If sec(t) = 3, how do you find the exact value of ${sin}^{2} (t)$ $sin^2 (t)$ ?

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If sec(t) = 3, how do you find the exact value of ${sin}^{2} (t)$ $sin^2 (t)$ ?

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Answer 1 · 2015-11-27T00:43:35

Find ${sin}^{2} t$ $sin^2 t$

Ans: ${sin}^{2} t = \frac{8}{9}$ $sin^2 t = 8/9$

Explanation:

$sec t = \frac{1}{cos t} = 3$ $sec t = 1/(cos t) = 3$ --> $cos t = \frac{1}{3}$ $cos t = 1/3$ --> ${cos}^{2} t = \frac{1}{9}$ $cos^2 t = 1/9$
${sin}^{2} t = 1 - {cos}^{2} t = 1 - \frac{1}{9} = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}$ $sin^2 t = 1 - cos^2 t = 1 - 1/9 = 9/9 - 1/9 = 8/9$

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If sec(t) = 3, how do you find the exact value of ${sin}^{2} (t)$ $sin^2 (t)$ ?

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Explanation:

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If sec(t) = 3, how do you find the exact value of ${sin}^{2} (t)$ $sin^2 (t)$ ?