SOLUTION: Im not sure how to solve for theta in: csc(θ+40)=sec(θ-20)

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Question 1109123: Im not sure how to solve for theta in: csc(θ+40)=sec(θ-20)
Found 2 solutions by Alan3354, ikleyn:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
csc(t+40)=sec(t-20)
1/sin(t+40)=1/cos(t-20)
sin(t+40)=cos(t-20)
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Similar to the one you solved earlier.
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solving for the variable: cos(2θ+50)=sin(2θ-20)
I changed it cosine to sine and got
90-2theta+50=2theta-20
40-2theta=2theta-20
60=4theta
theta=15
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Probably you, maybe someone else.

Answer by ikleyn(52908) About Me  (Show Source):
You can put this solution on YOUR website!
.
csc(t+40°) = sec(t-20°)     <=== ===>

sin(t+40°) = cos(t-20°)     <=== ===>

(t+40°) + (t-20°) = 90°


2t + 20° = 90°  ====>  2t = 90° - 20° = 70°  ====>


There are TWO solutions in the interval  0 <= t < 360°:


    1)  t = 70%5Eo%2F2 = 35°,     and

    2)  t = 70%5Eo%2F2 + 180° = 215°.




Plots y = 1/csc(t+40) (red) and y = 1/sce(t-20) (green)

      (horizontal axis in degrees)