Prove that cot x + tan x = csc x * sec x whenever both sides are defined.
I tried rewriting all sides in terms of sin x but it got all complicated and I can't make head or tail of it. Could someone assist me? Is it all just a pound-out algebra problem in essence? How do you simplfy sin x * sin x?
cot + tan
1/tan + tan
(1+tan^2)/tan
sec^2/tan
1/cos^2 * cos/sin
1/(sin*cos)
sec*csc
Thanks a lot!!!
To prove that cot x + tan x = csc x * sec x, start with the expressions for cot x and tan x in terms of sine and cosine:
cot x = cos x / sin x
tan x = sin x / cos x
Now, let's simplify the expression cot x + tan x using these definitions:
cot x + tan x = (cos x / sin x) + (sin x / cos x)
To work with this expression more easily, we can find a common denominator:
cot x + tan x = (cos x * cos x + sin x * sin x) / (sin x * cos x)
Now, recall the Pythagorean identity: sin^2 x + cos^2 x = 1. We can use this identity to simplify the numerator:
cot x + tan x = 1 / (sin x * cos x)
Now, the right side of the equation is csc x * sec x. Recall that csc x is the reciprocal of sin x, and sec x is the reciprocal of cos x:
csc x * sec x = (1 / sin x) * (1 / cos x)
Taking the reciprocal of a product gives us:
csc x * sec x = 1 / (sin x * cos x)
We can see that the right side of the equation matches the simplified expression for cot x + tan x. Therefore, we have proved that cot x + tan x = csc x * sec x whenever both sides are defined.
When simplifying sin x * sin x, you just multiply the two terms together:
sin x * sin x = sin^2 x
So, sin x * sin x simplifies to sin^2 x.