Solve each equation for 0 180 a sec 3cos tan b 2 sec 2 2 3tan2 1 c sec 4 2 6tan 2 d 2 cot 2 2 7 cosec 2 2 ...

a Rearranging the equation we get sec tan 3cos Substituting sec 1cos and tan sincos we get 1cos sincos 3cos Multiplying both sides by cos we get 1 sin ...

Solve each equation for 0 < 180. a. sec = 3cos tan b. 2 sec 2

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Solve each equation for 0° < 180°.
a. secθ = 3cosθ − tanθ

b. 2 sec²2θ = 3tan2θ + 1

c. sec⁴θ + 2 = 6tan²θ

d. 2 cot² 2θ + 7 cosec 2θ = 2

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Cambridge International AS & A Level Mathematics Pure Mathematics 2 & 3 Coursebook

ISBN: 9781108407199

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Authors: Sue Pemberton, Julianne Hughes, Julian Gilbey

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Solve each equation for 0 < 180. a. sec = 3cos tan b. 2 sec 2

Question:

Step by Step Answer:

a Rearranging the equation we get sec tan 3cos Substituting sec 1cos and tan sincos we get 1cos sincos 3cos Multiplying both sides by cos we get 1 sin ...View the full answer

Cambridge International AS & A Level Mathematics Pure Mathematics 2 & 3 Coursebook

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