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Question 30
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If sec θ + tan θ = p, then find the value of cosec θ

Last updated at Oct. 1, 2019 by Teachoo

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**
Question 30
**

If sec θ + tan θ = p, then find the value of cosec θ

Transcript

Question 30 If sec θ + tan θ = p, then find the value of cosec θ Given sec θ + tan θ = p Converting sec θ and tan θ in sin θ and cos θ 1/cos𝜃 + sin𝜃/cos𝜃 = p (1 + sin𝜃)/cos𝜃 = p We need to find cosec θ, To find cosec θ , we convert everything into sin θ So, we square both sides ((1 + sin𝜃)/cos𝜃 )^2 = p2 (1^2 + sin^2𝜃 + 2 sin𝜃)/cos^2𝜃 = p2 1 + sin2 θ + 2 sin θ = p2 × cos2 θ Now, sin2 θ + cos2 θ = 1 cos2 θ = 1 – sin2 θ 1 + sin2 θ + 2 sin θ = p2 × (1 – sin2 θ) 1 + sin2 θ + 2 sin θ = p2 – p2 sin2 θ Let x = sin θ 1 + x2 + 2x = p2 – p2 x2 x2 + p2 x2 + 2x + 1 – p2 = 0 (1 + p2) x2 + 2x + (1 – p2) = 0 Comparing with ax2 + bx + c a = (1 + p2) , b = 2, c = (1 – p2) Now, x = (−𝑏 ± √(𝑏^2 − 4𝑎𝑐))/2𝑎 x = (−2 ± √(2^2 − 4 × (1 + 𝑝^2 ) × (1 − 𝑝^2 ) ))/2(1 + 𝑝^2 ) x = (−2 ± √(4 − 4 × (1 + 𝑝^2 ) × (1 − 𝑝^2 ) ))/2(1 + 𝑝^2 ) x = (−2 ± √4 √(1 − (1 + 𝑝^2 ) × (1 − 𝑝^2 ) ))/2(1 + 𝑝^2 ) x = (−2 ± 2 √(1 − (1 + 𝑝^2 ) × (1 − 𝑝^2 ) ))/2(1 + 𝑝^2 ) x = (−1 ± √(1 − (1 + 𝑝^2 ) × (1 − 𝑝^2 ) ))/((1 + 𝑝^2 ) ) x = (−1 ± √(1 − (1^2 −(𝑝^2 )^2 ) ))/((1 + 𝑝^2 ) ) x = (−1 ± √(1 − (1 − 𝑝^4 ) ))/((1 + 𝑝^2 ) ) x = (−1 ± √(1 − 1 + 𝑝^4 ))/((1 + 𝑝^2 ) ) x = (−1 ± √(𝑝^4 ))/((1 + 𝑝^2 ) ) x = (−1 ± 𝑝^2)/((1 + 𝑝^2 ) ) Putting x = sin θ sin θ = (−1 ± 𝑝^2)/((1 + 𝑝^2 ) ) For + sin θ = (−1 + 𝑝^2)/((1 + 𝑝^2 ) ) sin θ = (𝑝^2 − 1)/(𝑝^2 + 1) Therefore, cosec θ = 1/sin𝜃 = (𝒑^𝟐 + 𝟏)/(𝒑^𝟐 − 𝟏) For – sin θ = (−1 − 𝑝^2)/((1 + 𝑝^2 ) ) sin θ = (−(1 + 𝑝^2))/((1 + 𝑝^2 ) ) sin θ = –1 Therefore, cosec θ = 1/sin𝜃 = 1/(−1) = –1

CBSE Class 10 Sample Paper for 2019 Boards

Paper Summary

Question 1

Question 2 (Or 1st)

Question 2 (Or 2nd)

Question 3 (Or 1st)

Question 3 (Or 2nd)

Question 4

Question 5

Question 6

Question 7 (Or 1st)

Question 7 (Or 2nd)

Question 8 (Or 1st)

Question 8 (Or 2nd)

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16 (Or 1st)

Question 16 (Or 2nd)

Question 17 (Or 1st)

Question 17 (Or 2nd)

Question 18

Question 19 (Or 1st)

Question 19 (Or 2nd)

Question 20

Question 21 (Or 1st)

Question 21 (Or 2nd)

Question 22

Question 23 (Or 1st)

Question 23 (Or 2nd)

Question 24

Question 25

Question 26

Question 27 (Or 1st)

Question 27 (Or 2nd)

Question 28 (Or 1st)

Question 28 (Or 2nd)

Question 29

Question 30 You are here

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.