1 + cot^2x =
csc^2x
There are different ways to approach this question, but one possible method is:
Recall the trigonometric identity: 1 + tan^2x = sec^2x.
Divide both sides of this identity by cos^2x:
(1/cos^2x) + (sin^2x/cos^2x) = (1/cos^2x)sec^2x.
Note that cot^2x is the same as (cos^2x/sin^2x), so we can substitute this into the left-hand side:
1 + (cos^2x/sin^2x) = (1/cos^2x)sec^2x.
Multiply both sides by sin^2x:
sin^2x + cos^2x = sin^2x(sec^2x/cos^2x).
Recall that sec^2x is the reciprocal of cos^2x, so we can substitute this:
sin^2x + cos^2x = sin^2x/cos^2x.
Now we can use the identity: sin^2x + cos^2x = 1.
Therefore, we have:
1 = (sin^2x/cos^2x) – 1/cos^2x.
Simplifying this equation, we get:
1 = (sin^2x – 1)/cos^2x.
Adding 1 to both sides and taking reciprocals, we obtain:
cot^2x = 1/(sin^2x – 1).
Therefore, we can substitute this into the original expression to get:
1 + cot^2x = 1 + 1/(sin^2x – 1) = sin^2x/(sin^2x – 1).
Hence, 1 + cot^2x = sin^2x/(sin^2x – 1).
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