How do you prove #(sinx+cosx)/sinx - (cosx-sinx)/cosx = sec x csc x#?

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Answer 1 · 2018-04-24T16:14:13

Please see below.

We know that,

#color(red)((1)cos^2x+sin^2x=1#

#color(blue)((2)1/cosx=secx and 1/sinx=cscx#

Here,

#(sinx+cosx)/sinx-(cosx-sinx)/cosx=secxcscx#

We take,

#LHS=(sinx+cosx)/sinx-(cosx-sinx)/cosx#

#=(cosx(sinx+cosx)-sinx(cosx-sinx))/(sinxcosx)#

#=(cancel(sinxcosx)+cos^2x-cancel(sinxcosx)+sin^2x)/(sinxcosx#

#=color(red)((cos^2x+sin^2x))/(sinxcosx)...tocolor(red)(Apply(1)#

#=color(red)(1)/(sinxcosx)#

#=color(blue)(1/cosx*1/sinx...tocolor(blue)(Apply(2)#

#=color(blue)(secxcscx#

#=RHS#