Ex 3.3, 9 - Chapter 3 Class 11 Trigonometric Functions

Last updated at April 16, 2024 by

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Transcript

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) × [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

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