Prove the identit: cosxcotx+sinx=cscx
1) rewrite cosx (cosx/sinx) + sinx = csc x
2) cosx^2/sinx + sinx = csc x
3) multiply sinx by (sinx/sinx) to get same denominator
this gives you cosx^2/sinx + sinx^2/sinx = cscx
4) combine your fractions
(cosx^2 + sinx^2)/sinx = cscx
5) rewrite numerator
1/sinx= csc x
and there you go it is proven (you have to know your trig identity to do this so look them up if you don't know them)
Thank You! :D
no problem lol this one almost had me stump for alittle
Ohh it was hard me :/ .Do you think you can answer my other questions?
To prove the identity cos(x)cot(x) + sin(x) = csc(x), we need to start with one side of the equation and manipulate it until we arrive at the other side. Let's start with the left side of the equation, cos(x)cot(x) + sin(x).
To simplify the equation, we need to express both cos(x) and cot(x) in terms of sin(x) and cos(x).
Recall that cot(x) is the reciprocal of tan(x), which is sin(x)/cos(x). Therefore, cot(x) = cos(x)/sin(x).
Substituting cot(x) = cos(x)/sin(x), we get:
cos(x) * (cos(x)/sin(x)) + sin(x)
Next, we can simplify this expression by multiplying cos(x) by cos(x)/sin(x):
(cos^2(x))/sin(x) + sin(x)
To combine the terms, we need to find a common denominator. The common denominator here is sin(x), so we can rewrite (cos^2(x))/sin(x) as cos^2(x)/sin(x) * sin(x)/sin(x):
(cos^2(x) + sin^2(x))/sin(x)
Using the Pythagorean identity cos^2(x) + sin^2(x) = 1, we can further simplify the expression:
1/sin(x)
Finally, since csc(x) is the reciprocal of sin(x), we can rewrite 1/sin(x) as csc(x):
csc(x)
Therefore, both sides of the equation are the same, and we have proved the identity cos(x)cot(x) + sin(x) = csc(x).