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

Last updated date: 20th Jun 2024
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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.