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In the figure above, find tan \[p{\text{ }}-\]cot R.
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(A) 5
(B) \[\dfrac{1}{{13}}\]
(C) 0
(D) none of these

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Last updated date: 20th Jun 2024
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Answer
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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 \[90^\circ .\]
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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:
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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.