Question #4d95b

Question

969 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-08T22:47:10

#(sec x + csc x)/(sin x + cosx) * sin 2x = 2#

Taking the left hand side,

#(sec x + csc x)/(sin x + cosx) * sin 2x#

= #(1/cosx + 1/sin x)/(sin x + cos x) * sin 2x#

= #((sin x + cos x)/(sin x cos x))/(sinx + cosx) * sin 2x#

#sin x + cos x# cancel out
recall the identity #sin 2x = 2 sin x cos x#

= #1/(sin x cos x) * 2 sin x cos x#

= #2#