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