Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Ex 3.3

Ex 3.3, 1
Important

Ex 3.3, 2 Important

Ex 3.3, 3 Important

Ex 3.3, 4

Ex 3.3, 5 (i) Important

Ex 3.3, 5 (ii)

Ex 3.3, 6 Important

Ex 3.3, 7

Ex 3.3, 8 Important

Ex 3.3, 9 Important You are here

Ex 3.3, 10

Ex 3.3, 11 Important

Ex 3.3, 12

Ex 3.3, 13 Important

Ex 3.3, 14

Ex 3.3, 15

Ex 3.3, 16 Important

Ex 3.3, 17

Ex 3.3, 18 Important

Ex 3.3, 19

Ex 3.3, 20

Ex 3.3, 21 Important

Ex 3.3, 22 Important

Ex 3.3, 23 Important

Ex 3.3, 24

Ex 3.3, 25

Chapter 3 Class 11 Trigonometric Functions

Serial order wise

Last updated at June 22, 2023 by Teachoo

Ex 3.3, 9 Prove cos (3π/2+𝑥) cos (2π + 𝑥)[cot (3π/2−𝑥) + cot (2π + 𝑥)] =1 Solving L.H.S. Now, cos (𝟑𝝅/𝟐 "+ " 𝒙) = sin x cos (2π + x) = cos x cot (2π + x) = cot x cot (𝟑𝝅/𝟐−𝒙) = tan x Now putting values in equation cos (3π/2+𝑥) cos (2π + 𝑥)[cot (3π/2−𝑥) + cot (2π + 𝑥)] = (sin x) × (cos x) × [tan x + cot x] = (sin x cos x) × [cot x + tan x] = (sin x cos x) × [𝒄𝒐𝒔𝒙/𝒔𝒊𝒏𝒙 + 𝒔𝒊𝒏𝒙/𝒄𝒐𝒔𝒙 ] = (sin x cos x) × [(〖(cos〗𝑥) × 〖(cos〗𝑥)+〖 (sin〗𝑥) × 〖(sin〗𝑥))/(sin𝑥 × 〖(cos〗𝑥))] = (sin x cos x) × [(𝐜𝐨𝐬𝟐𝒙 +〖 𝐬𝐢𝐧𝟐〗𝒙)/(𝒔𝒊𝒏𝒙 × 〖(𝒄𝒐𝒔〗𝒙))] = cos2𝑥 +〖 sin2〗𝑥 = 1 = R.H.S Hence proved