# If $x,y \in (0,\frac{\pi}{2})$ then expression $\sin x +\cos y +\tan^2y+\cot^2x+5>\ldots?$

Problem :

If $x,y \in (0,\frac{\pi}{2})$ then expression $\sin x +\cos y +\tan^2y+\cot^2x+5$ is always greater than :

(a) $\ 7$

(b) $\ 8$

(c) $\ 9$

(d) $\$none of these

Solution :

We can write the given expression $\sin x +\cos y +\tan^2y+\cot^2x+5$ as $\sin x +\cot^2x +\cos x +\tan^2y +5$

$\Rightarrow \sin x +\cot^2x +\cos y +\tan^2y +5 = \sin x + \csc^2x -1 +\cos x +\sec^2y-1+5$

$\Rightarrow \sin x + \csc^2x +\cos x +\sec^2y+3$

Please guide me how to proceed further. Thanks.

I think you are doing a mistake here. $\sin(x)+\cos(y)+\tan^2(y)+\cot^2(x)+5 = (\sin(x)+ \csc^2(x)) + (\cos(y)+ \sec^2(y))+ 3$ . Not, what you have written. Considering that to be a typo, Notice that the things in the bracket are independent, so we just find minimum value of each separately.

Consider $f(x)=\sin(x)+\csc^2(x) \ge \sin(x)+\csc(x) \ge 2$. (By AM-GM inequality) . Similarly consider $g(x)=\cos(y)+\sec^2(y)$

We get $7$ as the answer .

• How do you know that it need not be true that the expression is always greater than 9? Are you assuming that there is exactly 1 right answer in that MCQ? Jul 12, 2014 at 19:18

I don't think the simplifying is necessary. We can just analyze the functions within the open interval $(0,\pi/2)$. First, we can examine the minimum value of $\sin x+\cot^2x$. When $x=\pi/2$, sine is $1$ and cotangent is $0$. Cotangent increases without bound as we move towards $0$ from $\pi/2$. So, we can assume $\sin x+\cot^2x$ is always greater than $1$ on the interval $(0,\pi/2)$. The same reasoning applies to $\cos y+\tan^2 y$, however the minimum will be at $y=0$ now, and the minimum is again $1$. So, just add these together with $5$, and we see that the entire function is always greater than $7$.

• Cotangent increases, but sine decreases... so your argument, as it stands, doesn't work. You need a reason why the square of cotangent increases more from zero than sine decreases from 1. Jul 12, 2014 at 18:31
• The definition of $\cot=\cos/\sin$. Sine decreases as cosine increases. And visa versa. We don't have to talk about the squared Jul 12, 2014 at 18:50
• You've missed the point. Cotangent does increase without bound from zero as $x$ moves from $\frac{\pi}{2}$ to $0$, and therefore its square does too, but sine decreases from 1 as $x$ goes in the same direction. So you cannot claim $\sin x+\cot^2x$ is always at least 1 without providing a reason as I described above. Jul 12, 2014 at 19:04
• Yeah I know what you are saying. Sorry, my last comment didn't describe my thoughts very well. Was trying to say that since cotangent is defined that way, it will increase "faster" than sine will decrease. A better way to look at it is with derivatives maybe. The derivative of $\sin x + \cot x$ is always negative thus, the function is decreasing on the open interval or increasing when we look at it from the other direction. Jul 12, 2014 at 19:24