irisgm94 said:
I am having trouble..i get stumped at the first line of my problem basically. The directions are to verify it identity.
sinx + cosx cotx = cscx
tanx cosx + cosx cotx = cscx
*and this is as far as i can get..and im not even confident with this much
It is OFTEN (but not ALWAYS) a good idea to write each term on the most complicated side of the equation (clearly the LEFT side in your problem) in terms of sin x and cos x:
sin x + [cos x * (cos x/ sin x)] = csc x
Or,
sin x + [cos^2 x / sin x] = csc x
Since the right side has just a single term, it might be a good idea to see if we can combine the two terms on the LEFT side into a single term.
(sin x / 1) + [cos^2 x / sin x] = csc x
On the left side, we have two fractions we'd like to add. But, we can only add fractions if the denominators are the same. The least common denominator for those two fractions is sin x. Multiply numerator and denominator of the first fraction by sin x:
[sin x * sin x / 1*sin x] + [cos^2 x / sin x] = csc x
[sin^2 x / sin x] + [cos^2 x/sin x] = csc x
The denominators of the fractions on the left side of the equals sign are the same...add the numerators, and put the result over the common denominator:
[sin^2 x + cos^2 x] / sin x = csc x
Oh! And from the Pythagorean identities, sin^2 x + cos^2 x = 1...so....
1/sin x = csc x
Isn't 1/sin x DEFINED to be csc x?
So,
csc x = csc x
Done.
