2}{4}} = \sqrt{\frac{4-(x+1)^2}{4}} \).
03

Simplify the expression

Now, substitute the sine and cosine values into the expression: \( \tan \left(\cos^{-1} \frac{x+1}{2}\right) = \frac{\sin{\left(\cos^{-1} \frac{x+1}{2}\right)}}{\cos{\left(\cos^{-1} \frac{x+1}{2}\right)}} = \frac{\sqrt{\frac{4-(x+1)^2}{4}}}{\frac{x+1}{2}} \). Simplify the expression to get \( \frac{\sqrt{4-(x+1)^2}}{(x+1)} \), which is the same as the right-hand side of the original identity.

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