**Finding the value of sinX cosX when acosX bcosX=c**

To find the value of sinX cosX when acosX bcosX=c, we need to use trigonometric identities and algebraic manipulation.

**Using Trigonometric Identities**

First, we can use the identity sin²X + cos²X = 1 to rewrite the expression as:

sinX cosX = √(1 - cos²X) √(1 - sin²X)

Next, we can use the identity sin²X = 1 - cos²X to rewrite sinX as:

sinX = √(1 - cos²X)

Substituting this into the previous expression, we get:

sinX cosX = √(1 - cos²X) √(1 - (1 - cos²X))

Simplifying this expression, we get:

sinX cosX = √(2cos²X - 1)

**Algebraic Manipulation**

Another way to approach this problem is to use algebraic manipulation.

From the given equation, we can write:

acosX bcosX = c

Dividing both sides by bcosX, we get:

acosX = c/bcosX

Taking the cosine of both sides, we get:

cos(acosX) = cos(c/bcosX)

Using the identity cos(acosX) = X, we get:

X = cos(c/bcosX)

Substituting the expression for cosX, we get:

X = cos(c/b√(1 - X²))

Squaring both sides, we get:

X² = cos²(c/b√(1 - X²))

Using the identity sin²X + cos²X = 1, we can write:

sin²(c/b√(1 - X²)) + cos²(c/b√(1 - X²)) = 1

Simplifying this expression, we get:

sin²(c/b√(1 - X²)) = 1 - cos²(c/b√(1 - X²))

Substituting this into the expression for sinX cosX, we get:

sinX cosX = √(1 - cos²(c/b√(1 - X²))) √(sin²(c/b√(1 - X²)))

Simplifying this expression, we get:

sinX cosX = √(1 - X²)

Therefore, the value of sinX cosX when acosX bcosX=c is √(2cos²X - 1) or √(1 - X²), depending on the approach used.

Easy squre the both side and change cos into sin