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**Hint:**To solve this question, we must have an idea about Pythagoras theorem for right-angled triangles.

Here the above picture is a right-angled triangle.

The sides of right-angled triangle named as base, perpendicular, and hypotenuse \[\left[ {longest{\text{ }}side} \right]\] The angle of opposite side of hypotenuse is \[90^\circ .\]

According to Pythagoras theorem,

\[{\left( {hypotenuse} \right)^2} = {\text{ }}{\left( {Perpendicular} \right)^2} + {\text{ }}{\left( {Base} \right)^2}\]

\[ \Rightarrow \]hypotenuse \[ = \sqrt {{{\left( {perpendicular} \right)}^2} + {{\left( {Base} \right)}^2}} \]

\[ \Rightarrow \] we know that it \[\theta \] is the angle of a right-angled triangle,

Then, \[\tan \theta = \dfrac{{perpendicular}}{{base}}\]

\[\cot \theta = \dfrac{1}{{\tan \theta }} = \dfrac{{base}}{{perpendicular}}\]

**Complete step by step answer:**

Given that

\[PQ{\text{ }} = {\text{ }}12{\text{ }}cm\]

\[PR{\text{ }} = {\text{ }}13{\text{ }}cm\]

\[\angle PQR = 90^\circ \]

At first, we have to find out \[QR\], the base of \[\Delta PQR\]

Here, \[PQ{\text{ }} = \] perpendicular to the triangle.

\[QR{\text{ }} = \] base of the triangle.

\[PR{\text{ }} = \] hypotenuse of the triangle.

By Pythagoras theorem,

\[ \Rightarrow {\left( {hypotenuse} \right)^2} = {\text{ }}{\left( {perpendicular} \right)^2} + {\text{ }}{\left( {Base} \right)^2}\]

\[ \Rightarrow \left( {P{R^2}} \right) = {\left( {PQ} \right)^2} + {\left( {QR} \right)^2}\]

\[ \Rightarrow {\left( {QR} \right)^2} = {\left( {PR} \right)^2} - {\left( {PQ} \right)^2}\]

\[ \Rightarrow QR = \sqrt {{{\left( {PR} \right)}^2} - {{\left( {PQ} \right)}^2}} \]

\[ = \sqrt {{{\left( {13} \right)}^2} - {{\left( {12} \right)}^2}} \] \[[{\text{ }}since,\;PR{\text{ }} = {\text{ }}13{\text{ }}cm\]

\[ \Rightarrow PQ{\text{ }} = {\text{ }}12{\text{ }}cm]\]

\[ = \sqrt {169 - 144} \]

\[ = \sqrt {25} = 5\]

\[\therefore \] base \[QR{\text{ }} = {\text{ }}5{\text{ }}cm.\]

Now, we calculate the value of tan p & cot R.

\[ \Rightarrow \tan P = \dfrac{{perpendicular}}{{Base}}\]

\[ = \dfrac{{QR}}{{PQ}} = \dfrac{5}{{12}}\]

[for angle P, the opposite side of angle is perpendicular \[\left( {QR} \right).\]Base is \[PQ\]. Opposite side of right angle is hypotenuse]

\[\cot R = \dfrac{{Base}}{{perpendicular}}\]

\[ = \dfrac{{QR}}{{PQ}} = \dfrac{5}{{12}}\] [for here angle R, the of angle opposite sides is perpendicular \[\left( {PR} \right)\] base is \[QR]\]

\[\therefore \,\,\tan \,P - \cot R = \dfrac{5}{{12}} - \dfrac{5}{{12}} = 0\]

**\[\therefore \] option (c) is right.**

**Note:**A right-angled triangle’s base is one of the sides that adjoins the 90-degree angle.

- The three main functions in trigonometry are sine, cosine, and tangent. It \[\theta \] is the angle of right-angled triangle then

\[ \Rightarrow \sin \theta = \dfrac{{perpendicular}}{{hypotenuse}},\cos \theta = \dfrac{{base}}{{hypotenuse}}\] when we solve this type of question, we need to remember all the formulas of trigonometry.

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