In Exercises 17–20, θ is an acute angle and sin θ and cos θ are g...

Question

In Exercises 17–20, θ is an acute angle and sin θ and cos θ are given. Use identities to find tan θ, csc θ, sec θ, and cot θ. Where necessary, rationalize denominators.sin θ = 3/5, cos θ = 4/5

Accepted Answer

Hello everybody. I hope you're doing. All right. Today, we are going to be looking at this question that states an acute angle, alpha and trigonometric functions sign of alpha and cosine alpha are given. And they're asking us to find the remaining trigonometric functions. Tangents of alpha, cotangent alpha, C kins alpha and coins alpha by using identities. Now they tell us that sin of alpha is equal to five divided by 13 and cosine of alpha is 12 divided by 13. Now using the identities, let's first tackle tangent. So we know that tangent of alpha or any angle is equal to sign of that angle. In this case, alpha divided by cosine of the same angle in this case alpha once again. So that will be equal to, they give us both sine of alpha and cosine of alpha. So we can plug that in and have open parentheses. Five divided by 13, closed parentheses, divided by open parentheses, 12, divided by 13, closed parentheses. And in this scenario, the thirteens will cancel out and will be left with five divided by 12. So just by using that simple relationship between tangents sign and cosine, we are able to obtain our tangent of alpha and we can move on to coins. So we have cos alpha now is just one divided by sign of the angle. So in this case, sin of alpha. So when we have one divided by s sign of alpha, our sign of alpha is a fraction. So all we have to do is flip that fraction and we'll be left with 13 divided by five. And that would be our answer for alpha. So of alpha is 13 divided by five and we can move on to alpha. Now, now similar to cos C can will be one divided by cosine of alpha in this case. And we do the same thing since our cosine of alpha is a fraction and we have one divided by a fraction. All we have to do is flip that fraction and we'll be left with divided by 12 as our answer for C can of alpha. And finally, we can move on to cotangent of alpha and similar to co C cans and C cans. Co tangent of alpha will be one divided by tangent alpha. And once again, our tangent alpha was a fraction and we have one divided by a fraction. All you do is flip that fraction and we'll have 12 divided by five as our answer for cotangents of alpha. And just like that, we've been able to solve for all the trigonometric functions by using simple relationships between all of them. So, I hope that was helpful and until next time.

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In Exercises 17–20, θ is an acute angle and sin θ and cos θ are given. Use identities to find tan θ, csc θ, sec θ, and cot θ. Where necessary, rationalize denominators.sin θ = 3/5, cos θ = 4/5

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