How do you find the value of #cot(theta/2)# given #tantheta=-7/24# and #(3pi)/2<theta<2pi#?

1 Answer
Ratnaker Mehta
Jun 22, 2016

#cot(theta/2)=-7.#

Explanation:

We know that,
#tantheta=(2tan(theta/2))/{1-tan^2(theta/2)}#

So, letting #tan(theta/2)=t#, & using given value of #tantheta=-7/24# in the LHS, we get the eqn.,

#-7/24=(2t)/(1-t^2),# or,#-7+7t^2=48t,# i.e., #7t^2-48t-7=0.#
#:.(t-7)(7t+1)=0,.# which gives #t=tan(theta/2)=7, or, -1/7.#

Clearly, #cot(theta/2)=1/7, or, -7.#

Given that #3pi/2#<#theta#<#2pi.# #rArr# #3pi/4#<#(theta/2)#<#pi,# meaning, #theta/2# lies in the IInd Quadrant, where cot is #-ve.#

Finally, #cot(theta/2)!=1/7, but, =-7.#

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