Find the values of the trigonometric functions of θ if csc θ = −9/5 and θ is in Quadrant IIIsin=cos=tan=cot=sec=. Created by Gerald. science-mathematics-en - mathematics-en

I can't write theta, so I'll use A instead. A is in Q3so sinA & cosA <0, tanA>0. cscA = -9/5 so sinA = -5/9; cosA = 1 - sin^2A - 1 - 25/81 = 56/81 so cosA = -2rt14/9; tanA = sinA/cosA = [5/9}/{2rt14/9) = 5/2rt14 = 5rt14/2; cotA = 1/tanA = 2rt14/5; secA = 1/cosA = - 9/2rt14 = - 9rt14/2Source(s): Retired Maths Teacher00Still have questions? Get your answers by asking now.Ask Question+100Join Yahoo Answers and get 100 points today.JoinTrending QuestionsTrending QuestionsConvert the whole number to a percent: 58?18 answersThe sum of three consecutive integers is 36. What is the largest of the three integers? The correct answer is 13.?9 answersHelp with this math question?7 answersAm I right?6 answersHow do you solve this math word problem? PLEASE HELP ME AND EXPLAIN STEP BY STEP!!!?7 answersAnswer QuestionsAnswer QuestionsSolve the equation. (Find all solutions of the equation in the interval [0, 2𝜋). 2tan(2x)-4cos(x).?Write a system of two equations in two variables to solve the problem?9. Write a system of two equations in two variables to solve the problem?Calculus velocity problem?statistics probability?Terms・Privacy・AdChoices・RSS・HelpAbout Answers・Community Guidelines・Leaderboard・Knowledge Partners・Points & LevelsSend Feedback・International Sites

Find the values of the trigonometric functions of θ if csc θ = −9 5 and θ is in Quadrant III?

Gerald @Gerald

May 2021 1 90 Report

Find the values of the trigonometric functions of θ if csc θ = −9 5 and θ is in Quadrant III?

Find the values of the trigonometric functions of θ if csc θ = −9/5 and θ is in Quadrant III

sin=

cos=

tan=

cot=

sec=

Science & Mathematics / Mathematics

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John

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I can't write theta, so I'll use A instead. A is in Q3so sinA & cosA <0, tanA>0. cscA = -9/5 so sinA = -5/9; cosA = 1 - sin^2A - 1 - 25/81 = 56/81 so cosA = -2rt14/9; tanA = sinA/cosA = [5/9}/{2rt14/9) = 5/2rt14 = 5rt14/2; cotA = 1/tanA = 2rt14/5; secA = 1/cosA = - 9/2rt14 = - 9rt14/2

Source(s): Retired Maths Teacher

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