How do you use $csc θ = 5$ $csctheta=5$ to find $sec (90^{\circ} - θ)$ $sec(90^circ-theta)$ ?

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How do you use $csc θ = 5$ $csctheta=5$ to find $sec (90^{\circ} - θ)$ $sec(90^circ-theta)$ ?

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Answer 1 · 2017-02-20T04:35:13

sec (90 - t) = 5

Explanation:

$csc t = \frac{1}{sin t} = 5$ $csc t = 1/(sin t) = 5$ --> $sin t = \frac{1}{5}$ $sin t = 1/5$
Unit circle and property of complement arcs -->
$cos (90 - t) = sin t$ $cos (90 - t) = sin t$
There for
$sec (90 - t) = \frac{1}{cos (90 - t)} = \frac{1}{sin t} = 5$ $sec (90 - t) = 1/(cos (90 - t)) = 1/(sin t) = 5$

Answer 2 · 2018-02-27T13:30:31

$sec (90 - θ) = 5$ $sec (90-theta) = 5$

Explanation:

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$sec (90 - θ) = csc θ$ $sec(90-theta) = csc theta$

$G i v e n : csc θ = 5$ $Given : csc theta = 5$

Using cofunctions,

$:. sec(90-theta) csc theta = 5$

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How do you use $csc θ = 5$ $csctheta=5$ to find $sec (90^{\circ} - θ)$ $sec(90^circ-theta)$ ?

2 Answers

Explanation:

Explanation:

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How do you use csctheta=5cscθ=5csctheta=5 to find sec(90^circ-theta)sec(90∘−θ)sec(90^circ-theta)?

2 Answers

Explanation:

Explanation:

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How do you use $csc θ = 5$ $csctheta=5$ to find $sec (90^{\circ} - θ)$ $sec(90^circ-theta)$ ?