


Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Examples
Example 2 Important
Example 3
Example 4
Example 5 Important
Example 6 Important
Example 7 Important
Example 8
Example 9 Important
Example 10
Example 11 Important
Example 12
Example 13
Example 14
Example 15
Example 16 Important
Example 17 Important You are here
Example 18 Important
Example 19
Example 20 Important
Example 21 Important
Example 22 Important
Question 1 Deleted for CBSE Board 2024 Exams
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Last updated at May 29, 2023 by Teachoo
Example 17 Prove that sin〖5x − 〖2sin 3x +〗sinx 〗/𝑐𝑜𝑠〖5x − 𝑐𝑜𝑠x 〗 = tan x Taking L.H.S. sin〖5x + 〖sin x − 〗2sin3x 〗/𝑐𝑜𝑠〖5x − 𝑐𝑜𝑠x 〗 = 〖(sin〗〖5x + 〖sin x) − 〗〖2 sin〗3x 〗/𝑐𝑜𝑠〖5x − 𝑐𝑜𝑠x 〗 Solving numerator and denominator separately sin 5x + sin x = 2 sin ((5𝑥 + 𝑥)/2) cos ((5𝑥 − 𝑥)/2) = 2 sin (6𝑥/2) cos (4𝑥/2) = 2 sin 3x cos 2x sin 5x + sin x = 2 sin ((5𝑥 + 𝑥)/2) cos ((5𝑥 − 𝑥)/2) = 2 sin (6𝑥/2) cos (4𝑥/2) = 2 sin 3x cos 2x sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 5x & y = x sin x + sin y = 2 sin (𝑥 + 𝑦)/2 cos (𝑥 − 𝑦)/2 Putting x = 5x & y = x cos x – cos y = –2 sin (𝑥 + 𝑦)/2 sin (𝑥 − 𝑦)/2 Putting x = 5x & y = x Solving R.H.S 𝐬𝐢𝐧〖𝟓𝐱 + 〖𝐬𝐢𝐧 𝐱 − 〗2sin3x 〗/𝒄𝒐𝒔〖𝟓𝐱 − 𝒄𝒐𝒔𝐱 〗 Putting values = (2 sin3𝑥 cos2𝑥 − 2 sin3𝑥)/(−2 sin〖3𝑥 sin2𝑥 〗 ) = (2 sin3𝑥 (cos〖2𝑥 − 1)〗)/(−2 sin〖3𝑥 sin2𝑥 〗 ) = ( (cos〖2𝑥 − 1)〗)/(−sin2𝑥 ) = ( −(cos〖2𝑥 −1) 〗)/sin2𝑥 = (〖1 − 𝐜𝐨𝐬〗𝟐𝒙 )/𝒔𝒊𝒏𝟐𝒙 "Using cos 2x = 1 – 2sin2 x" "& sin 2x = 2 cos x sin x" = (1 − (𝟏 − 𝟐 𝐬𝐢𝐧𝟐𝒙 ) )/(𝟐 𝒄𝒐𝒔𝒙 𝒔𝒊𝒏𝒙 ) = (1 − 1 + 2 sin2𝑥)/(2 cos〖𝑥 〗 sin𝑥 ) = (0 + 2 sin2𝑥)/(2 cos〖𝑥 〗 sin𝑥 ) = (2 sin2𝑥)/(2 cos〖𝑥 〗 sin𝑥 ) = sin〖𝑥 〗/cos〖𝑥 〗 = tan x = R.H.S. Hence L.H.S. = R.H.S. Hence proved