Asked by krista
Show that: sec (A-B)= (cos(A+B))/(cos^2 A-Sin^2 B)
Answers
Answered by
Reiny
LS
= 1/cos(A-B)
= 1/(cosAcosB + sinAsinB)
multiply top and bottom by cosAcosB - sinAsinB
= (cosAcosB - sinAsinB)/(cos^2Acos^2B - sin^2Asin^2B)
= (cosAcosB - sinAsinB)/((1-sin^2A)(1-sin^2B) - sin^2A(sin^2B))
= (cosAcosB - sinAsinB)/(1 - sin^2B - sin^2A + sin^2Asin^2B - sin^2Asin^2B)
= cos(A+B)/(cos^2A - sin^2B)
= RS
= 1/cos(A-B)
= 1/(cosAcosB + sinAsinB)
multiply top and bottom by cosAcosB - sinAsinB
= (cosAcosB - sinAsinB)/(cos^2Acos^2B - sin^2Asin^2B)
= (cosAcosB - sinAsinB)/((1-sin^2A)(1-sin^2B) - sin^2A(sin^2B))
= (cosAcosB - sinAsinB)/(1 - sin^2B - sin^2A + sin^2Asin^2B - sin^2Asin^2B)
= cos(A+B)/(cos^2A - sin^2B)
= RS
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