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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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( \

Vedclass · Accepted Answer

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

Vedclass · Answer

(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?

Similar Questions

The range of $\frac{1}{\sin^2 x + 3 \sin x \cos x + 5 \cos^2 x}$ is

What is the maximum value of the function $f(x) = \sin x + \cos x$?

$2^{\sin \theta} + 2^{\cos \theta}$ is greater than

If the range of $f(\theta) = \frac{\sin^4 \theta + 3 \cos^2 \theta}{\sin^4 \theta + \cos^2 \theta}$,$\theta \in R$ is $[\alpha, \beta]$,then the sum of the infinite $G.P.$,whose first term is $64$ and the common ratio is $\frac{\alpha}{\beta}$,is equal to...........

If $\alpha+\beta+\gamma=\pi$,then the expression $\sin^2 \alpha+\sin^2 \beta-\sin^2 \gamma$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module