Prove? #cotx/(cscx-1)=(cscx+1)/cotx#

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Answer 1 · 2017-01-17T04:39:58

Remember that #1+cot^2x=csc^2x#.

#cotx/(cscx-1)=(cscx+1)/cotx#

Remember that #1+cot^2x=csc^2x#. This becomes useful if we multiply the terms with #cscx# to get them squared:

#cotx/(cscx-1)((cscx+1)/(cscx+1))=(cscx+1)/cotx#

#(cotxcscx+cotx)/(csc^2x-1)=(cscx+1)/cotx#

We can now use #csc^2x-1=cot^2x#

#(cotxcscx+cotx)/(cot^2x)=(cscx+1)/cotx#

#(cscx+1)/(cotx)=(cscx+1)/cotx#