Answer:

Given : x= cosecA+cosA
y= cosecA-cosA

So, value of x+y= 2 cosecA
and x-y = 2 cosA

Now, {2/(x+y)}²+{(x-y)/2}-1 = (2/2cosecA)²+(2cosA/2)²-1
{2/(x+y)}²+{(x-y)/2}-1 = sin²A+cos²A-1
{2/(x+y)}²+{(x-y)/2}-1 = 1-1
{2/(x+y)}²+{(x-y)/2}-1 = 0
Hence proved.

Hope its help you.

Given:
x = cosecA cosA
y = cosecA - cosA

To prove:
(2/x y)^2 (x-y/2)^2 - 1 = 0

Proof:

Step 1: Rewrite the given expressions
x = 1/sinA cosA (since cosecA = 1/sinA)
y = 1/sinA - cosA

Step 2: Simplify the expressions
x = cosA/sinA (multiply numerator and denominator by cosA)
y = (1 - sinA cosA)/sinA (common denominator)

Step 3: Substitute the values of x and y in the given expression
(2/x y)^2 (x - y/2)^2 - 1 = 0

Substituting the values of x and y,
(2/(cosA/sinA)(1 - sinA cosA)/sinA)^2 [(cosA/sinA) - (1 - sinA cosA)/2sinA]^2 - 1 = 0

Step 4: Simplify the expression inside the brackets
[(2sinA)/(cosA(1 - sinA cosA))]^2 [(2cosA - (1 - sinA cosA))/(2sinA)]^2 - 1 = 0

Simplifying further,
[(2sinA)/(cosA - sin^2AcosA)]^2 [(2cosA + sinA cosA - 1)/(2sinA)]^2 - 1 = 0

Step 5: Simplify the expression
[(2sinA)/(cosA - sin^2AcosA)]^2 [(2cosA + sinA cosA - 1)/(2sinA)]^2 - 1 = 0

Since the numerator and denominator of the first term are squared, they cancel each other out,
(2cosA + sinA cosA - 1)^2 - (cosA - sin^2AcosA)^2 = 0

Step 6: Simplify further
Expanding the squares,
(4cos^2A + 4sinA cosA - 2cosA - 2sinA cos^2A - 2sin^2AcosA + 1) - (cos^2A - 2sin^2AcosA + sin^4Acos^2A) = 0

Simplifying the expression,
4cos^2A + 4sinA cosA - 2cosA - 2sinA cos^2A - 2sin^2AcosA + 1 - cos^2A + 2sin^2AcosA - sin^4Acos^2A = 0

Combining like terms,
3cos^2A + 4sinA cosA - cosA + 2sin^2A - sin^4Acos^2A + 1 = 0

Step 7: Rearrange the terms
Rearranging the terms in descending powers of sinA and cosA,
-sin^4Acos^2A + 3cos^2