If x = cosec A + cos A

y = cosec A - cos A

then, \( { \left(\frac{2}{x+y}\right) }^{2 } + { \left(\frac{x-y}{2}\right) }^{2 } \) - 1 = ?

0

⇒ \( { \left(\frac{2}{x+y}\right) } + { \left(\frac{x-y}{2}\right) }\) = \(\frac{2}{(cosecA+cosA+cosecA-cosA)}+ \frac{(cosec A+cosA-cosecA+cosA)}{2}\)

⇒ \( { \left(\frac{2}{x+y}\right) }^{2 } + { \left(\frac{x-y}{2}\right) }^{2 } \) - 1 = \({ \left(\frac{2}{2cosecA}\right) }^{2 } + { \left(\frac{2cosA}{2}\right) }^{2 }\) - 1

= \({ \left(\frac{1}{cosecA}\right) }^{2 }\) + cos²A - 1

= sin²A + cos²A - 1

= 1 - 1 = 0