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Answer 1 · 2017-08-22T19:02:29

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

Explanation:

Since

$sec θ = \frac{1}{cos θ}$ $sectheta=1/costheta$

and

$csc θ = \frac{1}{sin θ}$ $csctheta=1/sintheta$

the left side of the equation is:

$⎡ ⎣ \frac{1}{\frac{1}{{cos}^{2} θ} - {cos}^{2} θ} + \frac{1}{\frac{1}{{sin}^{2} θ} - {sin}^{2} θ} ⎤ ⎦ {sin}^{2} θ {cos}^{2} θ$ $[1/(1/cos^2theta-cos^2theta)+1/(1/sin^2theta-sin^2theta)]sin^2thetacos^2theta$

Let's simplify the fractions and get:

$[\frac{{cos}^{2} θ}{1 - {cos}^{4} θ} + \frac{{sin}^{2} θ}{1 - {sin}^{4} θ}] {sin}^{2} θ {cos}^{2} θ$ $[cos^2theta/(1-cos^4theta)+sin^2theta/(1-sin^4theta)]sin^2thetacos^2theta$

that's, by summing the fractions:

$[\frac{{cos}^{2} θ (1 - {sin}^{4} θ) + {sin}^{2} θ (1 - {cos}^{4} θ)}{(1 - {cos}^{4} θ) (1 - {sin}^{4} θ)}] {sin}^{2} θ {cos}^{2} θ$ $[(cos^2theta(1-sin^4theta)+sin^2theta(1-cos^4theta))/((1-cos^4theta)(1-sin^4theta))]sin^2thetacos^2theta$

Let's expand the numerator and factorize the denominator: since

$a^{2} - b^{2} = (a - b) (a + b)$ $color(red)(a^2-b^2=(a-b)(a+b))$

we get:

$\frac{({cos}^{2} θ - {cos}^{2} θ {sin}^{4} θ + {sin}^{2} θ - {sin}^{2} θ {cos}^{4} θ)}{(1 - {cos}^{2} θ) (1 + {cos}^{2} θ) (1 - {sin}^{2} θ) (1 + {sin}^{2} θ)} {sin}^{2} θ {cos}^{2} θ$ $[(cos^2theta-cos^2thetasin^4theta+sin^2theta-sin^2thetacos^4theta))/color(red)((1-cos^2theta)(1+cos^2theta)(1-sin^2theta)(1+sin^2theta)]sin^2thetacos^2theta$

Since

${sin}^{2} θ + {cos}^{2} θ = 1$ $color(blue)(sin^2theta+cos^2theta=1)$

and

$1 - {cos}^{2} θ = {sin}^{2} θ$ $1-cos^2theta=sin^2theta$

and

$1 - {sin}^{2} θ = {cos}^{2} θ$ $1-sin^2theta=cos^2theta$

we get:

$[(1-cos^2thetasin^4theta-sin^2thetacos^4theta))/(cancel(sin^2theta)(1+cos^2theta)cancel(cos^2theta)(1+sin^2theta)]cancel(sin^2theta)cancel(cos^2theta)$

Let's factorize at the numerator and expand the denominator:

$(1-sin^2thetacos^2theta(color(blue)(sin^2theta+cos^2theta)))/((1+color(blue)(sin^2theta+cos^2theta)+sin^2thetacos^2theta)$

that's

$(1-sin^2thetacos^2theta)/((1+1+sin^2thetacos^2theta)$

or the right side of the given equation

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