Question

cosA +sinA=under root 2cosA; show that cosA-sinA=under root2sinA
​

Answer 1

Answer:

cosA +sinA= root2cosA =LHS

square on botb side

(cosA+sinA)^2=(root2cosA)^2

cos^2A+sin^2A+2sinAcosA =2cos^A

2cosAsinA=2cos^2A-cos^2A-sin^A

2cosAsinA=cos^2A-sin^2A

2cosAsinA=(cosA-sinA)(cosA+sinA)

cosA-sinA=2cosAsinA/cosA+sinA

cosA-sinA=2cosAsinA/root2cosA

cosA-sinA=root2sinA =RHS

hope you understand

and follow me