Question #f207b

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Answer 1 · 2018-02-12T19:07:49

Please see a Proof in the Explanation.

#secxcscx-cosxcscx#.

#=cscx(secx-cosx)#,

#=1/sinx{1/cosx-cosx}#,

#=1/sinx{(1-cos^2x)/cosx}#,

#=1/cancel(sinx){sin^cancel(2)x/cosx}#,

#=sinx/cosx#,

#=tanx," as desired!"#

Answer 2 · 2018-02-12T19:10:38

See below

We will use the following identities:

#secx := 1/cosx#

#cscx := 1/sinx#

#tanx = sinx/cosx#

We now start the proof.

#secxcscx-cosxcscx#

#=1/(cos xsinx)-cosx/sinx#

#=1/(cosxsinx)-cos^2x/(cosxsinx)#

#=(1-cos^2x)/(cosxsinx)#

#=sin^2x/(cosxsinx)#

#=sinx/cosx#

#=tanx#, as required. #square#