Question

If we have a logarithmic equation as ${\log _{\cos x}}\tan x + {\log _{\sin x}}\cot x = 0$, then the most general solution $x$ is
$A)2n\pi - \dfrac{{3\pi }}{4},n \in Z$
$B)2n\pi + \dfrac{\pi }{4},n \in Z$
$C)n\pi + \dfrac{\pi }{4},n \in Z$
$D)$ None of these

Answer 1

First. We need to know about the concepts of trigonometric functions and logarithm operations

The trigonometric functions are useful whenever trigonometric functions are involved in an expression or an equation and these identities are useful whenever expressions involving trigonometric functions need to be simplified.
$\log {x^m} = m\log x$ and since ${\log _{\cos x}}\tan x,{\log _{\sin x}}\cot x$ are composite functions and always positive
Formula used:
Using the logarithm law, ${\log _y}x = \dfrac{{\log x}}{{\log y}}$
$\dfrac{{\sin x}}{{\cos x}} = \tan x,\dfrac{{\cos x}}{{\sin x}} = \cot x$
$\log (\dfrac{x}{y}) = \log x - \log y$
${\tan ^{ - 1}}(1) = {\cot ^{ - 1}}(1) = 2n\pi + \dfrac{\pi }{4}$
${\log _{\cos x}}(\cos x),{\log _{\sin x}}(\sin x) = 1$ (because the logarithm base and values are the same and cancel each other with the operation ${\log _y}x = \dfrac{{\log x}}{{\log y}}$)

Complete step-by-step solution:
Since from the given that we have, ${\log _{\cos x}}\tan x + {\log _{\sin x}}\cot x = 0$ we need its generalized form of the variable x.
Now concert the given values using the logarithm operation, that is $\log (\dfrac{x}{y}) = \log x - \log y$ and $\dfrac{{\sin x}}{{\cos x}} = \tan x,\dfrac{{\cos x}}{{\sin x}} = \cot x$
Then we get ${\log _{\cos x}}\tan x + {\log _{\sin x}}\cot x = 0 \Rightarrow {\log _{\cos x}}\dfrac{{\sin x}}{{\cos x}} + {\log _{\sin x}}\dfrac{{\cos x}}{{\sin x}} = 0$
$\Rightarrow {\log _{\cos x}}\dfrac{{\sin x}}{{\cos x}} + {\log _{\sin x}}\dfrac{{\cos x}}{{\sin x}} = 0 \Rightarrow [{\log _{\cos x}}(\sin x) - {\log _{\cos x}}(\cos x)] + [{\log _{\sin x}}(\cos x) - {\log _{\sin x}}(\sin x)] = 0$, where $\log (\dfrac{x}{y}) = \log x - \log y$
Since we know that ${\log _{\cos x}}(\cos x),{\log _{\sin x}}(\sin x) = 1$, substituting the values in the above equation we get $[{\log _{\cos x}}(\sin x) - {\log _{\cos x}}(\cos x)] + [{\log _{\sin x}}(\cos x) - {\log _{\sin x}}(\sin x)] = 0 \Rightarrow [{\log _{\cos x}}(\sin x) - 1] + [{\log _{\sin x}}(\cos x) - 1] = 0$
Further solving we have $[{\log _{\cos x}}(\sin x) - 1] + [{\log _{\sin x}}(\cos x) - 1] = 0 \Rightarrow {\log _{\cos x}}(\sin x) + {\log _{\sin x}}(\cos x) = 2$
Since in the logarithm laws, we have ${\log _y}x = \dfrac{{\log x}}{{\log y}}$ and substituting these values we get ${\log _{\cos x}}(\sin x) + {\log _{\sin x}}(\cos x) = 2 \Rightarrow \dfrac{{\log \sin x}}{{\log \cos x}} + \dfrac{{\log \cos x}}{{\log \sin x}} = 2$
Now cross multiply we get $\dfrac{{\log \sin x}}{{\log \cos x}} + \dfrac{{\log \cos x}}{{\log \sin x}} = 2 \Rightarrow {\log ^2}\sin x + {\log ^2}\cos x = 2\log \cos x \log \sin x$
Turning the left values into the right side and applying the algebraic formula ${(a - b)^2} = {a^2} + {b^2} - 2ab$ then we get ${\log ^2}\sin x + {\log ^2}\cos x - 2\log \cos x\sin x = 0 \Rightarrow {(\log \sin x - \log \cos x)^2} = 0$
Taking root on both sides, $(\log \sin x - \log \cos x) = 0$ which also equals to $(\log \sin x - \log \cos x) = 0 \Rightarrow \log \sin x = \log \cos x$ then by division operation we have $\log \sin x = \log \cos x \Rightarrow \dfrac{{\log \sin x}}{{\log \cos x}} = 1,\dfrac{{\log \cos x}}{{\log \sin x}} = 1$
Thus, we get $\log \tan x = 1,\log \cot x = 1 \Rightarrow x = {\tan ^{ - 1}}(1) = {\cot ^{ - 1}}(1) = 2n\pi + \dfrac{\pi }{4}$
Therefore, the option $B)2n\pi + \dfrac{\pi }{4},n \in Z$ is correct.

Note: No matter on what side we divide, $\log \sin x = \log \cos x \Rightarrow \dfrac{{\log \sin x}}{{\log \cos x}} = 1,\dfrac{{\log \cos x}}{{\log \sin x}} = 1$ we will only get the same answer as above.
In the trigonometric function, the value of the inverse of tangent and cot with number one is the same.
The logarithm function we used $\log {x^m} = m\log x$ and logarithm derivative function can be represented as $\dfrac{d}{{dx}}\log x = \dfrac{1}{x}$
We will first understand what the logarithmic operator represents in mathematics. A logarithm function or log operator is used when we have to deal with the powers of a number, to understand it better which is $\log {x^m} = m\log x$