Question #54eff

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Question #54eff

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Answer 1 · 2017-05-15T00:30:23

Use definition of tangent and Pythagorean's identity
Answer: ${sin}^{2} x$ $sin^2x$

Explanation:

Simplify $1 - \frac{{sin}^{2} x}{{tan}^{2} x}$ $1-sin^2x/tan^2x$

Note that $tan x = \frac{sin x}{cos x}$ $tanx=sinx/cosx$

We can substitute ${tan}^{2} x = \frac{{sin}^{2} x}{{cos}^{2} x}$ $tan^2x=sin^2x/cos^2x$
$= 1 - \frac{{sin}^{2} x}{\frac{{sin}^{2} x}{{cos}^{2} x}}$ $=1-sin^2x/(sin^2x/cos^2x)$
$= 1 - {sin}^{2} x \cdot \frac{{cos}^{2} x}{{sin}^{2} x}$ $=1-sin^2x*cos^2x/sin^2x$
$= 1 - {cos}^{2} x$ $=1-cos^2x$

Consider the Pythagorean identity:
${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$
we can rearrange to get:
${sin}^{2} x = 1 - {cos}^{2} x$ $sin^2x=1-cos^2x$

Therefore,
$= {sin}^{2} x$ $=sin^2x$

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Question #54eff

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