How do you find the remaining trigonometric ratios if $sec (θ) = - 3.5$ $sec(theta) = -3.5$ and $\frac{π}{2} < θ < π$ $pi/2< theta < pi$ ?

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Answer 1 · 2015-05-30T02:48:29

$sec x = \frac{1}{cos x} = - 3.5 \to cos x = \frac{1}{- 3.5} = - 0.286$ $sec x = 1/cos x = -3.5 -> cos x = 1/-3.5 = -0.286$

${sin}^{2} x = 1 - {(0.286)}^{2} = 0.92 \to sin x = 0.96$ $sin^2 x = 1 - (0.286)^2 = 0.92 -> sin x = 0.96$

$tan x = \frac{0.96}{- 0.286} = - 3.36$ $tan x = 0.96/-0.286 = -3.36$

$cot x = \frac{1}{- 3.36} = 0.30$ $cot x = 1/ -3.36 = 0.30$

$csc x = \frac{1}{sin x} = \frac{1}{0.96} = 1.04$ $csc x = 1/(sin x) = 1/0.96 = 1.04$

How do you find the remaining trigonometric ratios if sec(θ)=−3.5sec(theta) = -3.5 and π2<θ<πpi/2< theta < pi?