Question #14fc5

1 Answer
Narad T.
Apr 24, 2017

See proof below

Explanation:

We need

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

#(a+b)(a-b)=a^2-b^2#

#cotx=cosx/sinx#

#cscx=1/sinx#

#tanx=sinx/cosx#

#secx=1/cosx#

#LHS=(cotx+cscx)(tanx-sinx)#

#=(cosx/sinx+1/sinx)(sinx/cosx-sinx)#

#=((1+cosx))/cancel(sinx)*cancel(sinx)((1-cosx))/cosx#

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

#=1/cosx-cos^2x/cosx#

#=secx-cosx#

#=RHS#

#QED#

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