Simplify?

#secx/sinx#-#sinx/cosx#

3 Answers
Aviv S.
Mar 4, 2018

Write everything in terms of #sinx# and #cosx#. These are the identities you need to use:

#secx=1/cosx#

#cotx=cosx/sinx#

#cos^2x+sinx^2=1#

#=>cos^2x=1-sin^2x#

Here's the actual problem:

#color(white)=secx/sinx-sinx/cosx#

#=(1/cosx)/sinx-sinx/cosx#

#=((1/cosx)*cosx)/(sinx*cosx)-(sinx*sinx)/(sinx*cosx)#

#=((1/color(red)cancelcolor(black)cosx)*color(red)cancelcolor(black)cosx)/(sinx*cosx)-(sinx*sinx)/(sinx*cosx)#

#=1/(sinx*cosx)-(sin^2x)/(sinx*cosx)#

#=(1-sin^2x)/(sinx*cosx)#

#=cos^2x/(sinx*cosx)#

#=cos^color(red)cancelcolor(black)2x/(sinx*color(red)cancelcolor(black)cosx)#

#=cosx/sinx#

#=cotx#

Hriman
Mar 4, 2018

#secx/sinx-sinx/cosx=#

#1/(cosxsinx)-sinx/cosx=#

#1/(cosxsinx)-(sinx(sinx))/(cosx(sinx))=#

#(1-sin^2x)/(cosxsinx)=#

#cos^2x/(cosxsinx)=#

#cosx/sinx=#

#cotx#

:)
Mar 4, 2018

The answer is #cot(x)#.

Explanation:

#sec(x)/sin(x)-sin(x)/cos(x)#

Using #sec(t)=1/cos(t)#, transform the expression.
#=1/cos(x)/sin(x)-sin(x)/cos(x)#

Simplify the complex fraction.
#=1/cos(x)sin(x)-sin(x)/cos(x)#

Write all numerators above the least common denominator #cos(x)sin(x)#.
#=1-sin(x)^2/cos(x)sin(x)#

Using #1-sin(t)^2=cos(t)^2#, simplify the expression.
#=cos(x)^2/cos(x)sin(x)#

Reduce the fraction with #cos(x)#
#=cos(x)/sin(x)#

Using #cos(t)/sin(t)=cot(t)#, transform the expression.
#=cot(x)#

I know this is freaking messy, sorry :)

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