How do you prove #(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)#?

Question

5293 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T18:52:35

See Below

#LHS : (tan x+cot x)/(sec x + csc x)#

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

#=((sin^2x+cos^2x)/(sinxcosx))/((sinx+cosx)/(sinxcosx))#->common denominator

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

#=(sin^2x+cos^2x)/cancel(sinxcosx) *cancel(sinxcosx)/ (sinx+cosx)#

#=(sin^2x+cos^2x)/(sinx+cosx)#->use property #sin^2x+cos^2x=1#

#=1/(sinx+cosx)#

#=RHS#