4cot^2x-4/tanx+cosxsecx
factorin algenraic single trig function
To factorize the expression 4cot^2x-4/tanx+cosxsecx, we need to determine a common trigonometric function that can be factored out of both the numerator and the denominator.
Let's start by simplifying the numerator, which is 4cot^2x - 4. We can rewrite cot^2x as 1/tan^2x, which gives us:
4cot^2x - 4 = 4(1/tan^2x) - 4 = 4/tan^2x - 4.
Next, we need to find a common denominator for these two terms. The common denominator for tan^2x is cos^2x, so we can rewrite the expression as follows:
4/tan^2x - 4 = 4/cos^2x - 4 = 4 - 4cos^2x/cos^2x.
Now let's simplify the denominator, tanx + cosxsecx. We can rewrite secx as 1/cosx:
tanx + cosxsecx = tanx + cosx(1/cosx) = tanx + 1.
Now, we can rewrite the original expression as:
(4 - 4cos^2x/cos^2x) / (tanx + 1).
To factorize this expression, we find the common factor, which is cos^2x. We can separate it out as follows:
(4 - 4cos^2x) / cos^2x * 1 / (tanx + 1).
Now, we can simplify the expression even further:
4(1 - cos^2x) / cos^2x * 1 / (tanx + 1).
Using the Pythagorean identity sin^2x + cos^2x = 1, we can replace (1 - cos^2x) with sin^2x:
4sin^2x / cos^2x * 1 / (tanx + 1).
Finally, we can simplify this expression into a single trigonometric function. The ratio sin^2x/cos^2x is equivalent to tan^2x, so we have:
4tan^2x / (tanx + 1).
Therefore, the factored form of the expression 4cot^2x-4/tanx+cosxsecx is 4tan^2x / (tanx + 1).