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Chapter 5: Problem 47

Verify each identity. $$\frac{\sec t+1}{\tan t}=\frac{\tan t}{\sec t-1}$$

Short Answer

Expert verified
Yes, \( \frac{\sec t+1}{\tan t} = \frac{\tan t}{\sec t-1} \) is a valid identity.

Step by step solution

01

Rewrite Using Sine and Cosine

On the left side of the equation, replace \(\sec t\) with \(\frac{1}{\cos t}\) and \(\tan t\) with \(\frac{\sin t}{\cos t}\). This will transform the equation into \(\frac{1/\cos t +1}{\sin t/\cos t}\). Simplify this to get \(\frac{1+\cos t}{\sin t}\). On the other hand, the right side of the equation should be rewritten as \(\frac{\sin t/\cos t}{1/\cos t-1}\), which simplifies to \(\frac{\sin t}{1-\cos t}\)
02

Check Both Sides

Now, both sides of the equation are \(\frac{1+\cos t}{\sin t} = \frac{\sin t}{1-\cos t}\). Although these seem different, they are in fact the same. They are the reciprocals of each other.
03

Establish the Identity

An identity is shown if both sides of the equation are the same. We've found that both sides of the equation are reciprocals of each other. Therefore, we can multiply both sides of the equation by their denominators to confirm they're the same. Doing so yields, \( (1+\cos t) * (1- \cos t) = (\sin t) * (\sin t) \) which simplifies to \(1 - (\cos t)^2 = (\sin t)^2 \), which holds true since \((\sin t)^2 + (\cos t)^2 = 1\). Therefore, the original equation is indeed an identity.

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Most popular questions from this chapter

Use this information to solve. When throwing an object, the distance achieved depends on its initial velocity, \(v_{0}\) and the angle above the horizontal at which the object is thrown, \(\theta\) The distance, \(d\), in feet, that describes the range covered is given by $$d=\frac{v_{0}^{2}}{16} \sin \theta \cos \theta$$ where \(v_{0}\) is measured in feet per second. You and your friend are throwing a baseball back and forth. If you throw the ball with an initial velocity of \(v_{0}=90\) feet per second, at what angle of elevation, \(\theta,\) to the nearest degree, should you direct your throw so that it can be easily caught by your friend located 170 feet away?

Graph each side of the equation in the same viewing rectangle. If the graphs appear to coincide, verify that the equation is an identity. If the graphs do not appear to coincide, this indicates the equation is not an identity. In these exercises, find a value of \(x\) for which both sides are defined but not equal. $$\cos \left(x+\frac{\pi}{4}\right)=\cos x+\cos \frac{\pi}{4}$$

Will help you prepare for the material covered in the first section of the next chapter. Solve each equation by using the cross-products principle to clear fractions from the proportion: If \(\frac{a}{b}=\frac{c}{d},\) then \(a d=b c,(b \neq 0 \text { and } d \neq 0)\) Round to the nearest tenth. $$\text { Solve for } a: \frac{a}{\sin 46^{\circ}}=\frac{56}{\sin 63^{\circ}}$$

Use a graphing utility to approximate the solutions of each equation in the interval \([0,2 \pi) .\) Round to the nearest hundredth of a radian. $$\sin x+\sin 2 x+\sin 3 x=0$$

Graph each side of the equation in the same viewing rectangle. If the graphs appear to coincide, verify that the equation is an identity. If the graphs do not appear to coincide, this indicates the equation is not an identity. In these exercises, find a value of \(x\) for which both sides are defined but not equal. $$\sin x=-\cos x \tan (-x)$$
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