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

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Answer 1 · 2016-01-10T12:26:21

8

${sin}^{2} t + {cos}^{2} t = 1$ $sin^2t+cos^2t=1$

If we divide this identity by ${cos}^{2} t$ $cos^2t$ we get:

${tan}^{2} t + 1 = {sec}^{2} t$ $tan^2t+1=sec^2t$

If $sec t = 3$ $sect=3$ then:

${tan}^{2} t + 1 = 9$ $tan^2t+1=9$

${tan}^{2} t = 8$ $tan^2t=8$

If sec(t) = 3, how do you find the exact value of tan^2 (t)tan2(t)tan^2 (t)?