How do you find the value of #cot2theta# given #cottheta=4/3# and #pi<theta<(3pi)/2#?

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Answer 1 · 2016-07-21T06:07:34

#cot2theta=7/24#

#cot2theta=(cot^2theta-1)/(2cottheta)#

= #((4/3)^2-1)/(2xx4/3)#

= #(16/9-1)/(8/3)#

= #((16-9)/9)/(8/3)#

= #7/9xx3/8#

= #7/(3cancel9)xx(1cancel3)/8#

= #7/24#