Question

Prove this trig identity.

Show all working please. :)

Cosx / Secx - Tanx = 1 + Sinx

Answer 1

\[\frac{ cosx }{ secx-tanx } = 1+ sinx\]

Answer 2

write sec x as 1/cos x and tan x as sin x/cos x

Answer 3

then write cos^2 x as 1- sin^2x = (1-sin x)(1+sin x)

Answer 4

so could u prove it ?

Answer 5

Nope, I need to refer to my notes but I left my book at home (at school atm)

Answer 6

\(\large \frac{cos x}{1/cos x-sin x/cos x}\quad =\frac{cos^2x}{1-sin x}=\frac{(1+sin x)(1-sin x)}{1-sin x}\)
now its clear ?

Answer 7

Yep, the 1-sines cancel. :)

Answer 8

\(\huge\frac{cos x}{\frac{1-sin x}{cos x}}=cos x .\frac{cos x}{1-sin x}\)

Answer 9

Oh right, you're multiplying the reciprocal!

Answer 10

yup :)

Answer 11

One more stupid question, if that's the case what happened to the original 1/cosine ?

Answer 12

\[\frac{ 1 }{ cosx } and \frac{ sinx }{ cosx }\]

Answer 13

cosx = cosx/1 so you're left with 1-sine/cosine on the bottom

Answer 14

\[cosx . \frac{ \cos }{ 1 } - \frac{ cosx }{ sinx } \]

Answer 15

or is it \[cosx.\frac{ cosx }{ 1 } - \frac{ sinx }{ cosx }\]

Answer 16

I understand where the cos^2x comes from but there are 3 cosines and I don't know where one of them disappeared to :]

Answer 17

sec x = 1/ cos x
tan x = sin x / cos x
so your denominator is:
1/cos x -sin x/cos x = (1-sin x)/cos x

Answer 18

Ahh, I see so the common denominator rule is applied, therefore the 2 cosines "merge" ?

Answer 19

yup, a/b-c/b=(a-c)/b

Answer 20

eg \[\frac{ 1 }{ 3 } + \frac{ 2 }{ 3 } = \frac{ 3 }{ 3 }\]

Answer 21

Awesome! Thank you so much for your patience and time, you've been utmost helpful. :)

Answer 22

Would it be alright if I asked you another one of these later on?

Answer 23

yes, sure :)

Answer 24

Great! much obliged. :)