Question

Verify the identity (1 + tan2u)(1 - sin2u) = 1

Answer 1

you want to prove it?

Answer 2

notice the trig pythagorean identities -> http://web.mit.edu/wwmath/trig/eq1.gif

Answer 3

Just need to play around with this known identity:\[\Large \cos^2(x)+\sin^2(x) = 1\]

Answer 4

Yeah, I don't understand identities very well, so how is it proven?

Answer 5

can you solve \(\bf \cos^2(x)+\sin^2(x) = 1\) for \(\bf cos^2(x)?\)

what do you get?

Answer 6

But how do you get from cos and sin to tan?
Like my problem up top?

Answer 7

\[\cos ^{2}x = 1 - \sin ^{2}x\] but how does it get to \[\tan ^{2}\]?

Answer 8

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Answer 9

Can you please explain how it works? @jdoe0001

Answer 10

You seem panicked :)
see what happens when the entire equation \[\Large \cos^2(x)+\sin^2(x) = 1\] is divided by \(\large \cos^2(x)\)...
\[\Large \frac{\cos^2(x)}{\cos^2(x) }+ \frac{\sin^2(x)}{\cos^2(x)}= \frac1{\cos^2(x)}\]

Answer 11

Obviously, this red part is just 1.
\[\Large \color{red}{\frac{\cos^2(x)}{\cos^2(x)} }+ \color{blue}{\frac{\sin^2(x)}{\cos^2(x)}}=\color{green}{ \frac1{\cos^2(x)}}\]

This blue part is \(\large \tan^2(x)\)\[\Large \color{red}{1 }+ \color{blue}{\frac{\sin^2(x)}{\cos^2(x)}}=\color{green}{ \frac1{\cos^2(x)}}\]
And this green part is just \(\large \sec^2(x)\)
\[\Large \color{red}1+\color{blue}{\tan^2(x)}= \color{green}{\sec^2(x)}\]

Answer 12

thank you!

Answer 13

No problem.