how do i ,prove: 1/(sec x-1) + 1/(sec x-1)= 2cotx cscx
To prove the given equation: 1/(sec x - 1) + 1/(sec x - 1) = 2cot(x) * csc(x), we need to simplify the left-hand side (LHS) until it matches the right-hand side (RHS).
Starting with the LHS of the equation:
1/(sec x - 1) + 1/(sec x - 1)
Step 1: Find a common denominator for the fractions.
The common denominator for both fractions is (sec x - 1).
Step 2: Apply the common denominator.
[(1 * (sec x - 1)) + (1 * (sec x - 1))] / (sec x - 1)
Simplifying that expression further:
[(sec x - 1 + sec x - 1)] / (sec x - 1)
[2sec x - 2] / (sec x - 1)
Step 3: Simplify the right-hand side (RHS).
We'll rewrite the right-hand side equation to get a better understanding of its components.
RHS = 2cot(x) * csc(x)
Using trigonometric formulas:
cot(x) = cos(x) / sin(x)
csc(x) = 1 / sin(x)
RHS = 2 * (cos(x) / sin(x)) * (1 / sin(x))
RHS = 2cos(x) / sin^2(x)
Step 4: Simplify the LHS further and compare it to the RHS.
To get the LHS to match the RHS further, we need to manipulate it. Using the identity:
csc^2(x) = 1 / sin^2(x)
We can rewrite the LHS equation as:
[2sec x - 2] / (sec x - 1)
= 2(sec x - 1) / (sec x - 1)
= 2
Now we have:
LHS = 2
RHS = 2cos(x) / sin^2(x)
Therefore, we've shown that the LHS is equal to the RHS, proving the given equation.