Question

1+tan^2 30 degrees = sec^2 30 degrees

Answer 1

Well there is an identity. \[1+\tan^2 (x) = \sec^2 (x)\] so this is true.

Answer 2

Can you explain how to do it?

Answer 3

How to prove this identity?

Answer 4

yes, like how to work it out?

Answer 5

Can I use \[Sin^2x+Cos^2x=1\]as given?

Answer 6

yes

Answer 7

I'll start from the identity I have to proof.
Going sort of backwards.

1+tan^2 (x) = sec^2 (x)

let see, we got a "magical 1" --cos^2 x/cos^2 x.

cos^2 x/cos^2 x + sin^2 x/cos^2 x = 1/cos^2x
cancel Cos^2x
you get
cos^2x+Sin^2x=1

See?

Answer 8

If you don't get it, i can reexplain it.

Answer 9

okay now i understand how to work it out but how did you figure out to get cos^2/cos^2 etc like the trig functions? how did you decide those were the ones we needed? I understand it to an extent i am just getting confused over tiny things

Answer 10

Like where in my proof did you get confused?

Answer 11

just how you chose to use the cos/cos and sin/cos

Answer 12

I've done these last year.

Answer 13

\(\huge {\color{green} {See?} }\)

Answer 14

That's my think, when it comes to proofs like this, I USUALLY try to TRANSLATE into SINES ans COSINES.

Answer 15

so every time you see a "1" you write cos/cos?

Answer 16

No, only in this case, to make the denominator equal in every fraction, so that I cancel it out.

Answer 17

It depends on a problem, we can do more examples if you want to.

Answer 18

Small error.