Hey does anyone know these trig idenities??
((Cosx +1)/(secx-tanx)) - ((1-cosx)/secx+tanx)) = 2(1+tanx)
sec(-x) / sin(180-x) = tan(270-x) + tan(180+x)
(tanx-sinx) / (sin(cubed)x) = (sec) / (1+cosx)
Hello Giggly2If you mean do I know how to prove them, yes.
For this one, you need to know that if you divide both sides of by , you get:((Cosx +1)/(secx-tanx)) - ((1-cosx)/secx+tanx)) = 2(1+tanx)
Then just put both the fractions on the LHS over a common denominator, multiply out the brackets (be careful with the minus signs!) and you'll get . I'll start you off:
Can you complete it now?
You need to know the following identities:sec(-x) / sin(180-x) = tan(270-x) + tan(180+x)
- So
Then
Now
- start from the RHS using the last two identities above
- express and in terms of sine and cosine
- add the two fractions
- use
and you're there.
Again, I'll start you off:(tanx-sinx) / (sin(cubed)x) = (sec) / (1+cosx)
Now you need to:
- factorise the numerator
- divide top and bottom by the common factor
- use
- divided top and bottom by a common factor again
You should end up with
Can you finish them off now?
Grandad