Question

More Trigonometric Proofs
\[2\sec^2=\frac{ 1-\sin t }{ \cos^2 t} + \frac{ 1 }{ 1-sint }\]

Answer 1

Do I multiply by the conjugate?

Answer 2

HI (again)

Answer 3

i would add

Answer 4

Hi again xD

Answer 5

So find common denominator?

Answer 6

not really "find" one, just multiply the denominators together

Answer 7

Yeah same thing

Answer 8

oh so one of the fractions reduces to 1

Answer 9

I think

Answer 10

Oh wait, oops

Answer 11

here is what i do
it is too confusing to do the algebra with sines and cosines , so i would replace cosine by \(a\) and sine by \(b\) and add up
\[\frac{1-b}{a^2}+\frac{1}{1-b}\]

Answer 12

so now I have \[\frac{ \cos^2(1-\sin t) }{ \cos^2t(1-sint) } + \frac{ \cos t }{ \cos^2t(1-sint)}\]

Answer 13

yeah but it is a lot easier to write
\[\frac{(1-b)(1-b)+a^2}{a^2(1-b)}\]

Answer 14

Sorry left side on numerator is cos^2(1-sin^2t)

Answer 15

yeah i know but it is annoying to write
\[\frac{ \cos^2(1-\sin t) }{ \cos^2t(1-sint) } + \frac{ \cos^2 t }{ \cos^2t(1-sint)}\]

Answer 16

Yeah okay

Answer 17

So then combine

Answer 18

\[\frac{(1-b)(1-b)+a^2}{a^2(1-b)}\] is a lot easier
now lets work from there

Answer 19

multiply out and get
\[\frac{1-2b+b^2+a^2}{a^2(1-b)}\]

Answer 20

now comes one trig step, namely \(a^2+b^2=1\) because \(\cos^2(x)+\sin^2(x)=1\)

Answer 21

you get \[\frac{2-2b}{a^2(1-b)}\]

Answer 22

Oh okay, so now it becomes 2-2b

Answer 23

Okay

Answer 24

factor out the 2, cancel the \(1-b\) top and bottom and voila

Answer 25

ah okay! Thank you!

Answer 26

\[\color\magenta\heartsuit\]

Answer 27

@satellite73 hii can you please come continue with me after youre done here? its lagging so i dont kow if youre getting my notifications