If $x = cosecA + cos A$ and $y = cosecA - cosA$, then find the value of $(\frac{2}{x+y})^2 + (\frac{x-y}{2})^2 - 1 $.
Answer & explanation
Correct answer: option 2
Given :-
x = cosecA + cosA and y = cosecA - cosA
x + y = 2 cosecA
x - y = 2 cosA
Now,
(\(\frac{2 }{x+y}\))² + (\(\frac{x-y }{2}\))² - 1
= (\(\frac{2 }{2cosecA}\))² + (\(\frac{2cosA }{2}\))² - 1
= (sinA)² + (cosA)² - 1
= sin²A + cos²A - 1
= 1 - 1 { sin²A + cos²A = 1 }
= 0