The problem is as follows:


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


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


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

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?.


1 Answer 1


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

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


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

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 $\cos y >0 \Rightarrow |\cos y| = \cos y$.


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

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 $|x|=−x$

  • $\begingroup$ Thanks for the clarification. However, can you include a version of your answer without using $\operatorname{sgn} x \cdot \left | x \right |$? Don't take me wrong, I'm just beginning to learn this, and I have not been introduced to such concept yet. The rest you wrote based on that concept, I do understand because its logical (but I had to read Wikipedia entry first. So if you don't mind, can you please add an alternate solution without using $\operatorname{sgn}$? $\endgroup$ Jan 22, 2021 at 2:23
  • $\begingroup$ @Chris I've edited. $\endgroup$
    – 19aksh
    Jan 22, 2021 at 2:25
  • $\begingroup$ Thanks!. I've accepted your answer, but can you explain me why? $\left | \sin x \right | = -\sin x$ as $\sin x < 0$ ?. It doesn't seem logical that something inside of an absolute value yields a negative sign?. Why is this happening or am I getting the wrong picture?. I'm stuck with this, please help. $\endgroup$ Jan 22, 2021 at 2:37
  • 1
    $\begingroup$ Consider a numerical value. Let $x = 1$, now $|x| = 1 = x$ . Again let $x = -1$. Now what is it's absolute value? It's again $1$, right? $1= -(-1)$ or $|x| = -x$ $\endgroup$
    – 19aksh
    Jan 22, 2021 at 2:39
  • $\begingroup$ Gee How come I didn't noticed that!. Crystal clear. Thanks. You may want to move the earlier comment to be part of your answer as a reminder for anyone who wants to look at this question in the future. $\endgroup$ Jan 22, 2021 at 2:48

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