Question

If ${\log }_{\cos \left(x\right)}\tan \left(x\right)+{\log }_{\sin \left(x\right)}cot\left(x\right)=0$, then the most general solution of $x$ is

• A. $2n\mathrm{\pi }-\frac{3\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}}$
• B. $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}}$
• C. $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}}$
• D. None of these

Answer 1

The correct option is B

$2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}}$

Explanation for the correct option

Given that, ${\log }_{\cos \left(x\right)}\tan \left(x\right)+{\log }_{\sin \left(x\right)}cot\left(x\right)=0$

$\Rightarrow$ ${\log }_{\cos \left(x\right)}\left(\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)+{\log }_{\sin \left(x\right)}\left(\frac{\cos \left(x\right)}{\sin \left(x\right)}\right)=0$ $\left[\because \tan \left(x\right)=\frac{\sin \left(x\right)}{\cos \left(x\right)},cot\left(x\right)=\frac{\cos \left(x\right)}{\sin \left(x\right)}\right]$

$\Rightarrow$${\log }_{\cos \left(x\right)}\sin \left(x\right)-{\log }_{\cos \left(x\right)}\cos \left(x\right)+{\log }_{\sin \left(x\right)}\cos \left(x\right)-{\log }_{\sin \left(x\right)}\sin \left(x\right)=0$ $\left[\because {\log }_a\left(\frac{m}{n}\right)={\log }_{a}m-{\log }_{a}n\right]$

$\Rightarrow$ ${\log }_{\cos \left(x\right)}\sin \left(x\right)+{\log }_{\sin \left(x\right)}\cos \left(x\right)-1-1=0$ $\left[\because {\log }_{a}a=1\right]$

$\Rightarrow$ ${\log }_{\cos \left(x\right)}\sin \left(x\right)+{\log }_{\sin \left(x\right)}\cos \left(x\right)=2$ $\left[\because {\log }_{a}b=\frac{\log b}{\log a}\right]$

$\Rightarrow$ $\frac{\log \sin \left(x\right)}{\log \cos \left(x\right)}+\frac{\log \cos \left(x\right)}{\log \sin \left(x\right)}=2$

$\Rightarrow$ ${\left(\log \sin \left(x\right)\right)}^2+{\left(\log \cos \left(x\right)\right)}^2=2\left(\log \cos \left(x\right)\right)\left(\log \sin \left(x\right)\right)$

$\Rightarrow$ ${\left(\log \sin \left(x\right)\right)}^2-2\left(\log \cos \left(x\right)\right)\left(\log \sin \left(x\right)\right)+{\left(\log \cos \left(x\right)\right)}^2=0$

$\Rightarrow$ ${\left(\left(\log \sin \left(x\right)\right)-\left(\log \cos \left(x\right)\right)\right)}^2=0$

$\Rightarrow$ $\left(\left(\log \sin \left(x\right)\right)-\left(\log \cos \left(x\right)\right)\right)=0$

$\Rightarrow$ $\log \sin \left(x\right)-\log \cos \left(x\right)=0$ $\left[\because {\log }_{a}m-{\log }_{a}n={\log }_a\left(\frac{m}{n}\right)\right]$

$\Rightarrow$ $\log \left(\frac{\sin \left(x\right)}{\cos \left(x\right)}\right)=0$

$\Rightarrow$ $\log \tan \left(x\right)=0$

Since no base is given we assume base to be $10$

$\Rightarrow$$\tan \left(x\right)={10}^0$ $...\left[\because {\log }_{a}b=x\Rightarrow b=a^x\right]$

$\Rightarrow$$\tan \left(x\right)=1$

The general solution of $\tan \left(x\right)=1$is $n\mathrm{\pi }+\frac{\mathrm{\pi }}{4}$ but, for odd values of $n$, $\cos \left(x\right),\sin \left(x\right)<0$

But $\cos \left(x\right),\sin \left(x\right)$ should not be negative as they are bases in the given logarithm

Therefore, the general solution of $\tan \left(x\right)=1$ is $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}}$.

Hence, option(B) i.e. $2n\mathrm{\pi }+\frac{\mathrm{\pi }}{4},n\in \mathrm{\mathbb{Z}}$ is the correct answer.