Prove that # sin^2x + sin^2xcot^2x -= 1# is an identity?

Question

Dylan Arnold · Accepted Answer

We will use the following standard trigonometric identities: # 1+cot^2theta -= csc^2 theta # So then: # sin^2x + sin^2xcot^2x -= sin^2x(1+cot^2x) #
# "                            " -= sin^2x(csc^2x) #
# "                            " -= sin^2x*1/(sin^2x) #
# "                            " -= 1 # QED

Asher Friedman · Answer

undefined To prove the identity $ \sin^2(x) + \sin^2(x)\cot^2(x) = 1$, we can start with the left-hand side (LHS) of the equation and manipulate it until it is equivalent to the right-hand side (RHS), which is 1.

Starting with LHS:
$$ \sin^2(x) + \sin^2(x) \cot^2(x) $$

Using the identity $ \cot(x) = \frac{1}{\tan(x)} = \frac{\cos(x)}{\sin(x)} $:
$$ \sin^2(x) + \sin^2(x) \left(\frac{\cos(x)}{\sin(x)}\right)^2 $$

Expanding and simplifying the expression:
$$ \sin^2(x) + \sin^2(x) \left(\frac{\cos^2(x)}{\sin^2(x)}\right) $$
$$ \sin^2(x) + \frac{\sin^2(x)\cos^2(x)}{\sin^2(x)} $$
$$ \sin^2(x) + \cos^2(x) $$

Using the Pythagorean identity $ \sin^2(x) + \cos^2(x) = 1 $, we find:
$$ \sin^2(x) + \cos^2(x) = 1 $$

Thus, we have shown that the left-hand side (LHS) is equal to the right-hand side (RHS), proving the identity $ \sin^2(x) + \sin^2(x)\cot^2(x) = 1 $.