Question #75a89

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Question #75a89

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Answer 1 · 2017-11-20T19:30:34

$see explanation$ $"see explanation"$

Explanation:

$using the trigonometric identities$ $"using the "color(blue)"trigonometric identities"$

$∙ x 1 + {tan}^{2} x = {sec}^{2} x$ $•color(white)(x)1+tan^2x=sec^2x$

$∙ x tan x = \frac{sin x}{cos x}$ $•color(white)(x)tanx=sinx/cosx$

$∙ x sec x = \frac{1}{cos x}$ $•color(white)(x)secx=1/cosx$

$consider the LHS$ $"consider the LHS"$

$\Rightarrow \frac{tan x}{1 + {tan}^{2} x}$ $rArr(tanx)/(1+tan^2x)$

$= \frac{\frac{sin x}{cos x}}{{sec}^{2} x}$ $=(sinx/cosx)/(sec^2x)$

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

$=sinx/cancel(cosx) xxcancel(cos^2x)^(cosx)$

$=sinxcosx="RHS"rArr"proven"$

Answer 2 · 2017-11-20T19:33:23

See below.

Explanation:

We're trying to prove $tan x / (1+tan^2 x) = sin x cos x$ .

Let's manipulate the left side since it's more complicated.

There is a trigonometric identity that states $1 + tan^2 x = sec^2x$ . (It's an alternative form of $sin^2x + cos^2x = 1$ ; simply divide the whole equation by $cos^2x$ .)

Thus,

$tan x / (1+tan^2 x)$

$= tan x / (sec^2 x)$

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

$=sin x / cancel(cos x) * cancel(cos^2 x)^cos x /1$

$=sin x cos x$

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Question #75a89

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