**
Question 14
**

Find the value of:

sin (2 tan
^{
-1
}
1/4) + cos (tan
^{
-1
}
2√2)

Last updated at Nov. 1, 2018 by Teachoo

**
Question 14
**

Find the value of:

sin (2 tan
^{
-1
}
1/4) + cos (tan
^{
-1
}
2√2)

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

Paper Summary

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

Class 12

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About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.