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 = ?
Answer & explanation
Correct answer: option 1
⇒ \( { \left(\frac{2}{x+y}\right) } + { \left(\frac{x-y}{2}\right) }\) = \(\frac{2}{(cosecA+cosA+cosecA-cosA)}+ \frac{(cosec A+cosA-cosecA+cosA)}{2}\)
⇒ \( { \left(\frac{2}{x+y}\right) }^{2 } + { \left(\frac{x-y}{2}\right) }^{2 } \) - 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