(cos^4x -sin ^4 x)/(sin^2x)=cot^2x -1
How to prove true?
(cos^4x -sin ^4 x)/(sin^2x)
(cos^2x-sin^2x)(cos^2x+sin^2x)/sin^2x
(cos^2x-sin^2x)(1)/sin^2x
cos^2x/sin^2x - sin^2x/sin^2x
cot^2x - 1
To prove the given equation true, we can simplify both sides of the equation and then show that they are equal.
Let's start with the left-hand side (LHS) of the equation:
LHS = (cos^4x - sin^4x) / sin^2x
Using the identity cos^2x + sin^2x = 1, we can rewrite cos^4x as (cos^2x)^2 and sin^4x as (sin^2x)^2:
LHS = [(cos^2x)^2 - (sin^2x)^2] / sin^2x
Now, let's use the difference of squares factorization:
LHS = [(cos^2x + sin^2x)(cos^2x - sin^2x)] / sin^2x
Using the identity cos^2x - sin^2x = cos(2x), we can simplify further:
LHS = [(cos^2x + sin^2x)cos(2x)] / sin^2x
Since cos^2x + sin^2x = 1, we have:
LHS = cos(2x) / sin^2x
Now, let's simplify the right-hand side (RHS) of the equation:
RHS = cot^2x - 1
Using the identity cot^2x = 1 + cosec^2x, we can rewrite the RHS:
RHS = (1 + cosec^2x) - 1
Simplifying further:
RHS = cosec^2x
Now, let's compare the LHS and RHS:
LHS = cos(2x) / sin^2x
RHS = cosec^2x
Using the identity cosec^2x = 1/sin^2x, we can rewrite the RHS:
RHS = 1/sin^2x
Therefore, we can see that the LHS is equal to the RHS:
LHS = RHS
Hence, the given equation (cos^4x - sin^4x) / sin^2x = cot^2x - 1 is proven to be true.