Question #185da

1 Answer
Andy Y.
May 14, 2017

This identity is true for any value of #x in RR#, except where the denominators are zero.

Explanation:

Start by cross multiplying
#cos^2(x)=(1-sin(x))(1+sin(x))#

Factor the right hand side
#cos^2(x)=1+sin(x)-sin(x)-sin^2(x)#
#cos^2(x)=1-sin^2(x)#

Adding #sin^2(x)# both sides gives the identity
#sin^2(x)+cos^2(x)=1#

This identity is true for any value of #x# in the real numbers #RR#

However, recall that denominators cannot be zero, so the answer is all #RR# except wherever

#cos(x)=0#, so when #x=pi/2(2n-1)#, where #ninZZ#
and #sin(x)=1#, so when #x=pi/2(4n+1)#, where #ninZZ#

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