Another one (1 Viewer)

[DarkDude]

How do you prove that (tanx + 1/sec x) - (cosec x/sec^2x) is a constant?

 

[haboozin]

How do you prove that (tanx + 1/sec x) - (cosec x/sec^2x) is a constant?

derivative = 0

then put in any value for x and get the constant value...

 

[MarsBarz]

I know the gradient of a horizontal line is 0 since rise/run=0/run=0 but what is the gradient of a vertical line? (rise/run=rise/0=undefined?)

 

[word.]

Um.
It isn't a constant...

 

[MarsBarz]

According to what I've just read a vertical line has infinite gradient which makes sense since rise/run=rise/0

Therefore I do not understand how getting derivative=0 will solve this thing. Can you explain this?

At the moment I am guessing that you have to expand it and somehow the x variable cancels out and you get a constant. Getting the derivative = 0 cannot solve this. You must expand it and cancel the x.

 

[acmilan]

When i graph it i dont get that its a constant?

 

[DarkDude]

i have tried both ways and i am still lost. just to clarify its (tanx+1)/sec x.

 

[acmilan]

Well, that function simplifies to


cosx - cosecx + sinx, which still isnt a constant.

 

[DarkDude]

i am sorry guys i think i must have copied down the question wrongly. i went back and found out the question actually read
prove:
(tanx + 1/sec x) - (cotx+ 1/cosecx)

 

[acmilan]

y
= ((tanx + 1)/sec x) - ((cotx+ 1)/cosecx)
= cosx(tanx + 1) - sinx(cotx + 1)
= cosx(sinx/cosx + 1) - sinx(cosx/sinx + 1)
= sinx + cosx - cosx - sinx
= 0

Which is a constant

 

[DarkDude]

Thanks acmilan.