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Need help: Find the exact value of sec(2tan^-1 (3/4)
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hmmm lemme take a poke
$$\bf sec\left(2tan^{-1}\left(\cfrac{3}{4}\right)\right)\\ sec(2\theta) \implies \cfrac{1}{cos(2\theta)}\\ cos(2\theta) = 1- 2 sin^2(\theta)\\ \text{now let's check above}\\ \color{green}{ tan^{-1}\left(\cfrac{3}{4}\right) \implies \cfrac{b}{a}\\ c = \sqrt{4^2+3^2} = 5 } $$
so \(\bf cos(2\theta) = 1- 2 sin^2(\theta) \implies cos(2\theta) = 1- 2\left(\cfrac{3}{5}\right)^2\)
so, that'd give you the cosine, flip it around to the secant
gives me 7/25, so the secant will be 25/7
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