how do i ,prove: 1/(sec x-1) + 1/(sec x-1)= 2cotx cscx
LS , using common denominator
= ( secx - 1 + secx + 1)/(sec^2 -1)
= 2secx/tan^2x
= 2/cosx / [(sin^2 x)/cos^2 x]
= 2cosx/sin^2x
= 2cosx/sinx / sinx
= 2cotx cscx
= RS
To prove the given equation: 1/(sec x-1) + 1/(sec x-1) = 2cotx cscx, we need to express both sides of the equation in the same form and then test their equality.
Let's start by simplifying the left-hand side (LHS) of the equation:
1/(sec x-1) + 1/(sec x-1)
To simplify this expression, we can find a common denominator, which is (sec x - 1) in this case. Multiply the first term by (sec x - 1)/(sec x - 1):
[(1/(sec x - 1)) * (sec x - 1)] / (sec x - 1) + 1/(sec x - 1)
Now, simplify:
(sec x - 1)/(sec x - 1) + 1/(sec x - 1)
= (sec x - 1 + 1)/(sec x - 1)
= sec x / (sec x - 1)
Now, simplify the right-hand side (RHS) of the equation:
2cotx cscx
Recall that cot x = cos x / sin x and csc x = 1 / sin x. Substitute these values:
2 * (cos x / sin x) * (1 / sin x)
Simplify:
2cos x / (sin^2 x)
Next, we need to simplify the LHS and RHS so that they are expressed in the same form. We can achieve this by multiplying the LHS by (sin^2 x / sin^2 x):
(sec x / (sec x - 1)) * (sin^2 x / sin^2 x)
= (sin^2 x * sec x) / ((sec x - 1) * sin^2 x)
Now, we can compare this expression to the RHS:
2cos x / (sin^2 x)
As both sides have the same denominator, we can compare their numerators:
sin^2 x * sec x = 2cos x
To simplify further, we can rewrite sec x as 1/cos x:
sin^2 x * (1/cos x) = 2cos x
(sin^2 x / cos x) = 2cos x
To further simplify, recall that sin^2 x = 1 - cos^2 x:
(1 - cos^2 x) / cos x = 2cos x
Multiply both sides of the equation by cos x:
1 - cos^2 x = 2cos^2 x
Rearrange the equation to solve for cos^2 x:
3cos^2 x = 1
cos^2 x = 1/3
cos x = ±√(1/3)
Now, substitute this value back into the original equation to find the solution(s) for x and prove the equation.