Question 14

Find the value of:

sin⁡ (2 tan -1 ⁡ 1/4) + cos ⁡(tan -1 ⁡ 2√2)

Find the value of:  sin⁡ (2 tan^-1⁡ 1/4) + cos ⁡(tan^-1⁡ 2√2)

Question 14 - CBSE Class 12 Sample Paper for 2019 Boards - Part 2
Question 14 - CBSE Class 12 Sample Paper for 2019 Boards - Part 3 Question 14 - CBSE Class 12 Sample Paper for 2019 Boards - Part 4 Question 14 - CBSE Class 12 Sample Paper for 2019 Boards - Part 5 Question 14 - CBSE Class 12 Sample Paper for 2019 Boards - Part 6

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Transcript

Question 14 Find the value of: sin⁡(2 tan^(−1)⁡〖1/4〗 ) + cos⁡(tan^(−1)⁡〖2√2〗 ) We solve sin⁡(2 tan^(−1)⁡〖1/4〗 ) & cos⁡(tan^(−1)⁡〖2√2〗 ) separately Solving 𝒔𝒊𝒏⁡(𝟐 〖𝒕𝒂𝒏〗^(−𝟏)⁡〖𝟏/𝟒〗 ) Let 〖𝑡𝑎𝑛〗^(−1)⁡〖1/4〗=𝜃 1/4 = tan θ tan θ = 1/4 Now, 𝒔𝒊𝒏⁡(𝟐 〖𝒕𝒂𝒏〗^(−𝟏)⁡〖𝟏/𝟒〗 ) = sin 2θ Writing sin 2θ in terms of tan θ = (2 tan⁡𝜃)/(1 + tan^2⁡𝜃 ) = (2 × 1/4)/(1 +(1/4)^2 ) = (1/2)/(1 + 1/16) = (1/2)/((16 + 1)/16) = (1/2)/(17/16) = 1/2×16/17 = 8/17 ∴ 𝑠𝑖𝑛⁡(2 〖𝑡𝑎𝑛〗^(−1)⁡〖1/4〗 ) = 8/17 Now, Solving 𝒄𝒐𝒔⁡(〖𝒕𝒂𝒏〗^(−𝟏)⁡〖𝟐√𝟐〗 ) Let tan^(−1)⁡〖2√2〗 = ϕ ∴ tan ϕ = 2√2 And, we need to find 𝑐𝑜𝑠⁡(〖𝑡𝑎𝑛〗^(−1)⁡〖2√2〗 ) = cos ϕ We know that 1 + tan2 ϕ = sec2 ϕ 1 + (2√2)^2 = sec2 ϕ 1 + 〖2^2 (√2)〗^2 = sec2 ϕ 1 + 4 × 2 = sec2 ϕ 1 + 8 = sec2 ϕ 9 = sec2 ϕ sec2 ϕ = 9 1/cos^2⁡𝜙 = 9 And, we need to find 𝑐𝑜𝑠⁡(〖𝑡𝑎𝑛〗^(−1)⁡〖2√2〗 ) = cos ϕ We know that 1 + tan2 ϕ = sec2 ϕ 1 + (2√2)^2 = sec2 ϕ 1 + 〖2^2 (√2)〗^2 = sec2 ϕ 1 + 4 × 2 = sec2 ϕ 1 + 8 = sec2 ϕ 9 = sec2 ϕ sec2 ϕ = 9 1/cos^2⁡𝜙 = 9 1/cos^2⁡𝜙 = 9 cos2 ϕ = 1/9 cos ϕ = ± √(1/9) cos ϕ = ± 1/3 Since value of tan ϕ is positive, we take positive value of cos ϕ ∴ cos ϕ = 1/3 So, 𝑐𝑜𝑠⁡(〖𝑡𝑎𝑛〗^(−1)⁡〖2√2〗 ) = 1/3 Now, sin⁡(2 tan^(−1)⁡〖1/4〗 ) + cos⁡(tan^(−1)⁡〖2√2〗 ) = 8/17 + 1/3 = (8 × 3 + 1 × 17)/(17 × 3) = (24 + 17)/51 = 𝟒𝟏/𝟓𝟏

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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.