Trig: Sec^2 18 - Tan^2 18?

Question

Could somebody work this problem for me
Sec^2 18 - Tan^2 18 =
The exact answer, not using a calculator

A short explanation of how it was found would also be helpful, as the process has slipped my mind tonight.

Answers tonight will be far more helpful than those later, thanks

Answer 1

sec^2(18) - tan^2(18)

First off, remember that there is an identity that relates sec^2(x) with tan^2(x); it goes

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

Therefore,

sec^2(18) = tan^2(18) + 1. Substituting as such,

[tan^2(18) + 1] - tan^2(18)

Removing the brackets,

tan^2(18) + 1 - tan^2(18)

The tan^2(18) will cancel out with the -tan^2(18), leaving just

0 + 1

1

Answer 2

Sec ^2 (x) = 1 + tan^2 (x) . Therefore sec^21(8) - tan^2 (18 )= 1 + tan^2(18) - tan^2 (18) = 1

Answer 3

At first blush this looks moderately difficult. But you have to know your trig indentities:
1 + (tan t)^2 = (sec t)^2
So just plug it in and
(sec 18)^2 - (tan 18)^2 = [1+(tan 18)^2] - (tan 18)^2 = 1

Answer 4

Sec^2 18 - Tan^2 18
1/cos^2 18-sin^2 18/cos^2 18
(1-sin^2 18)/cos^2 18 from sin^2 x + cos^2 x=1
cos^2 18/cos^2 18
1

Answer 5

Sec^2 x - Tan^2 x = 1 ["identically" (that is for all values of x)]

so for your case too, it is 1 !!

Answer 6

sec²x - tan²x = 1, when x is in the domain of
both functions. So the answer to your problem is 1.