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 can rewrite the equation as $(a + 1)^2 + 1 + \cot^2 \left( \frac{\pi}{2}(a + x) ight) = 1$.Subtracting $1$ from both sides,we get $(a + 1)^2 + \cot^2 \left( \frac{\pi}{2}(a + x) ight) = 0$.Since both terms are squares of real numbers,their sum can be zero only if each term is zero individually.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 $\cot \left( \frac{\pi}{2}(x - 1) ight) = 0$.This occurs when the argument is an odd multiple of $\frac{\pi}{2}$,i.e.,$\frac{\pi}{2}(x - 1) = (2n + 1) \frac{\pi}{2}$ for some integer $n$.$x - 1 = 2n + 1 \Rightarrow x = 2n + 2 = 2(n + 1)$.Thus,$\frac{x}{2} = n + 1$,which is an integer $(I)$.

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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