Tan×+cosx÷1+sinx?

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Tan×+cosx÷1+sinx?

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Answer 1 · 2018-05-16T22:17:49

$tan x + \frac{cos x}{1 + sin x} = sec x = \frac{1}{cos x}$ $tanx+cosx/(1+sinx)=secx=1/cosx$

Explanation:

I'm assuming you meant $cos x \div (1 + sin x)$ $cosx-:(1+sinx)$ , as $cos x \div 1$ $cosx-:1$ is trivial.

When we have $1 + sin x$ $1+sinx$ in the denominator, we want to amplify the fraction by $1 - sin x$ $1-sinx$ to get $1 - {sin}^{2} x = {cos}^{2} x$ $1-sin^2x=cos^2x$ .

$:. E(x)=tanx+cosx/(1+sinx)*color(red)((1-sinx)/(1-sinx))$

$E(x) = tanx+(cosx(1-sinx))/(1-sin^2x)=tanx+(cosx(1-sinx))/cos^2x$

Simplifying the $cosx$ :

$E(x)=tanx+(1-sinx)/cosx$

We know, by definition, that $tanx=sinx/cosx$ .

$:. E(x) = sinx/cosx+(1-sinx)/cosx=(sinx+1-sinx)/cosx=1/cosx$

$=>color(red)(E(x)=secx)$

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Tan×+cosx÷1+sinx?

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