Prove the identity
(Cosxsinx) ÷ tanx = 1 - sin²x
To prove the identity, we will start with the left-hand side (LHS) and manipulate it to obtain the right-hand side (RHS).
LHS: (cos(x)sin(x))/tan(x)
Recall that tan(x) = sin(x)/cos(x). Substituting this in the LHS, we get:
LHS: (cos(x)sin(x))/(sin(x)/cos(x))
Now, let's simplify:
LHS: (cos(x)sin(x)) * (cos(x)/sin(x))
Cancel out sin(x) from the numerator and denominator:
LHS: cos²(x)
Recall the Pythagorean identity: sin²(x) + cos²(x) = 1. Solve for cos²(x) by subtracting sin²(x) from both sides:
cos²(x) = 1 - sin²(x)
Therefore, we have:
LHS = 1 - sin²(x)
which is the same as the RHS.
Hence, we have proved the identity:
(cos(x)sin(x))/tan(x) = 1 - sin²(x)
To prove the identity (Cosxsinx) ÷ tanx = 1 - sin²x, we will simplify the left side of the equation and show that it is equal to the right side.
First, let's rewrite the left side of the equation using the trigonometric identities:
(Cosxsinx) ÷ tanx = [(Cosx)(sinx)] / (sinx / cosx)
Next, we can simplify by multiplying the numerator by cosx and the denominator by sinx:
[(Cosx)(sinx)] / (sinx / cosx) = [(Cosx)(sinx)(cosx)] / [(sinx)(sinx)]
Now, we can simplify further by canceling out the common factors:
[(Cosx)(sinx)(cosx)] / [(sinx)(sinx)] = [(Cosx)(cosx)] / sinx
Using the trigonometric identity cos²x = 1 - sin²x, we can rewrite the numerator:
[(Cosx)(cosx)] = cos²x = 1 - sin²x
Finally, substituting this value back into the equation, we get:
[(Cosx)(cosx)] / sinx = (1 - sin²x) / sinx
This is equivalent to the right side of the given identity, 1 - sin²x.
Therefore, the identity (Cosxsinx) ÷ tanx = 1 - sin²x is proven to be true.