Question

Can you show me step by step, I'm having trouble with these types of expressions.

Simplify the expression.

quantity one minus cotangent of x divided by quantity tangent of x minus one

Answer 1

\[1 - \cot x \over \tan x -1\]

Answer 2

we know:
\[\cot x = \frac{ \cos x }{ \sin x }\]
and
\[\tan x = \frac{ \sin x }{ \cos x }\]

next, change cot and tan with them.

\[= \frac{ 1 - \frac{ \cos x }{ \sin x } }{ \frac{ \sin x }{ \cos x } -1 }\]

\[ 1 - \frac{ \cos x }{ \sin x } = \frac{ \sin x - \cos x }{ \sin x } \]
\[{\frac{ \sin x }{ \cos x } -1} = \frac{ \sin x - \cos x }{ \cos x } \]

so,

\[= \frac{ \frac{ \sin x - \cos x }{ \sin x } }{ \frac{ \sin x - \cos x }{ \cos x } }\]

\[=\frac{ \frac{ 1 }{ \sin x } }{ \frac{ 1 }{ \cos x } }\]

\[= \frac{ \cos x }{ \sin x }\]

\[= ....\] hehe :)

Answer 3

how did you get 1= cosx/sinx ?

Answer 4

I meant 1 -*

Answer 5

oh okay, you know \[1- \frac{ 2 }{ 5 }\] is same with \[\frac{ 5 }{ 5 } - \frac{ 2 }{ 5 } = \frac{ 3 }{ 5 }\]?

is same with \[1-\frac{ \sin x }{ \cos x }\] so 1 is same with \[\frac{ \cos x }{ \cos x }\] so we can calculate to \[\frac{ \cos x }{ \cos x } - \frac{ \sin x }{ \cos x }\] it will be same with \[\frac{ \cos x - \sin x }{ \cos x }\] right? hehehe :)

Answer 6

in my answers is \[1 - \frac{ \cos x }{ \sin x }\] it mean 1 is same with \[\frac{ \sin x }{ sin x }\] so, \[\frac{ \sin x }{ \sin x } - \frac{ \cos x }{ \sin x } = \frac{ \sin x - \cos x }{ \sin x }\]

Answer 7

Wow thank you, lol. Can you help me with two more problems?
\[\tan \theta \over \cot \theta \]
I got to \[\frac{ \sin }{ \cos } \over \frac{ \cos }{ \sin }\] and then I am kind of stuck from here

Answer 8

first, thanks for medal :)

of course,hehe i think the answers is 1.

we know cot is 1/tan x.

so \[\tan x \frac{ 1 }{ \tan x } = 1\]

Answer 9

Thanks and your welcome

Answer 10

sorry i was make mistake with second question

Answer 11

the answers is tan^2 x

Answer 12

it must \[tanx \frac{ 1 }{ cotx }\] \[tanx \frac{ 1 }{ \frac{ 1 }{ tanx } }\] so, tanx go up, then multiply tanx with tanx = \[\tan^{2}x\]