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Example 4 In a right triangle ABC, right-angled at B, if tan A = 1, then verify that 2 sin A cos A = 1. In a right angle triangle ABC tan A = 1 (๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘ก๐‘œ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด)/(๐‘†๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘๐‘’๐‘›๐‘ก ๐‘ก๐‘œ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด) = 1 ๐ต๐ถ/๐ด๐ต = 1 AB = BC Let AB = BC = k Where k is a positive number. Finding AC by pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 AC2 = AB2 + BC2 Putting AB = BC = k AC2 = k2 + k2 AC2 = 2k2 AC = โˆš2๐‘˜2 AC = โˆš๐Ÿ "k" Now, cos A = (๐‘ ๐‘–๐‘‘๐‘’ ๐‘Ž๐‘‘๐‘—๐‘Ž๐‘›๐‘๐‘’๐‘›๐‘ก ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด)/๐ป๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’ cos A = ๐ด๐ต/๐ด๐ถ cos A = ๐‘˜/(๐‘˜โˆš2) cos A = ๐Ÿ/โˆš๐Ÿ sin A = (๐‘ ๐‘–๐‘‘๐‘’ ๐‘œ๐‘๐‘๐‘œ๐‘ ๐‘–๐‘ก๐‘’ ๐‘Ž๐‘›๐‘”๐‘™๐‘’ ๐ด)/๐ป๐‘ฆ๐‘๐‘œ๐‘ก๐‘’๐‘›๐‘ข๐‘ ๐‘’ sin A = ๐ต๐ถ/๐ด๐ถ sin A = ๐‘˜/(๐‘˜โˆš2) sin = ๐Ÿ/โˆš๐Ÿ We have to find 2 sin A cos A Substituting the value of sin A and cos A = 2 ร—1/โˆš2ร—1/โˆš2 = ๐Ÿ/(โˆš๐Ÿ ร— โˆš๐Ÿ) = 2/(โˆš2 )^2 = 2/2 = 1

  1. Chapter 8 Class 10 Introduction to Trignometry
  2. Serial order wise

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo