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Chapter 6: Problem 77

Verify the identity. $$\tan \left(\sin ^{-1} \frac{x-1}{4}\right)=\frac{x-1}{\sqrt{16-(x-1)^{2}}}$$

Short Answer

Expert verified
The given identity \(\tan \left(\sin ^{-1} \frac{x-1}{4}\right) = \frac{x-1}{\sqrt{16-(x-1)^{2}}}\) holds true. After forming the triangle from the inverse sine function, applying Pythagorean theorem to find the third side and then evaluating the tangent, both sides of the identity match perfectly.

Step by step solution

01

Create the Triangle for the Inverse Sine Function

To create the triangle for the inverse sine function, label the angles and sides according to the given function. In this case, \(\sin^{-1}\frac{x-1}{4}\) means the angle \(\theta\) of triangle ABC satisfies \(\sin\theta = \frac{x-1}{4}\), where AB is \(x-1\) and BC is 4.
02

Find the Missing Side AC

By using the Pythagorean theorem, the third side AC can be derived. \[AC = \sqrt{BC^{2}-AB^{2}} = \sqrt{16-(x-1)^{2}}\].
03

Evaluate the Tangent of the Angle

The tangent of any angle is given by \(\frac{{opposite}}{{adjacent}}\), i.e., \[ \tan \theta = \frac{AB}{AC} = \frac{x-1}{\sqrt{16-(x-1)^2}}\].
04

Connect the Result with Initial Expression

The tangent of the angle is equal to the tangent of the angle given in the expression. Therefore, \(\tan \left(\sin ^{-1} \frac{x-1}{4}\right) = \frac{x-1}{\sqrt{16-(x-1)^{2}}}\)

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