Ex 8.1, 2 - Chapter 8 Class 10 Introduction to Trignometry

Last updated at August 2, 2026 by

Ex 8.1, 2 - In fig, find tan P - cot R - Chapter 8 Class 10 - Teachoo - Ex 8.1

part 2 - Ex 8.1, 2 - Ex 8.1 - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry
part 3 - Ex 8.1, 2 - Ex 8.1 - Serial order wise - Chapter 8 Class 10 Introduction to Trignometry

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Ex 8.1, 2 In figure , find tan P – cot R. Finding sides of triangle In right βˆ† 𝑷𝑸𝑹, Using Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 PR2 = PQ2 + QR2 132 = 122 + QR2 169 = 144 + QR2 169 – 144 = QR2 25 = QR2 QR2 = 25 QR = βˆšπŸπŸ“ QR = √(5^2 ) QR = 5 Thus, QR = 5 cm Finding tan P tan P = (𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ βˆ π‘ƒ)/(𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œβˆ π‘ƒ) = 𝑄𝑅/𝑃𝑄 = πŸ“/𝟏𝟐 Finding cot R For cot R , Lets first find tan R tan R = (𝑠𝑖𝑑𝑒 π‘œπ‘π‘π‘œπ‘ π‘–π‘‘π‘’ π‘‘π‘œ βˆ π‘…)/(𝑠𝑖𝑑𝑒 π‘Žπ‘‘π‘—π‘Žπ‘π‘’π‘›π‘‘ π‘‘π‘œβˆ π‘…) = 𝑃𝑄/𝑄𝑅 = 𝟏𝟐/πŸ“ Hence, cot R = 1/tan⁑〖 𝑅〗 = πŸ“/𝟏𝟐 Now, tan P – cot R = 5/12βˆ’5/12 = 0

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