Question

Find the simplest form of ${\tan ^{ - 1}}\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right|$, if it is given that $\dfrac{3}{4}\tan x > - 1$.

Answer 1

We need to convert the expression given inside ${\tan ^{ - 1}}\left| {} \right|$ in the form of tangents of an angle (say $\theta$). Such a conversion leads to the form of ${\tan ^{ - 1}}\left( {\tan \left( \theta \right)} \right)$. Simplify further to get the required result.

Complete step-by-step answer:
We can convert the expression in ${\tan ^{ - 1}}\left( {\tan \left( \theta \right)} \right)$ form if we can express $\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right|$ in the form of $\tan \left( {a + b} \right)$ or $\tan \left( {a - b} \right)$.
We know that $\tan \left( {a + b} \right) = \dfrac{{\tan a + \tan b}}{{1 - \tan a\tan b}}$.
Also, we know that $\tan \left( {a - b} \right) = \dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$ .
There is difference of two terms in the numerator and sum of two terms in the denominator of $\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right|$.
It is similar to $\dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$.
Hence, let us try to express $\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right|$ as $\tan \left( {a - b} \right)$ where, $a - b = \theta$.
We need to express the denominator in the form of $1 + \tan a\tan b$.
First, let us divide both the numerator and the denominator by $4\cos x$.
Also, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
Hence, substitute $\tan x$ for $\dfrac{{\sin x}}{{\cos x}}$ in the calculation.
$\begin{array}{c}{\tan ^{ - 1}}\left| {\dfrac{{\dfrac{{3\cos x - 4\sin x}}{{4\cos x}}}}{{\dfrac{{4\cos x + 3\sin x}}{{4\cos x}}}}} \right| = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{{3\cos x}}{{4\cos x}} - \dfrac{{4\sin x}}{{4\cos x}}}}{{\dfrac{{4\cos x}}{{4\cos x}} + \dfrac{{3\sin x}}{{4\cos x}}}}} \right|\\\ = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \tan x}}{{1 + \dfrac{3}{4}\tan x}}} \right|\end{array}$
Let us assume that $\tan \phi = \dfrac{3}{4}$.
We also know that ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta$.
As we have to express the dfraction in the form $\dfrac{{\tan a - \tan b}}{{1 + \tan a\tan b}}$, let us substitute $\tan \phi$ for $\dfrac{3}{4}$ and use the property ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta$ to simplify the expression.
$\begin{array}{c}\tan \phi = \dfrac{3}{4}\\\\{\tan ^{ - 1}}\left( {\tan \phi } \right) = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)\\\\\phi = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)\end{array}$
From the above simplification, we get $\phi = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$.
After substitution, we get the following expression.
$\begin{array}{c}{\tan ^{ - 1}}\left| {\dfrac{{\tan \phi - \tan x}}{{1 + \tan \phi \tan x}}} \right| = {\tan ^{ - 1}}\left( {\tan \left( {\phi - x} \right)} \right)\\\ = \phi - x\end{array}$
Hence, we get $\theta = \phi - x$.
Now, we will back substitute ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)$ for $\phi$ and we will get the simplified form of the given expression.
${\tan ^{ - 1}}\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right| = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) - x$

Hence, the simplest form of ${\tan ^{ - 1}}\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right|$ is ${\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) - x$.

Note: We can also solve the question in a different way. We will repeat the first step.
Let us divide both the numerator and the denominator by $4\cos x$.
Also, we know that $\tan x = \dfrac{{\sin x}}{{\cos x}}$.
Hence, substitute $\tan x$ for $\dfrac{{\sin x}}{{\cos x}}$ in the calculation.
$\begin{array}{c}{\tan ^{ - 1}}\left| {\dfrac{{\dfrac{{3\cos x - 4\sin x}}{{4\cos x}}}}{{\dfrac{{4\cos x + 3\sin x}}{{4\cos x}}}}} \right| = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{{3\cos x}}{{4\cos x}} - \dfrac{{4\sin x}}{{4\cos x}}}}{{\dfrac{{4\cos x}}{{4\cos x}} + \dfrac{{3\sin x}}{{4\cos x}}}}} \right|\\\ = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \tan x}}{{1 + \dfrac{3}{4}\tan x}}} \right|\end{array}$

After that we get, ${\tan ^{ - 1}}\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right| = {\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \tan x}}{{1 + \dfrac{3}{4}\tan x}}} \right|$.
We will use the formula ${\tan ^{ - 1}}\left( {\dfrac{{a - b}}{{1 + ab}}} \right) = {\tan ^{ - 1}}a - {\tan ^{ - 1}}b$ to further simplify the expression.
Let us assume that $\dfrac{3}{4} = a$ and $\tan x = b$.
Let us substitute $\dfrac{3}{4}$ for $a$ and $\tan x$ for $b$ in the formula.
We will get, ${\tan ^{ - 1}}\left| {\dfrac{{\dfrac{3}{4} - \tan x}}{{1 + \dfrac{3}{4}\tan x}}} \right| = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) - {\tan ^{ - 1}}\left( {\tan x} \right)$.
As we know that ${\tan ^{ - 1}}\left( {\tan \theta } \right) = \theta$, let us substitute $x$ for ${\tan ^{ - 1}}\left( {\tan x} \right)$.
Hence, we get ${\tan ^{ - 1}}\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right| = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) - x$.