Prove that

Question

sec(A+B)=[cos(A-B)]/[cos^2A-sin^2B]

Answer 1

Given trigonometric identity is sec(A+B)=[cos(A-B)]/[cos^2A-sin^2B]

Start from left side

Substitute

Substitute formula : cos(A+B) = cosAcosB - sinAsinB

Rationalize with cosAcosB+sinAsinB

Substitute formula : sin²A = 1-cos²A and cos²B = 1- sin²B

Substitute formula : cos(A+B) = cosAcosB-sinAsinB

Solution :

The Trigonometric identity is sec(A+B)=[cos(A-B)]/[cos^2A-sin^2B] is established