Question #1a94c

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Question #1a94c

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Answer 1 · 2017-05-14T00:25:14

See below.

Explanation:

Prove: $\frac{{tan}^{2} x}{{tan}^{2} x + 1} = {sin}^{2} x$ $tan^2x/(tan^2x+1)=sin^2x$

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

Consider Pythagorean's identity:
${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

which we can divide by ${cos}^{2} x$ $cos^2x$ to get:
$\frac{{sin}^{2} x}{{cos}^{2} x} + {cos}^{2} x = \frac{1}{{cos}^{2} x}$ $sin^2x/cos^2x+cos^2x=1/cos^2x$

Note that $sec x = \frac{1}{cos x}$ $secx=1/cosx$ and $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ :
${tan}^{2} x + 1 = {sec}^{2} x$ $tan^2x+1=sec^2x$ , which is most applicable to this problem

We can substitute ${tan}^{2} x + 1 = {sec}^{2} x$ $tan^2x+1=sec^2x$ into the LHS:
$L H S = \frac{{tan}^{2} x}{{sec}^{2} x}$ $LHS=tan^2x/sec^2x$

Note that $sec x = \frac{1}{cos x}$ $secx=1/cosx$ and $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$ :
$L H S = \frac{{tan}^{2} x}{\frac{1}{{cos}^{2} x}}$ $LHS=tan^2x/(1/cos^2x)$

$= \frac{\frac{{sin}^{2} x}{{cos}^{2} x}}{\frac{1}{{cos}^{2} x}}$ $=(sin^2x/cos^2x)/(1/cos^2x)$

$= \frac{{sin}^{2} x}{{cos}^{2} x} \cdot {cos}^{2} x$ $=sin^2x/cos^2x*cos^2x$

$= {sin}^{2} x = R H S$ $=sin^2x=RHS$

Therefore, $\frac{{tan}^{2} x}{{tan}^{2} x + 1} = {sin}^{2} x$ $tan^2x/(tan^2x+1)=sin^2x$

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Question #1a94c

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