Prove: (cotx sinx)(secx-cosx)=sin^2(X)
(cosx/sinx X sinx)(1/cosx - cosx) = sin^2x
(cosx)(1-cos^2x) /cosx = sin^2x
1-cos^2x = sin^2x
sin^2x = sin^2x
PROVED
To prove the equation (cot(x)sin(x))(sec(x)-cos(x)) = sin^2(x), we will simplify the left-hand side (LHS) and the right-hand side (RHS) of the equation separately and then compare the results.
Starting with the LHS:
(cot(x)sin(x))(sec(x)-cos(x))
Using the reciprocal identities of cotangent and secant:
= [(cos(x)/sin(x))(sin(x))][1/cos(x)-cos(x)]
= cos(x) [1/cos(x) - cos(x)]
Now, let's simplify the right-hand side (RHS) of the equation:
sin^2(x)
= (sin(x))^2
Next, we will simplify the expression we obtained for the LHS:
cos(x) [1/cos(x) - cos(x)]
Using the distributive property, multiply cos(x) to each term:
= cos(x)/cos(x) - cos^2(x)
= 1 - cos^2(x)
Using the Pythagorean identity sin^2(x) = 1 - cos^2(x), we can rewrite the LHS as:
= sin^2(x)
Therefore, we have proven that (cot(x)sin(x))(sec(x)-cos(x)) = sin^2(x).