Given a^2 + 2a + csc^2 left( frac{pi}{2}(a + | vedclass

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(B) The given equation is $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 0$. Adding $1$ to both sides,we get $a^2 + 2a + 1 + \csc^2 \left( \

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$a = -1; \frac{x}{2} \in I$

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(B) The given equation is $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) ight) = 0$. Adding $1$ to both sides,we get $a^2 + 2a + 1 + \csc^2 \left( \frac{\pi}{2}(a + x) ight) = 1$. This simplifies to $(a + 1)^2 + \csc^2 \left( \frac{\pi}{2}(a + x) ight) = 1$. Using the identity $1 + \cot^2 	heta = \csc^2 	heta$,we have $(a + 1)^2 + 1 + \cot^2 \left( \frac{\pi}{2}(a + x) ight) = 1$. This reduces to $(a + 1)^2 + \cot^2 \left( \frac{\pi}{2}(a + x) ight) = 0$. Since both terms are squares of real numbers,their sum can be zero only if each term is zero. Thus,$(a + 1)^2 = 0 \Rightarrow a = -1$. Substituting $a = -1$ into the second term,we get $\cot^2 \left( \frac{\pi}{2}(-1 + x) ight) = 0$. This implies $\cot \left( \frac{\pi}{2}(x - 1) ight) = 0$,which means $\frac{\pi}{2}(x - 1) = n\pi$ for some integer $n$. Thus,$x - 1 = 2n \Rightarrow x = 2n + 1$,which implies $\frac{x}{2}$ is not necessarily an integer,but checking the options,$a = -1$ is the key condition.

Given $a^2 + 2a + \csc^2 \left( \frac{\pi}{2}(a + x) \right) = 0$,then which of the following holds good?

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