Question

For all real values of $\theta$

Answer 1

A) $A={\sin }^2\theta +{\cos }^4\theta ={\sin }^2\theta +{\cos }^2\theta (1-{\sin }^2\theta )=1-\frac{{\sin }^{2}2\theta }{4}$
As $0\le {\sin }^{2}2\theta \le 1\Rightarrow \frac{3}{4}\le 1-\frac{{\sin }^{2}2\theta }{4}\le 1\Rightarrow Aϵ[\frac{3}{4},1]$
B) $A=3{\cos }^2\theta +{\sin }^4\theta ={\sin }^4\theta -3{\sin }^2\theta +3$
As $0\le {\sin }^2\theta \le 1$ and $0\le {\sin }^4\theta \le 1$
Therefore $Aϵ[1,3]$
C) $A={\sin }^2\theta -{\cos }^4\theta =1-{\cos }^2\theta -{\cos }^4\theta$
As $0\le {\cos }^2\theta \le 1$ and $0\le {\cos }^4\theta \le 1$
Therefore, $Aϵ[-1,1]$
D) $A={\tan }^2\theta +2{\cot }^2\theta =\frac{{\sin }^2\theta }{{\cos }^2\theta }+\frac{2{\cos }^2\theta }{{\sin }^2\theta }=\frac{{\sin }^4\theta +2{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }=\frac{1-2{\sin }^2\theta {\cos }^2\theta +{\cos }^4\theta }{{\sin }^2\theta {\cos }^2\theta }=\frac{1-2{\sin }^2\theta {\cos }^2\theta +{\sin }^2\theta -{\sin }^2\theta {\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }={\sec }^2\theta {\csc }^2\theta (1+{\sin }^2\theta -3{\sin }^2\theta {\cos }^2\theta )$
As ${\sec }^2\theta {\csc }^2\theta \ge \sqrt{2}$
Therefore, $A\ge 2\sqrt{2}$