How do you verify $(sinA+cosA)^2=(2+secAcscA)/(secAcscA)$ ?

Question

4324 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-12T09:26:09

see below

$(sinA+cosA)^2=(2+secAcscA)/(secAcscA)$

$(sinA+cosA)(sinA+cosA)=(2+1/cosA*1/sinA)/(1/cosA*1/sinA)$

$sin^2A+2sinAcosA+cos^2A=((2sinAcosA+1)/(cosAsinA))/(1/(cosAsinA))$

$sin^2A+2sinAcosA+cos^2A=(2sinAcosA+1)/(cosAsinA)*(cosAsinA)/1$

$2sinAcosA+sin^2A+cos^2A=(2sinAcosA+1)/cancel(cosAsinA)*cancel(cosAsinA)/1$

$2sinAcosA+1=2sinAcosA+1$

$:.$ Left Hand Side=Right Hand Side

How do you verify (sinA+cosA)^2=(2+secAcscA)/(secAcscA)(sinA+cosA)^2=(2+secAcscA)/(secAcscA)?