Simplify the expression:? #(sin^2(pi/2+alpha)-cos^2(alpha-pi/2))/(tg^2(pi/2+alpha)-ctg^2(alpha-pi/2))#

Question

#(sin^2(pi/2+alpha)-cos^2(alpha-pi/2))/(tg^2(pi/2+alpha)-ctg^2(alpha-pi/2))#

4899 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T08:58:17

#(sin^2(pi/2+alpha)-cos^2(alpha-pi/2))/(tan^2(pi/2+alpha)-cot^2(alpha-pi/2))#

#=(sin^2(pi/2+alpha)-cos^2(pi/2-alpha))/(tan^2(pi/2+alpha)-cot^2(pi/2-alpha))#

#=(cos^2(alpha)-sin^2(alpha))/(cot^2(alpha)-tan^2(alpha))#

#=(cos^2(alpha)-sin^2(alpha))/(cos^2(alpha)/sin^2(alpha)-sin^2(alpha)/cos^2(alpha))#

#=(cos^2(alpha)-sin^2(alpha))/((cos^4(alpha)-sin^4(alpha))/(sin^2(alpha)cos^2(alpha)))#

#=(cos^2(alpha)-sin^2(alpha))/(cos^4(alpha)-sin^4(alpha))xx(sin^2(alpha)cos^2(alpha))/1#

#=(cos^2(alpha)-sin^2(alpha))/((cos^2(alpha)-sin^2(alpha))(cos^2(alpha)+sin^2(alpha))xx(sin^2(alpha)cos^2(alpha))/1#

#=sin^2(alpha)cos^2(alpha)#