How do you determine $tan θ$ $tantheta$ given $cos θ = \frac{1}{8}, \frac{3 π}{2} < θ < 2 π$ $costheta=1/8, (3pi)/2<theta<2pi$ ?

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Answer 1 · 2017-12-19T17:02:45

$- 3 \sqrt{7} .$ $-3sqrt7.$

$cos θ = \frac{1}{8} \Rightarrow sec θ = 8 .$ $costheta=1/8 rArr sectheta=8.$

$:. tan^2theta=sec^2theta-1=8^2-1=63=9*7.$

Hence, $tantheta=sqrt63=+-3sqrt7.$

But, given that, $3/2pi lt theta lt 2pi$ ,

$rArr tantheta=-3sqrt7.$

Answer 2 · 2017-12-19T17:22:48

$tantheta=-3sqrt7$

$costheta=1/8$

$or,cos^2theta=1/64$

$or,1-sin^2theta=1/64$ $color(blue)([[sin^2+cos^2theta=1,so,1-sin^2theta=cos^2theta]])$

$or,sin^2theta=1-1/64$

$or,sin^2theta=63/64$

$or,sintheta=+-sqrt63/8$

Here $sintheta=-sqrt63/8$ **,As in fourth quarter the value of $sintheta$ is negative **

$tantheta=sintheta/costheta=(-sqrt63/8/1/8=-sqrt63/cancel8xxcancel8)=-sqrt63$

So, $tantheta=-sqrt63=-3sqrt7$

How do you determine tanthetatanθtantheta given costheta=1/8, (3pi)/2<theta<2picosθ=18,3π2<θ<2πcostheta=1/8, (3pi)/2<theta<2pi?