Question

@AZ

Answer 1

should i first convert them to cos and sin stuff?

Answer 2

ye, I was just working it out and that would be the best first step

and also convert the 1 on the LHS to cos(x) / cos(x)

so that way you could add the fractions

Answer 3

oh, kay

Answer 4

[(sinx/cosx)-1][1/(sinx/cosx) + cosx/cosx)]
and then
[(sinx/cosx)-1][1/(sinx+cosx/cosx)]

Answer 5

[(sinx/cosx)-1]{[1/[{sinx+cosx)/cosx]}

Answer 6

more parenthesis lmao

Answer 7

what do i do with the negative 1 tho

Answer 8

maybe you should like draw it out mhmm



now replace the 1 with cos/cos and simplify the fractions in the numerator and denominator

Answer 9

(sinx-cosx)/cosx + (sinx+cosx)/cosx

Answer 10

I think you meant to have divided by not a + in the middle but good
I'm just going to drop all the x's so it's easier and faster to write but you shouldn't forget about them



now let's divide, remember

\(\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}} = \dfrac{a}{b} \div \dfrac{c}{d} = \dfrac{a}{b}\times \dfrac{d}{c} = \dfrac{ad}{bc}\)

Answer 11

[(sinx-cosx)cosx]/[(sinx+cosx)/cosx]

Answer 12

so
(sinx-cosx)/(sinx+cosx)

Answer 13

yes

Answer 14

it's been a while for me, I forget what kind of trig problem this is

technically you could do the same process with the right side and it also equals (sinx-cosx)/(sinx+cosx)

Answer 15

have to divide by sin in both the numerator and denominator now

we're just proving the identity xD

Answer 16

if you simplify it the top part and the bottom part, you'll get the RHS

\(\dfrac{a+b}{c} = \dfrac{a}{c} + \dfrac{b}{c}\)

and sin/sin = 1

and thus LHS = RHS

Answer 17

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh
thank you so much AZ and Angle xdddddddddddddddddddddddddddddd

Answer 18

no problem haha ;b