Question

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

Answer 1

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

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)`