Question #1a94c

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Answer 1 · 2017-05-14T00:25:14

See below.

Prove: #tan^2x/(tan^2x+1)=sin^2x#

Let us start on the left-hand side (LHS) and work toward the right-hand side (RHS).

Consider Pythagorean's identity:
#sin^2x+cos^2x=1#

which we can divide by #cos^2x# to get:
#sin^2x/cos^2x+cos^2x=1/cos^2x#

Note that #secx=1/cosx# and #tanx=sinx/cosx#:
#tan^2x+1=sec^2x#, which is most applicable to this problem

We can substitute #tan^2x+1=sec^2x# into the LHS:
#LHS=tan^2x/sec^2x#

Note that #secx=1/cosx# and #tanx=sinx/cosx#:
#LHS=tan^2x/(1/cos^2x)#

#=(sin^2x/cos^2x)/(1/cos^2x)#

#=sin^2x/cos^2x*cos^2x#

#=sin^2x=RHS#

Therefore, #tan^2x/(tan^2x+1)=sin^2x#