To find [tex]\(\tan(P - \varphi)\)[/tex], given [tex]\(\tan P = \left(\frac{3}{x-1}\right)^2\)[/tex] and [tex]\(\tan \varphi = \frac{2}{x+1}\)[/tex], we can use the tangent subtraction formula:
[tex]\[
\tan(P - \varphi) = \frac{\tan P - \tan \varphi}{1 + \tan P \cdot \tan \varphi}
\][/tex]
First, we are given:
[tex]\[
\tan P = \left(\frac{3}{x-1}\right)^2
\][/tex]
and
[tex]\[
\tan \varphi = \frac{2}{x+1}
\][/tex]
Now we should find [tex]\(\tan(P - \varphi)\)[/tex]. Plugging in the values into the tangent subtraction formula, we get:
[tex]\[
\tan(P - \varphi) = \frac{\left(\frac{9}{(x-1)^2}\right) - \left(\frac{2}{x+1}\right)}{1 + \left(\frac{9}{(x-1)^2}\right) \cdot \left(\frac{2}{x+1}\right)}
\][/tex]
Simplify the numerator:
[tex]\[
\tan(P - \varphi) = \frac{\frac{9}{(x-1)^2} - \frac{2}{x+1}}{1 + \frac{9 \cdot 2}{(x-1)^2 \cdot (x+1)}}
\][/tex]
This simplifies to:
[tex]\[
\tan(P - \varphi) = \frac{\frac{9}{(x-1)^2} - \frac{2}{x+1}}{1 + \frac{18}{(x-1)^2(x+1)}}
\][/tex]
To combine the terms in the numerator, we must find a common denominator, which would be [tex]\((x-1)^2(x+1)\)[/tex]:
[tex]\[
\tan(P - \varphi) = \frac{\left(\frac{9(x+1) - 2(x-1)^2}{(x-1)^2(x+1)}\right)}{1 + \frac{18}{(x-1)^2(x+1)}}
\][/tex]
Putting it together:
[tex]\[
\tan(P - \varphi) = \frac{\left(\frac{9(x+1) - 2(x-1)^2}{(x-1)^2(x+1)}\right)}{\frac{(x-1)^2(x+1) + 18}{(x-1)^2(x+1)}}
\][/tex]
Let's further simplify the numerator:
[tex]\[
9(x+1) - 2(x-1)^2 = 9x + 9 - 2(x^2 - 2x + 1) = 9x + 9 - 2x^2 + 4x - 2 = -2x^2 + 13x + 7
\][/tex]
Inserting back into place:
[tex]\[
\tan(P - \varphi) = \frac{-2x^2 + 13x + 7}{(x-1)^2(x+1) + 18}
\][/tex]
Returning the final expressions:
[tex]\[
\tan(P - \varphi) = \frac{\frac{9}{(x-1)^2} - \frac{2}{x+1}}{1 + \frac{18}{(x-1)^2(x+1)}}
\][/tex]
Therefore, the value of [tex]\(\tan(P - \varphi)\)[/tex] is:
[tex]\[
\tan(P - \varphi) = \frac{(-2/(x + 1) + 9/(x - 1)^2)}{1 + \frac{18}{((x - 1)^2(x + 1))}
\][/tex]
