Question 32 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard - Solutions of Sample Papers for Class 10 Boards

Transcript

Question 32 (OR 2nd question) Evaluate: (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + (cot 30° + sin 90°) × (tan 60° − sec 0°) (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + (cot 30° + sin 90°) × (tan 60° − sec 0°) = (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + (1/tan⁡〖30°〗 + sin 90°) × (tan 60° − 1/cos⁡〖0°〗 ) = (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + (1/(1/√3) + 1) × (√3 − 1/1) = (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + (√3 + 1) × (√3 − 1) = (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + ((√3)^2 – 12) = (cos^2⁡〖(45° + 𝜃)〗 + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 Writing cos θ = sin (90 – θ) = (sin^2⁡(90° − (45° + 𝜃)) + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 = (sin^2⁡(90° − 45° − 𝜃) + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 = (sin^2⁡(45° − 𝜃) + cos^2⁡〖(45° − 𝜃)〗)/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 Using sin2 x + cos2 x = 1 = 1/(〖tan 〗⁡(60° + 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 Writing tan θ = cot (90 – θ) = 1/(〖cot 〗⁡(90° − (60° + 𝜃)) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 = 1/(〖cot 〗⁡〖(30° − 𝜃)〗 × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 = 1/(1/tan⁡(30° − 𝜃) × 〖tan 〗⁡〖(30° − 𝜃)〗 ) + 2 = 1/1 + 2 = 1 + 2 = 3