NCERT Solutions
Class 10 - Mathematics - Chapter 8: INTRODUCTION TO TRIGONOMETRY - Exercise 8.1

Question 2

From Fig. 8.13: \(\triangle PQR\) is right-angled at \(Q\). Given \(PQ = 12\) cm and \(PR = 13\) cm (the slant side).

To find: \(\tan P - \cot R\).

Step 1: Identify the hypotenuse

The hypotenuse is the side opposite the right angle. Since \(\angle Q = 90^\circ\), the hypotenuse is \(PR\). So, \(PR = 13\) cm.

Step 2: Find the third side \(QR\) using Pythagoras theorem

Pythagoras theorem: \(PR^2 = PQ^2 + QR^2\)

Substitute the values:

\(13^2 = 12^2 + QR^2\)

\(169 = 144 + QR^2\)

\(QR^2 = 169 - 144 = 25\)

\(QR = \sqrt{25} = 5\) cm.

Student Note: Here the triangle is a common \(5\text{-}12\text{-}13\) right triangle, so the missing side comes out neatly.

Step 3: Calculate \(\tan P\)

For angle \(P\):

Opposite side = \(QR = 5\) cm, Adjacent side = \(PQ = 12\) cm.

So, \(\tan P = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{5}{12}\).

Step 4: Calculate \(\cot R\)

For angle \(R\):

Opposite side = \(PQ = 12\) cm, Adjacent side = \(QR = 5\) cm.

So, \(\cot R = \dfrac{\text{adjacent}}{\text{opposite}} = \dfrac{5}{12}\).

Step 5: Find \(\tan P - \cot R\)

\(\tan P - \cot R = \dfrac{5}{12} - \dfrac{5}{12} = 0\).

Quick Check: In a right triangle, angles \(P\) and \(R\) are complementary, so \(\tan P = \cot R\). That’s why the difference becomes \(0\).