Question

prove that tan^2θ+1sec^2θ for all θ and use 80 degrees to verify the identity

Answer 1

what is tan^2 and sec^2 equal to?

Answer 2

i will make a guess that this is
\[\tan^2(\theta)+1=\sec^2(\theta)\]

Answer 3

take your famous pythagorean identity and divide everything by \(\cos^2(\theta)\)

Answer 4

you got this?

Answer 5

Rewrite the left side in sines and cosines: sin^2 (theta) / cos^2 (theta) + 1 =

sin^2 (theta)/cos^2 (theta) + cos^2(theta)/cos^2 (theta) =
(sin^2 (theta) + cos^2 (theta)) / cos^2 (theta) =

1 / cos^2 (theta) =
sec^2 (theta)

At 80 degrees: tan^2(80) + 1 = (5.67128182)^2 + 1 = 33.16343748
sec^2 (80) = (5.758770483)^2 = 33.16343748

Answer 6

hope you can read it

Answer 7

do u have a brighter photo u cant really read it

Answer 8

yea just a sec though

Answer 9

Rewrite the left side in sines and cosines: sin^2 (theta) / cos^2 (theta) + 1 =

sin^2 (theta)/cos^2 (theta) + cos^2(theta)/cos^2 (theta) =
(sin^2 (theta) + cos^2 (theta)) / cos^2 (theta) =

1 / cos^2 (theta) =
sec^2 (theta)

At 80 degrees: tan^2(80) + 1 = (5.67128182)^2 + 1 = 33.16343748
sec^2 (80) = (5.758770483)^2 = 33.16343748
this is the answer i think so here u go fari321

Answer 10

this is as good as it gets

Answer 11

then just plug in the angles and they will equal 1

Answer 12

plug in 80 degrees?

Answer 13

yes, to verify the identity just say sin(80)^2 + cos(80)^2 =1