Practicing Success
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 $. |
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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 |