How do you prove #1+(Cosθ + Sinθ)(Cosθ-Sinθ)= 2/sec^2 θ#? Trigonometry Trigonometric Identities and Equations Proving Identities 1 Answer Lucy Apr 18, 2018 Below Explanation: #1+(costheta+sintheta)(costheta-sintheta)=2/(sectheta)^2# LHS #1+(costheta+sintheta)(costheta-sintheta)# =#1+((costheta)^2-sinthetacostheta+sinthetacostheta-(sintheta)^2)# =#1+(costheta)^2-(sintheta)^2# =#(costheta)^2+(sintheta)^2+(costheta)^2-(sintheta)^2# =#2(costheta)^2# RHS #2/(sectheta)^2# =#2/(1/(costheta)^2# =#2(costheta)^2# =LHS Therefore, #1+(costheta+sintheta)(costheta-sintheta)=2/(sectheta)^2# 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 3180 views around the world You can reuse this answer Creative Commons License