Prove (cos^2t+4cos^2+4)/(cost+2)=2sect+1)/sect
To prove the given equation,
(cos^2(t) + 4 * cos^2 + 4) / (cos(t) + 2) = (2 * sec(t) + 1) / sec(t)
we need to manipulate one side of the equation until it equals the other side. Let's begin by working on the left side:
(cos^2(t) + 4 * cos^2 + 4) / (cos(t) + 2)
Next, we can notice that the numerator can be factored as a perfect square trinomial:
(cos^2(t) + 4 * cos^2 + 4) = (cos(t) + 2)^2
So, now the equation becomes:
(cos(t) + 2)^2 / (cos(t) + 2)
Now, we can cancel out the common factor of (cos(t) + 2) from the numerator and denominator:
(cos(t) + 2)^2 / (cos(t) + 2) = cos(t) + 2
Now, let's simplify the right side of the equation:
(2 * sec(t) + 1) / sec(t)
We can rewrite sec(t) as 1/cos(t):
(2 * (1/cos(t)) + 1) / (1/cos(t))
Simplifying further:
(2/cos(t) + 1) * (cos(t)/1)
Now, we can multiply the numerators and denominators:
(2 + cos(t)) * cos(t)
Expanding:
2 * cos(t) + cos^2(t)
Now, the equation becomes:
2 * cos(t) + cos^2(t) = cos(t) + 2
Rearranging:
cos^2(t) + cos(t) - 2 = 0
Now, we have a quadratic equation. Factoring it:
(cos(t) - 1)(cos(t) + 2) = 0
This implies that either:
cos(t) - 1 = 0
which gives us cos(t) = 1, or:
cos(t) + 2 = 0
which gives us cos(t) = -2.
However, the range of the cosine function is -1 to 1. So, we can discard cos(t) = -2 as an extraneous solution.
Therefore, the only solution is cos(t) = 1, or t = 2πn, where n is an integer.
So, we have proved that the given equation is true for all values of t = 2πn, where n is an integer.