Given Tangent Theta= -4/5, Theta in quadrant 2 , how do you find Cos Theta?

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Answer 1 · 2016-07-12T18:50:11

$cos θ = - \frac{5}{\sqrt{41}}$ $costheta=-5/sqrt41$ .

Given that $tan θ = - \frac{4}{5}, θ \in Q_{I I}$ $tantheta=-4/5, theta in Q_(II)$

Recall that $cos θ = \frac{1}{sec θ}$ $costheta=1/sectheta$ and, $tan$ $tan$ and $sec$ $sec$ are related

by the Identity $: {sec}^{2} θ = 1 + {tan}^{2} θ .$ $: sec^2theta=1+tan^2theta.$

$:. sec^2theta=1+(-4/5)^2=1+16/25=41/25.$

$:. sectheta=+-sqrt41/5 rArr costheta=+-5/sqrt41$

But, $theta in Q_(II) rArr costheta$ is $-ve$ .

Hence, $costheta=-5/sqrt41$ .