Question

Verify that each equation is an identity. 1-cot^2 theta/1+cot^2 theta+1=2 sin^2 theta

Answer 1

The two sides aren't equal to one another. I just graphed the two and they don't match up. So there's a typo somewhere.

Answer 2

so it is

\[\Large \frac{1-\cot^2(\theta)}{1+\cot^2(\theta)}+1 = 2\sin^2(\theta)\]

right?

Answer 3

or no?

Answer 4

yes

Answer 5

Notice the denominator is 1+cot^2(theta)

Answer 6

we can write that "1" that is all by itself as (1+cot^2(theta)) over (1+cot^2(theta))

basically a fancy and complicated way of saying 1

Answer 7

from there, we can combine the fractions and simplify, like this

Answer 8

\[\Large \frac{1-\cot^2(\theta)}{1+\cot^2(\theta)}+1 = 2\sin^2(\theta)\]

\[\Large \frac{1-\cot^2(\theta)}{1+\cot^2(\theta)}+\frac{1+\cot^2(\theta)}{1+\cot^2(\theta)} = 2\sin^2(\theta)\]

\[\Large \frac{1-\cot^2(\theta)+1+\cot^2(\theta)}{1+\cot^2(\theta)} = 2\sin^2(\theta)\]

\[\Large \frac{2}{1+\cot^2(\theta)} = 2\sin^2(\theta)\]

\[\Large \frac{2}{\csc^2(\theta)} = 2\sin^2(\theta)\]

\[\Large \frac{2/1}{1/\sin^2(\theta)} = 2\sin^2(\theta)\]

\[\Large \frac{2}{1}*\frac{\sin^2(\theta)}{1} = 2\sin^2(\theta)\]

\[\Large 2\sin^2(\theta) = 2\sin^2(\theta)\]

Answer 9

One thing to point out: I did not alter the right side at all.