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

Hint: 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 \[{\tan ^{ - 1}}\left| {\dfrac{{3\cos x - 4\sin x}}{{4\cos x + 3\sin x}}} \right| = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) - x\].