Question #22205

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Answer 1 · 2018-01-26T16:18:20

See explanation.

Prove #sinx+cotx=cscx#:

working on the left-hand side:

using the ratio identity for #cotx#:

#sinx + cosx*cosx/sinx#

getting a common denominator:

# = (sin^2x+cos^2x)/sinx#

using the Pythagorean identity, #sin^2x+cos^2x=1#:

#=1/sinx#

using the reciprocal identity:

#=cscx#

Answer 2 · 2018-01-26T16:21:46

See the proof below

#"Reminder"#

#cotx=cosx/sinx#

#cscx=1/sinx#

#sin^2x+cos^2x=1#

Therefore,

#"LHS"=sinx+cosx*cotx#

#=sinx+cosx*(cosx/sinx)#

#=(sin^2x+cos^2x)/sinx#

#=1/sinx#

#=cscx#

#="RHS"#

#"QED"#