How do you determine #tantheta# given #costheta=1/8, (3pi)/2<theta<2pi#?

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Answer 1 · 2017-12-19T17:02:45

# -3sqrt7.#

#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#