Use the given information to find tan(s + t). See Ex...

Question

Use the given information to find tan(s + t). See Example 3.

cos s = -15/17 and sin t = 4/5, s in quadrant II and t in quadrant I

Accepted Answer

Hello, today we're going to be solving for the value of tangent of X plus Y by using the trigonometric identity. Before we jump into the problem, let's write down the trigonometric identity for tangent of X plus Y tangent of X plus Y is given to us as tangent of X plus tangent of Y divided by one minus tangent of X multiplied by tangent of Y. In order to solve for tangent of X plus Y, there are two values that we need, we need the tangent X value and we need the tangent Y value. The problem doesn't provide those values for us. However, we can solve for these values by using the Pythagorean identity for tangent of X and tangent of Y. The Pythagorean identity for tangent of X will equal to sine of X divided by cosine of X. And the Python identity for tangent of Y will be given to us as sine of Y divided by cosine of Y. Now, the problem gives us a cosine X value and a sine Y value. But what we are not given is a cosine Y value and assign X value. So in order to solve for tangent of X and tangent of Y. We'll first need to solve for sine of X and cosine of Y. Let's go ahead and start by solving for cosine of Y. In order to solve for cosine of Y, we'll be using the pythagorean identity sine squared of Y plus cosine squared of Y is equal to one. Let's go ahead and solve for cosine within this identity. If we subtract both sides of this equation by sine squared of Y, we can rewrite the equation as cosine squared of Y is equal to one minus sine squared of Y. By taking the square root of both sides of this equation will be left with cosine of Y is equal to the positive and negative square root of one minus sine squared of Y. Now we need to be careful because the square root produces a positive and negative value. So we need to figure out if we want the positive value or negative value of the square root. Keep in mind that we are using the angle Y and we are told that the angle Y lies within quadrant, one of the unit circle cosine is positive in quadrant one of the unit circle. So we'll be taking the positive version of the square root. Furthermore, we have the value of sine that we can plug into the square root. So we can rewrite the equation as cosine of Y is equal to the positive square root of one minus the quantity 84 divided by 85 squared. After algebraically simplifying will be left with 13 divided by 85 for the value of cosine of Y. Now we need to solve for the value of sin of X. Now we'll be using the same Pythagorean identity as before. But this time it will be with respect to the X angle. So the Pythagorean identity we'll be using is sine squared of X plus cosine squared of X is equal to one. Now, just like before we're going to solve for sine by subtracting both sides of this equation by cosine squared of X that will leave us with sine squared of X is equal to one minus cosine squared of X. By taking the square root of both sides, we will be left with sine of X is equal to the positive and negative square root of one minus cosine squared of X. Now, just like the cosine Y value, we'll need to figure out if we want to keep the positive or negative version of this square root. Well, keep in mind that the problem tells us that the X angle lies within quadrant two of the unit circle. Sine is positive in quadrant two of the unit circle. So we'll be taking the positive version of the square root. Furthermore, we know that the value of cosine is negative 112 divided by 113. So we can plug in that value to the square root and that will leave us with sin of X is equal to the positive square root of one minus negative 112 divided by 113. That quantity squared. After solving algebraically, we will be left with the fraction 15 divided by 113. Now that we have the missing cosine Y and sine X value, we can now solve for the tangent of X and tangent of Y values. Let's start by solving for tangent of X. Now again, tangent of X is equal two sine of X which we solved to be 15 divided by 113 divided by cosine of X, which is given to us as negative 112 divided by 113 by algebraically simplifying tangent of X will equal to negative 15 divided by 112. Now we'll need to solve for a tangent of why we'll be using the same Pythagorean identity tin of wine is equal to sin of Y which is given to us as 84 divided by 85 divided by cosine of Y which we solved to be 13 divided by 85. After algebraically simplifying the value of tangent of Y will give us 84 divided by 13. Now that we have the values for tangent of X and tangent of Y, we can plug in these values to solve for tangent of X plus Y So tangent of X plus Y is given to us as tangent of X which is equal to negative 15 divided by 112 plus tangent of Y which is 84 divided by 13. This quantity divided by one minus tangent of X which is negative 15 divided by 112 multiplied by tangent of Y which is 84 divided by 13. After algebraically simplifying this value will be left with 9213 divided by 2716. This is going to be the exact value of tangent of X plus Y. And with that being said, the answer to this problem is going to be D. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

Use the given information to find tan(s + t). See Example 3.
cos s = -15/17 and sin t = 4/5, s in quadrant II and t in quadrant I

Key Concepts

Trigonometric Identities for Sum of Angles

Determining Sine, Cosine, and Tangent from Given Information

Sign of Trigonometric Functions in Different Quadrants

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