How can I simplify this trigonometry expression? Thanks!

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How can I simplify this trigonometry expression? Thanks!

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Answer 1 · 2018-03-11T01:43:53

See below

Explanation:

Whenever I see two fractions being added or subtracted with different denominators, I automatically think we need a common denominator.

$\frac{1}{1 + sin θ} + \frac{1}{1 - sin θ} =$ $1/(1+sintheta)+1/(1-sintheta)=$
$\frac{1 (1 - sin θ)}{(1 + sin θ) (1 - sin θ)} + \frac{1 (1 + sin θ)}{(1 - sin θ) (1 + sin θ)} =$ $(1(1-sintheta))/((1+sintheta)(1-sintheta))+(1(1+sintheta))/((1-sintheta)(1+sintheta))=$
$\frac{2}{1 - {sin}^{2} θ}$ $2/(1-sin^2theta)$
Now that we successfully added the two fractions, we will apply a Pythagorean identity to simplify:
${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta+cos^2theta=1$
Therefore:
${cos}^{2} θ = 1 - {sin}^{2} θ$ $cos^2theta=1-sin^2theta$
Let's substitute ${cos}^{2} θ$ $cos^2theta$ in our denominator:
$\frac{2}{{cos}^{2} θ}$ $2/(cos^2theta)$
Now apply the reciprocal identity:
$sec θ = \frac{1}{cos θ}$ $Sectheta=1/costheta$
Final answer: $2 {sec}^{2} θ$ $2sec^2theta$

Answer 2 · 2018-03-11T01:49:44

$2 {sec}^{2} θ$ $2 sec^2 theta$

Explanation:

$\frac{1}{1 + sin θ} + \frac{1}{1 - sin θ}$ $1/(1 + sin theta) + 1/(1 - sin theta)$

multiplying each term by one in the form of a ratio of an appropriately chosen numerator and denominator:

$= \frac{1 - sin θ}{(1 + sin θ) (1 - sin θ)} + \frac{1 + sin θ}{(1 - sin θ) (1 + sin θ)}$ $= (1 - sin theta)/((1 + sin theta)(1 - sin theta)) + (1 + sin theta)/((1 - sin theta)(1 + sin theta))$

using the corollary of the difference of two squares for the terms in the denominators:

$= \frac{1 - sin θ}{1 - {sin}^{2} θ} + \frac{1 + sin θ}{1 - {sin}^{2} θ}$ $= (1 - sin theta)/(1 - sin^2theta) + (1 + sin theta)/(1 - sin^2theta)$

summing terms with the same denominator

$\frac{1 - sin θ + 1 + sin θ}{1 - {sin}^{2} θ}$ $(1 - sin theta + 1 + sin theta)/(1 - sin^2theta)$

using a rearrangement of the identity ${cos}^{2} θ + {sin}^{2} θ = 1$ $cos^2 theta + sin^2 theta = 1$ :

$= \frac{2}{{cos}^{2} θ}$ $= 2 / cos^2theta$

noting ${sec}^{2} θ = \frac{1}{{cos}^{2} θ}$ $sec^2 theta = 1/(cos^2 theta)$ :

$= 2 {sec}^{2} θ$ $= 2 sec^2 theta$

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How can I simplify this trigonometry expression? Thanks!

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