Question 32 (OR 2nd question) - CBSE Class 10 Sample Paper for 2020 Boards - Maths Standard
Last updated at Oct. 9, 2019 by Teachoo

Evaluate:
(cos
^{
2
}
(45° + θ) + cos
^{
2
}
(45° - θ))/(tan (60° + θ) × tan(30° - θ) ) + (cot 30° + sin 90°) × (tan 60° − sec 0°)

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

Show More