Get the value of $2\left |\csc x \right | \sin x + 3\left | \cos y\right|\sec y$ given two constrains

absolute valuealgebra-precalculustrigonometry

The problem is as follows:

Given:

$\sqrt{\cos x}\cdot \sin y > 0$

and,

$\tan x\cdot \sqrt{\cot y} < 0$

Find:

$B=2\left |\csc x \right | \sin x + 3\left | \cos y\right|\sec y$

The alternatives given in my book are as follows:

$\begin{array}{ll}
1.&\textrm{-1}\\
2.&\textrm{5}\\
3.&\textrm{1}\\
4.&\textrm{-5}\\
\end{array}$

I'm confused on exactly how to use the given clues to solve the problem?

To me the source of confusion is how to use the absolute value in the question?

From the first given expression I'm getting this, assuming squaring both sides of the equation will not modify its order:

$\left(\sqrt{\cos x}\cdot \sin y\right)^2 > 0^2$

$\cos x \cdot \sin^2 y > 0$

$\left(\tan x\cdot \sqrt{\cot y}\right )^2 < 0^2$

$\tan^2 x\cdot \cot y < 0^2$

But the thing is this where I'm stuck, where to go from here?. There isn't known a relationship between those angles. If so then I believe trigonometric identities could be used to simplify the expression. Therefore, should these expressions be divided or what?.}

Can someone help me here on what should be done? and more importantly why?. It would help me the most is an answer which would explain how does absolute value is used here?.

Best Answer

$a<0$ doesn't necessarily mean $a^2<0$.

For example, $-2<0$ but $(-2)^2 = 4 >0$.

Now,

$\begin{align}B &= 2|\csc x|\sin x + 3 |\cos y|\sec y = 2 \frac{\sin x}{|\sin x|}+3\frac{|\cos y|}{\cos y}\\ \\& = 2\text{ sgn}(\sin x) + 3\text{ sgn}(\cos y)\end{align}$

Then, $\sqrt{\cos x} \sin y > 0 \Rightarrow \sin y > 0$ and $\cos x > 0$ as $\cos$ is under the square root.

Similarly, $\cot y>0$ and $\tan x<0$.

So we have,

$\sin y >0, \cot y = \frac{\cos y}{\sin y} > 0 \Rightarrow \boxed{\cos y >0}$

$\cos x>0, \tan x = \frac{\sin x}{\cos x}<0 \Rightarrow \boxed{\sin x<0}$

Thus, $B = 2(-1) + 3(1) = 1$

Edit (based on comment)

As $\sin x<0 \Rightarrow |\sin x| = -\sin x$ and $\cos y >0 \Rightarrow |\cos y| = \cos y$.

So,

$\begin{align}B = 2 \frac{\sin x}{|\sin x|}+3\frac{|\cos y|}{\cos y} = 2 \frac{\sin x}{-\sin x}+3\frac{\cos y}{\cos y} = -2 + 3 = 1\end{align}$

Clarification for abs value:
Consider some numerical values. Let $x=1$, so $|x|=1=x$.
Again let $x=−1$. Now what is it's absolute value? It's again 1, right? $1=−(−1)$ or $|x|=−x$

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