How do you simplify #(2-sec^2x)/(sec^2x)#?

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Answer 1 · 2018-04-06T21:42:51

#(2-sec^2(x))/sec^2(x) = cos(2x)#

First, split the fraction.

#(2 - sec^2(x))/sec^2(x)#

#= 2/sec^2(x) - sec^2x/sec^2(x)#

Because #cos(x) = 1/sec(x)#, #color(red)(cos^2(x) = 1/sec^2(x))#

#= 2/color(red)(sec^2(x)) - sec^2(x)/sec^2(x)#

#= color(blue)(2cos^2(x) - 1)#

Observe the following:

#color(blue)cos(2x)#

#= cos^2(x) - sin^2(x)#

# = cos^2(x) - (1 - cos^2x)#

# = color(blue)(2cos^2(x) - 1)#

Therefore,

#(2-sec^2(x))/sec^2(x) = cos(2x)#

Answer 2 · 2018-04-06T21:47:49

It simplifies to #cos(2x)#.

Use these identites:

#cos(2x)=2cos^2x-1#

#secx=1/cosxqquadcolor(blue)=>qquadsec^2x=1/cos^2x#

First, split the fraction:

#color(white)=(2-sec^2x)/sec^2x#

#=2/sec^2x-sec^2x/sec^2x#

#=2/sec^2x-1#

#=2*1/sec^2x-1#

#=2*1/(1/cos^2x)-1#

#=2*cos^2x-1#

#=2cos^2x-1#

#=cos(2x)#

Hope this helped!