Question

tan^2x-sec^2x

Answer 1

http://www.gradeamathhelp.com/image-files/trigonometric-identities-pythagorean.gif <--- notice the trig identity at the bottom-left

Answer 2

is this the answer?

Answer 3

did you see the bottom-left identity, with the tangent function? can you isolate the tangent and secant on the left side? what would that give you?

Answer 4

im confusing. i really dont know this, can you give me an example? :)

Answer 5

ok... what about this... if you had .... say .... \(\bf 1+x^2=y^2\) how would you isolate the "1", what would that equals to?

Answer 6

\(\large \bf 1+x^2=y^2\implies \bbox[border: 1px solid black]{ \textit{ what goes here? }} =-1\)

Answer 7

\[x ^{2}+( y ^{2} \x ^{2})=1\]

Answer 8

im not sure of that :)

Answer 9

sorry that would be =-1

Answer 10

hhmmm... lemme put it differently... then

if you had say \(\large \bf 1+a=b\implies \bbox[border: 1px solid black]{ \textit{ what goes here? }} =-1 \)

Answer 11

its really confusing

Answer 12

a+(b/a)= -1 ????

Answer 13

well, not quite... you may want to revise your linear equation simplifications though -> http://www.youtube.com/watch?v=zNtqneXTSWM

Answer 14

tan^2x-sec^2x= tan^2x-sec^2=-1
tan^2x-(sec^2x/tan^2x)=-1
tan^4x-sec^2x/tan^2x
1/tan^2x
cot^2x ANS!

is it wrong? :(

Answer 15

still there?