Question

write the first expression in terms of the second sec^2t x sin^2t, cos t

Answer 1

This is the question stated in the correct format correct?

\[1st = \sec ^{2t}(x) \sin ^{2t}(x)\]

\[2nd \cos (t)\]

Answer 2

im sorryi was trying to write\[\sec ^{2}t \sin ^{2}t , \cos t\]

Answer 3

change sin^2(t) into [1- cos^2(t)] using --> identity: sin^2(t)+cos^2(t) = 1

Answer 4

and the sec^2 (t) ?
to 1/ cos^2(t)?

Answer 5

yes, and then distribute

Answer 6

so it would be [1- cos^2(t)] / cos^2 (t) ?

Answer 7

im sorry im confused what am i distributing?

Answer 8

hold on i need to do this on paper

Answer 9

okay thank you so much

Answer 10

[1- cos^2(t)] would change to sin^2(t)
then that would be sin^2 (t) / cos^2 (t) = tan^2 (t)

Answer 11

could you please rewrite the quesiton stating in full what it is asking of you?

Answer 12

im not sure how to get it to cos (t) though

Answer 13

using the equation editor

Answer 14

write the first expression in terms of the second
\[ \sec^2(t)\sin^2(t) \to \cos(t) \]

Answer 15

Yes,

\(\large \sec^2t \sin^2t \)

\(\large \frac{\sin^2t}{\cos^2t} \)

Answer 16

but how do i change that in cos (t)

Answer 17

next, u need to change numerator to cos term ok

Answer 18

\(\sin^2t = 1 - \cos^2t\)

simply use that in numerator

Answer 19

Yes,

\(\large \sec^2t \sin^2t \)

\(\large \frac{\sin^2t}{\cos^2t} \)

\(\large \frac{1-\cos^2t}{\cos^2t} \)


Now that everything in \(\cos\) terms, we're done.

Answer 20

so its ok if its left as cos^2 (t) ?

Answer 21

yes thats what the question is asking us to do i think from the way i see it

Answer 22

oh ok thank you so much!

Answer 23

np :)