Question 14
Find the value of:
sin (2 tan -1 1/4) + cos (tan -1 2√2)





Subscribe to our Youtube Channel - https://you.tube/teachoo
Last updated at Oct. 1, 2019 by Teachoo
Question 14
Find the value of:
sin (2 tan -1 1/4) + cos (tan -1 2√2)
Subscribe to our Youtube Channel - https://you.tube/teachoo
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 tan𝜃)/(1 + tan^2𝜃 ) = (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^2𝜙 = 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^2𝜙 = 9 1/cos^2𝜙 = 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
CBSE Class 12 Sample Paper for 2019 Boards
Question 1
Question 2
Question 3
Question 4 (Or 1st)
Question 4 (Or 2nd)
Question 5
Question 6
Question 7
Question 8 (Or 1st)
Question 8 (Or 2nd)
Question 9
Question 10 (Or 1st)
Question 10 (Or 2nd)
Question 11
Question 12 (Or 1st)
Question 12 (Or 2nd)
Question 13 (Or 1st)
Question 13 (Or 2nd)
Question 14 You are here
Question 15
Question 16 (Or 1st)
Question 16 (Or 2nd)
Question 17
Question 18
Question 19
Question 20
Question 21 (Or 1st)
Question 21 (Or 2nd)
Question 22
Question 23
Question 24 (Or 1st)
Question 24 (Or 2nd)
Question 25
Question 26 (Or 1st)
Question 26 (Or 2nd)
Question 27 (Or 1st)
Question 27 (Or 2nd)
Question 28
Question 29
About the Author