How do you find the exact value of cos2x, given that cotx = -5/3 with pi/2<x<pi?

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Answer 1 · 2018-06-20T18:18:29

#cos2x=8/17#

Here,

#cotx=-5/3 < 0and pi/2 < x < pi=>II^(nd)Quadrant#

#=>sinx>0,cscx>0,cosx<0,secx<0,tanx<0#

Now , #csc^2x=1+cot^2x=1+25/9=34/9#

#=>sin^2x=1/csc^2x=9/34#

We know that,

#cos2x=1-2sin^2x=1-2(9/34)#

#=>cos2x=1-9/17=8/17#
...........................................................................
Note:

#pi/2 < x < pi=>2*pi/2 < 2*pi < 2*pi#

#=>pi < 2x < 2pi=>III^(rd)Quadrant or IV^(th)Quadrant#

But , #cos2x=8/17 > 0=>IV^(th)Quadrant#

Answer 2 · 2018-06-20T18:45:30

# 8/17#.

#because cotx=-5/3, :. tanx=-3/5#.

Hence, #cos2x=(1-tan^2x)/(1+tan^2x)={1-(-3/5)^2}/{1+(-3/5)^2},#

# i.e., cos2x=16/34=8/17#, as Respected Maganbhai P. has

already obtained!