How do you prove $sec^2(csc^2) = sec^2 + csc^2$ ?

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How do you prove $sec^2(csc^2) = sec^2 + csc^2$ ?

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Answer 1 · 2016-05-04T19:29:15

see below

Explanation:

Right Side: $=sec^2x+csc^2x$

$=1/cos^2x + 1/sin^2 x$

$=(sin^2x+cos^2x)/(cos^2xsin^2x)$

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

$=1/cos^2x * 1/sin^2x$

$=sec^2xcsc^2x$

$=$ Left Side

Answer 2 · 2016-05-04T19:56:30

Note the following identity:

$csc^2theta = 1 + cot^2theta$

So, let's see how that works out.

$\mathbf(sec^2theta(csc^2theta) = sec^2theta + csc^2theta)$

$sec^2theta(1 + cot^2theta) = sec^2theta + csc^2theta$

$sec^2theta + sec^2thetacot^2theta = sec^2theta + csc^2theta$

Lastly, use the identities

$cot^2theta = cos^2theta/sin^2theta,$

$1/sin^2theta = csc^2theta,$

$1/cos^2theta = sec^2theta,$

to get:

$sec^2theta + sec^2theta(cos^2theta/sin^2theta) = sec^2theta + csc^2theta$

$sec^2theta + 1/cancel(cos^2theta)(cancel(cos^2theta)/sin^2theta) = sec^2theta + csc^2theta$

$sec^2theta + 1/sin^2theta = sec^2theta + csc^2theta$

$color(blue)(sec^2theta + csc^2theta = sec^2theta + csc^2theta)$

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How do you prove $sec^2(csc^2) = sec^2 + csc^2$ ?

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Explanation:

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