In Fig., Find \[tanP{\text{ }}-{\text{ }}cot{\text{ }}R.\]

Hint: We will have to apply Pythagoras theorem in triangle $PQR$ to get $QR$ and then we will apply the trigonometric formulas of $tan$ & $cot$ to get the required solution.

Complete step by step solution:
Given: In right angled triangle $PQR$, $PQ$ is perpendicular of length\[12{\text{ }}cm\], PR is Hypotenuse of length \[13{\text{ }}cm\]
To find: \[tanP{\text{ }}-{\text{ }}cot{\text{ }}R.\]
To find this we will have to apply Pythagoras theorem in triangle $PQR$, $\angle $PQR = 90$^\circ $
Pythagoras' Theorem states that, for any right-angled triangle, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two shorter sides.
By applying Pythagoras theorem in right angled $\vartriangle $ABC,
$ \Rightarrow (13)^2 = (12)^2+(QR)^2$
We know the formula of tan $\theta $ =$\dfrac{{perpendicular}}{{base}}$
So, in triangle PQR, QR is perpendicular & PQ is base with respect to P]
$ \Rightarrow $\[tan{\text{ }}P{\text{ }} = \] \[tan{\text{ }}P{\text{ }} = \]$\dfrac{{QR}}{{PQ}}$ = $\dfrac{5}{{12}}$
$ \Rightarrow $ $\dfrac{5}{{12}}$ $\theta $ = $\dfrac{{base}}{{perpendicular}}$
[In triangle $PQR$, $PQ$ is perpendicular & $QR$ is base]
$ \Rightarrow $ $cot R = \dfrac{{QR}}{{PQ}} = \dfrac{5}{{12}}$
So, the value of \[tanP{\text{ }}-{\text{ }}cot{\text{ }}R.\] will be
$tan P – cot R = \dfrac{5}{{12}}$ - $\dfrac{5}{{12}}$ = 0

$\therefore$The required value, \[tanP{\text{ }}-{\text{ }}cot{\text{ }}R=0.\]

Note:
Value of trigonometric functions ‘$\theta$’ should be taken with respect to angle concerned, so that perpendicular & base are taken correctly. Do the calculation to find the value of the ultimate answer carefully.