Question

Which of the following is/are positive?
A. ${\log }_{\sin 1}\tan 1$
B. ${\log }_{\cos 1}\left(1+\tan 3\right)$
C. ${\log }_{{\log }_{10}5}\left(\cos \theta +\sec \theta \right)$
D. ${\log }_{\tan {15}^o}\left(2\sin {18}^o\right)$

Answer 1

Hint: In this question, from the given functions, we have to tell the positive ones. We have that, in ${\log }_{a}x$ , if $a>1$ and $x>1$ or $a<1$ and $x<1$ then, ${\log }_{a}x>0$ , otherwise, it is negative.
$\tan \theta$ is always an increasing function, whereas, $\sin \theta$ and $\cos \theta$ are neither increasing nor decreasing.

Complete answer:
Given are four logarithmic functions.
To tell which of these are positive.
Firstly, we know, the graph of ${\log }_{a}x$ , when $a>1$ is given by

Then, ${\log }_{a}x>0$ , when $x>1$ .
And when $a<1$ , the graph is given by

Then, ${\log }_{a}x>0$ , when $x<1$ .
Now, consider the function, ${\log }_{\sin 1}\tan 1$ , we have, ${\displaystyle \frac{\pi }{2}}={\displaystyle \frac{3.14}{2}}>1$ , so, $\sin {\displaystyle \frac{\pi }{2}}>\sin 1$ which gives $\sin 1<1$ .
And, ${\displaystyle \frac{\pi }{4}}={\displaystyle \frac{3.14}{4}}<1$ , so, $\tan {\displaystyle \frac{\pi }{4}}<\tan 1$ which give $\tan 1>1$ . Now, $\sin 1<1$ and $\tan 1>1$ , hence, ${\log }_{\sin 1}\tan 1<0$ .
Now, consider the function, ${\log }_{\cos 1}\left(1+\tan 3\right)$ , $0<1<{\displaystyle \frac{\pi }{2}}$ , so, $\cos 0>\cos 1>\cos {\displaystyle \frac{\pi }{2}}$ , since, the function is decreasing for $0<\theta <\pi$ , hence, $\cos 1<1$
And, ${\displaystyle \frac{\pi }{2}}<3<\pi$ , which means it is in the second quadrant, and the second quadrant $\tan$ is negative, which means $1+\tan 3<1$ .
Now, $\cos 1<1$ and $1+\tan 3<1$ , hence, ${\log }_{\cos 1}\left(1+\tan 3\right)>0$ .
Now, consider the function, ${\log }_{\tan {15}^o}\left(2\sin {18}^o\right)$ , we know, $\sin {18}^o<\sin {30}^o$ , which means, $\sin {18}^o<{\displaystyle \frac{1}{2}}$ i.e., $2\sin {18}^o<1$ .
And, we know, $\tan {15}^o<\tan {45}^o$ , which gives, $\tan {15}^o<1$ . Now, $2\sin {18}^o<1$ and $\tan {15}^o<1$ , so, ${\log }_{\tan {15}^o}\left(2\sin {18}^o\right)>0$ .
Now, at last, consider the function, ${\log }_{{\log }_{10}5}\left(\cos \theta +\sec \theta \right)$ , we know, ${\log }_{10}5<{\log }_{10}10$ i.e., ${\log }_{10}5<1$ .
And, we know, the Arithmetic mean is always greater than the geometric mean, therefore, ${\displaystyle \frac{\left(\cos \theta +\sec \theta \right)}{2}}>{\left(\cos \theta \cdot \sec \theta \right)}^{{\displaystyle \frac{1}{2}}}$ , which gives, $\cos \theta +\sec \theta >2>1$ .
Now, ${\log }_{10}5<1$ and $\cos \theta +\sec \theta >2>1$ , hence, ${\log }_{{\log }_{10}5}\left(\cos \theta +\sec \theta \right)<0$ .
Thus, ${\log }_{\tan {15}^o}\left(2\sin {18}^o\right)$ and ${\log }_{\cos 1}\left(1+\tan 3\right)$ are positive.

Therefore, the correct option is B and C

Note: We know that, $\sec \theta$ is reciprocal of $\cos \theta$ .
“Arithmetic mean is always greater than the geometric mean” is true for every function or equation.
For checking, if a logarithmic function is positive or negative, we have to check if $a$ and $x$ are greater than or less than $1$ , if the sign of $a$ and $x$ are same i.e., either both greater than $1$ or both less than $1$, then, the logarithmic function is positive, and if the sign of $a$ and $x$ are not same, i.e., one is greater than $1$ and other is less than $1$, then the logarithmic function is negative.