In Exercises 7–14, use the given information to find the exact va...

Question

In Exercises 7–14, use the given information to find the exact value of each of the following:

c. tan 2θ

cot θ = 3, θ lies in quadrant III.

Accepted Answer

Hey, everyone in this problem, we're asked to determine the exact value of the expression using the provided data. We're asked to evaluate 10 of two beta where cotangent of beta is equal to four and beta lies in quadrant three. We're given four answer choices. Option A eight divided by 15. Option B 15, divided by 32. Option C 15, divided by eight and option D 16 divided by 15, we're gonna start with the expression we're trying to evaluate and that is of two beta. Now, the information we're given about this problem is about cotangent of beta. So what we wanna do is we want to use our double angle identities. We wanna write tangent of two beta in terms of some trick functions with the angle of just beta instead of two beta and recall your trick identity that says that tangent of two beta is equal to two tan beta divided by one minus 10 square be. So in order to calculate tangent of two beta, we just need to calculate tangent of beta and substitute it into this identity. And we're gonna go ahead and call this equation one. And now we're gonna find tangent of beta. Now, what we're given is cotangent of beta. Let's recall the relationship between code tangent and tangent and the cotangent of beta is just the reciprocal of tangent. So we have that cotangent of beta is equal to one divided by the tangent of beta, which tells us that tan beta is equal to one divided by the co tangent of beta. All right. So we know the code tangent of beta, we're gonna substitute in that value and we'll have tan beta exactly what we need to evaluate our expression. OK. So 10 beta is going to be equal to one divided by a cotangent of beta, which is four. So we have this tan beta value. Now, we can use that in our expression in equation one to solve for tan two beta. So we're gonna do that. Now we have that 10 of two beta is equal to two multiplied by 10 beta, which we've found to be one quarter divided by one minus 10 squared beta. So one minus one quarter squared. Now, we've dealt with all of the trig identities. We've dealt with these trig functions. Now, we just have to have a fraction that we need to simplify. And we know how to do that. The hard part is done. Let's go ahead and simplify this fraction. We have two multiplied by a quarter in the numerator that's gonna give us one half and that is divided by one minus one quarter squared. So that's one minus 1 16, we want a common denominator in the denominator so that we can simplify this into a single fraction. So we have one half divided by, we can write one as 16, divided by 16. And then we have minus 1/ this gives us one half divided by 15 16. And when we're dividing by a fraction, we can multiply by the reciprocal. So we can write this as one half multiplied by 16 15. Both the numerator and denominator are divisible by two. So if we divide the numerator by two, we can change that 16 into an eight. And then the denominator, that term of two is going to go away. And what we are left with is that tangent of two beta is equal to 8/15. And that is the exact value for the expression that we were looking for. If we compare this to our answer choices, we can see that this corresponds with answer choice. A thanks everyone for watching. I hope this video helped see you in the next one.

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In Exercises 7–14, use the given information to find the exact value of each of the following: c. tan 2θ cot θ = 3, θ lies in quadrant III.