How do you prove #(cos[theta]+cot[theta])/(csc[theta]+1)= cos[theta]#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer sente May 2, 2016 Using the definitions of #cot(theta)# and #csc(theta)#, for #sin(theta)!=0# and #sin(theta)!=-1#, we have #(cos(theta)+cot(theta))/(csc(theta)+1) = (cos(theta)+cos(theta)/sin(theta))/(1/sin(theta)+1)# #=cos(theta)(1+1/sin(theta))/(1+1/sin(theta))# #=cos(theta)*1# #=cos(theta)# Answer link Related questions What does it mean to prove a trigonometric identity? How do you prove #\csc \theta \times \tan \theta = \sec \theta#? How do you prove #(1-\cos^2 x)(1+\cot^2 x) = 1#? How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#? How do you prove that #(2sinx)/[secx(cos4x-sin4x)]=tan2x#? How do you verify the identity: #-cotx =(sin3x+sinx)/(cos3x-cosx)#? How do you prove that #(tanx+cosx)/(1+sinx)=secx#? How do you prove the identity #(sinx - cosx)/(sinx + cosx) = (2sin^2x-1)/(1+2sinxcosx)#? See all questions in Proving Identities Impact of this question 3739 views around the world You can reuse this answer Creative Commons License