Question

Verify the identity. Show your work.
1 + sec2xsin2x = sec2x

Answer 1

I like your pic lucy is it you???

Answer 2

\(\large\color{black}{ \displaystyle 1 + \sec^2x\sin^2x = \sec^2x }\)

Like this?

Answer 3

I would multiply everything times \(\color{black}{\cos^2x}\).
That would give me the pathegorean identity.

But, if you are only allowed to play with one side at a time,
I would factor out of \(\color{black}{\sec^2x}\) on the left side.

Answer 4

Hint:

\(\Large\color{black}{ \displaystyle \color{red}{1} + \sec^2x\sin^2x = \sec^2x }\)

\(\Large\color{black}{ \displaystyle \color{red}{\sec^2x\cos^2x} + \sec^2x\sin^2x = \sec^2x }\)

Look,

\(\Large\color{black}{ \displaystyle \color{blue}{\underline{\color{black}{\sec^2x}}}\cos^2x +\color{blue}{\underline{\color{black}{\sec^2x}}}\sin^2x = \color{green}{\underline{\color{black}{\sec^2x}}} }\)

Answer 5

i just dont get it