How do you prove #secx - cosx = tanx * sinx#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer sente May 2, 2016 Using the definitions #sec(x)=1/cos(x)# and #tan(x)=sin(x)/cos(x)# along with the identity #sin^2(x)+cos^2(x)=1 => sin^2(x)=1-cos^2(x)#, for #cos(x)!=0# we have #sec(x)-cos(x) = 1/cos(x)-cos^2(x)/cos(x)# #=(1-cos^2(x))/cos(x)# #=sin^2(x)/cos(x)# #=sin(x)/cos(x)*sin(x)# #=tan(x)*sin(x)# 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 2765 views around the world You can reuse this answer Creative Commons License