Question

@AZ hehe

Answer 1

hmm let's just foil and see what we're dealing with

\( (a+b)(c+d) = ac + ad + bc + bd\)

Answer 2

sinxtanx + sinxcotx + cosxtanx + cosxcotx

Answer 3

okay mhmm

now let's just re-write those all in terms of sin and cos and then simplify

remember
tan(x) = sin(x) / cos(x)

cot(x) = 1/ tan(x) = cos(x) / sin(x)

Answer 4

and once you get that you'll find that you can add the fractions that have the same denominators

and then you can use identities to simplify the numerators

and ta-da

Answer 5

so I got
sinx + cosx + sinx tanx + cosx cotx

Answer 6

so then
sinx + cosx + sinx(sinx/cosx) + cosx(cosx/sinx)

Answer 7

then, i'm confused ?_?

Answer 8

sinx + cosx + sinx(sinx^2/sinxcosx) + cosx(cosx^2/sinxcosx)

Answer 9

yep, i'm definitely still so confused

Answer 10

\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
sinxtanx + sinxcotx + cosxtanx + cosxcotx
\(\color{#0cbb34}{\text{End of Quote}}\)

the one you said more recently is incorrect, I'm not sure where you got rid of some terms but let's start form here

Answer 11

sinx(sinx/cosx) + sinx(cosx/sinx) + cosx(sinx/cosx) + cos(cosx/sinx)

Answer 12

ohhh I see now

Answer 13

am i still wrong? lol

Answer 14

no no, you're on the right track I was wrong with what I had said earlier

so this is correct
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
so then
sinx + cosx + sinx(sinx/cosx) + cosx(cosx/sinx)
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 15

kk

Answer 16

once you're here now,

you multiply the sin and cos into the parenthesis respectively

remember it's like
a*(b/c) = (ab) / c

Answer 17

whatever you had done afterwards like the following is all wrong because you multiplied both the numerator and denominator
\(\color{#0cbb34}{\text{Originally Posted by}}\) @snowflake0531
sinx + cosx + sinx(sinx^2/sinxcosx) + cosx(cosx^2/sinxcosx)
\(\color{#0cbb34}{\text{End of Quote}}\)

big no no

Answer 18

So
(sin^2 x)/cosx + cos^2 x)/sinx + sinx + cosx

Answer 19

Good! Now remember the identities

sin^2 x = 1 - cos^2 x
and
cos^2 x = 1 - sin^2 x

plug both of those respective ones in place of sin^2 x and cos^2 x and simplify the fractions

remember
\(\dfrac{a-b}{c} = \dfrac{a}{c} - \dfrac{b}{c}\)

Answer 20

those two identities are based off of
sin^2(x) + cos^2(x) = 1

Answer 21

1/cosx - cosx^2/cosx = secant x - cosx
1/sinx - sin^2x/sinx = cosecant x - sinx
sec x - cosx + cscx - sinx

Answer 22

and hten plus the sinx and cosx

Answer 23

and you'll see that the sin and cos will cancel out leaving you with the RHS

Answer 24

ohhhhhhh, kay, i finished writing it
thankssssssssssssssssssss xdxd

Answer 25

no problemo!!