MCQ
If $\text{x}^2<-4$ then the value of x is:
- A$(-2,2)$
- B$(2,\infty)$
- C$(-2,\infty)$
- D$\text{No solution}$
Solution:
Given, $\text{x}^2<-4$
$\Rightarrow\text{x}^2+4<0$
Which is not possible.
So, there is no solution.
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The value of $\cos^248^\circ-\sin^212^\circ$ is:
$\frac{\sqrt{5}+1}{8}$
$\frac{\sqrt{5}-1}{8}$
$\frac{\sqrt{5}+1}{5}$
$\frac{\sqrt{5}+1}{2\sqrt{2}}$
[Hint: Use $\cos^2\text{A}-\sin^2\text{B}=\cos(\text{A + B})\cos(\text{A}-\text{B})$]