#secx/sinx - sinx/cosx = cotx#
I'll be simplifying down the LHS of the equation to get to the RHS of the equation. The #color(blue)("blue color")# refers to what is being changed.
First, we know that #color(blue)(sinx/cosx)# is the same as #color(blue)(tanx)#:
#secx/sinx - color(blue)(tanx)#
Now, multiply #color(blue)(sinx/sinx)# to #color(blue)(tanx)# so that both expressions have the same denominator:
#secx/sinx - (tanxcolor(blue)(sinx))/color(blue)(sinx)#
Combine both expressions to one denominator:
#(secx - tanxsinx)/sinx#
We know that #color(blue)(secx = 1/cosx)# and #color(blue)(tanx = sinx/cosx)#:
#(color(blue)(1/cosx) - color(blue)(sinx/cosx)*sinx)/sinx#
Multiply #color(blue)sinx# with #color(blue)sinx#:
#(1/cosx - color(blue)(sin^2x)/cosx)/sinx#
Combine numerator and denominator (from the top)
#(color(blue)(1-sin^2x)/cosx)/sinx#
From the Pythagorean Identities, we know that #color(blue)(1-sin^2x = cos^2x)#:
#(color(blue)(cos^2x/cosx))/sinx#
Divide #color(blue)(cos^2x/cosx)#:
#(color(blue)(cosx/1))/sinx#
Simplify:
#color(blue)(cosx)/sinx#
We know that #color(blue)(cosx/sinx = cotx)#:
#color(blue)cotx#
Now, the left hand side is equivalent to the right hand side, so we have proven that #secx/sinx - sinx/cosx = cotx#.
Hope this helps!
