Check sibling questions

Ex 3.3, 9 - Prove cos (3pi/2+x) cos (2pi + x)[cot (3pi/2 - x)

Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 2
Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 3 Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions - Part 4

Get live Maths 1-on-1 Classs - Class 6 to 12


Transcript

Ex 3.3, 9 Prove cos (3π/2+𝑥) cos (2π + 𝑥)[cot (3π/2−𝑥) + cot (2π + 𝑥)] =1 Taking L.H.S. First we solve cos (𝟑𝝅/𝟐 "+ " 𝒙) Putting π = 180° = cos ((3 × 180°)/2 " + x" ) = cos ( 270° + x) = cos ( 360° – 90° + x) = cos (2π – 90° + x) = cos (2π + (x – 90°)) = cos (x – 90°) = cos (– (90°– x)) = cos (90°– x) = sin x Now cos (2π + x) = cos x & cot (2π + x) = cot x Now we solve cot (𝟑𝝅/𝟐 "− " 𝒙) Putting π = 180° = cot ((3 × 180°)/2 " − x" ) = cot (270° "−" x) = cot (360° – 90° "−" x) = cot (2π "−" 90° "−" x) = cot (2π "−" (x + 90°)) = "−"cot (x + 90°) = –(–tan 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) × [cos⁡𝑥/sin⁡𝑥 + sin⁡𝑥/cos⁡𝑥 ] = (sin x cos x) × [(〖(cos〗⁡𝑥) × 〖(cos〗⁡𝑥)+〖 (sin〗⁡𝑥) × 〖(sin〗⁡𝑥))/(sin⁡𝑥 × 〖(cos〗⁡𝑥))] = (sin x cos x) × [(cos2⁡𝑥 +〖 sin2〗⁡𝑥)/(sin⁡𝑥 × 〖(cos〗⁡𝑥))] = cos2⁡𝑥 +〖 sin2〗⁡𝑥 = 1 = R.H.S Hence proved

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.