How do you verify the identity #cosA-sinA=(cos2A)/(cosA+sinA)#? Trigonometry Trigonometric Identities and Equations Half-Angle Identities 1 Answer Narad T. Nov 12, 2016 See proof below Explanation: RHS #cos(A+B)=cosAcosB-sinAsinB# Let #B=A# Then #cos2A=cos^2A-sin^2A=(cosA+sinA)(cosA-sinA)# #(cos2A)/(cosA+sinA)=(cancel(cosA+sinA)(cosA-sinA))/cancel(cosA+sinA# #=cosA-sinA# #=# LHS #:. RHS=LHS# Answer link Related questions What is the Half-Angle Identities? How do you use the half angle identity to find cos 105? How do you use the half angle identity to find cos 15? How do you use the half angle identity to find sin 105? How do you use the half angle identity to find #tan (pi/8)#? How do you use half angle identities to solve equations? How do you solve #\sin^2 \theta = 2 \sin^2 \frac{\theta}{2} # over the interval #[0,2pi]#? How do you find the exact value for #sin105# using the half‐angle identity? How do you find the exact value for #cos165# using the half‐angle identity? How do you find the exact value of #cos15#using the half-angle identity? See all questions in Half-Angle Identities Impact of this question 9553 views around the world You can reuse this answer Creative Commons License