Question

topic: PROVING IDENTITIES - help me please. i really dont know how to do this (SEE COMMENTS)
(trigonometry)

1)

[(2+csc x)/(sec x)] - 2 csc x = cot x

2)

(1)/(sec x - tan x) = sec x + tan x

Answer 1

Are you sure about the first equation?

Answer 2

yep. that's what's given
@MIssCiku

why? what's wrong with it in your opinion?

Answer 3

Here is the answer to the second equation
http://symbolab.com/solver/trigonometric-identity-calculator/prove%20%5Cfrac%7B1%7D%7B%5Csec(x)-%5Ctan(x)%7D%3D%5Csec(x)%2B%5Ctan(x)

Answer 4

Was that link helpful? and I wrote the first wrongly.. still working it out

Answer 5

very helpful

okay

Answer 6

the second one is easy but the first one is not. There is something wrong with it, please check
for the second one , time both numerator and denominator with (sec +tan) , denominator = sec^2 -tan^2 =1 , you have the answer

Answer 7

Yeah. I thought the first one had a problem. Thanks

Answer 8

in joining the fractions part
how come there're a negative before the whole equation?
@MIssCiku

Answer 9

If the last term from the left side is 2 cos , the problem turns true. I mean
\[\frac{2+csc x}{secx}-\color{red}{2cosx}=cotx\]

Answer 10

oh i see @Loser66

if it turns out to be like that ^
how'll it be solved?

Answer 11

\[=\frac{2}{secx}+\frac{cscx}{secx}-2cosx\]
\[2cosx +\frac{cosx}{sinx}-2cosx=cot x\]

Answer 12

@MIssCiku : nevermind my last question "in joining the fractions part
how come there're a negative before the whole equation?"

i got it already

Answer 13

@Loser66

how did you get "2cosx+cosxsinx−2cosx=cotx"