How to prove using trigonometric identities?

Question

#(tanx+cotx)/(secx+cscx)=(1)/(cosx+sinx)#

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Answer 1 · 2018-03-03T04:59:57

Let's try manipulating the left side...

#(tanx+cotx)/(secx+cscx)=1/(cosx+sinx)#

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

#(sin^2x/(cosxsinx)+cos^2x/(cosxsinx))/(sinx/(cosxsinx)+cosx/(cosxsinx))=1/(cosx+sinx)#

#((sin^2x+cos^2x)/(cosxsinx))/((sinx+cosx)/(cosxsinx))=1/(cosx+sinx)#

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

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

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

Answer 2 · 2018-03-03T05:02:22

See below

To prove: #(tanx+cotx)/(secx + cscx) = 1/(cosx+sinx)#

#LHS = (tanx+cotx)/(secx + cscx) #

#LHS# Numerator #= sinx/cosx+cosx/sinx#

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

#=1/(sinxcosx)#

#LHS# Denominator #= 1/cosx+1/sinx#

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

#:. LHS = (1/(sinxcosx))/((sinx+cosx)/(sinxcosx)) = (sinxcosx)/(sinxcosx(sinx+cosx))#

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

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

#= RHS#