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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 Nov. 1, 2018 by Teachoo

**
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 …(1 We know that 1 + tan2 θ = sec2 θ 1 = sec2 θ – tan2 θ sec2 θ – tan2 θ = 1 Using a2 – b2 = (a + b) (a – b) (sec θ + tan θ) (sec θ – tan θ) = 1 Putting sec θ + tan θ = p p × (sec θ – tan θ) = 1 (sec θ – tan θ) = 1/𝑝 Now, our equations are sec θ + tan θ = p …(1) (sec θ – tan θ) = 1/𝑝 …(2) Adding (1) & (2) (sec θ + tan θ) + (sec θ – tan θ) = p + 1/𝑝 2 sec θ = p + 1/𝑝 2 sec θ = (𝑝^2 + 1)/𝑝 sec θ = (𝑝^2 + 1)/2𝑝 1/cos𝜃 = (𝑝^2 + 1)/2𝑝 Cross multiplying cos θ = 2𝑝/(𝑝^2 + 1) Now, to find cosec θ We first find sin θ We know that cos2 θ + sin2 θ = 1 Putting value of cos θ from (3) (2𝑝/(𝑝^2 + 1))^2 + sin2 θ = 1 (4𝑝^2)/(𝑝^2 + 1)^2 + sin2 θ = 1 sin2 θ = 1 – (4𝑝^2)/(𝑝^2 + 1)^2 sin2 θ = ((𝑝^2 + 1)^2 − 4𝑝^2)/(𝑝^2 + 1)^2 sin2 θ = ((𝑝^2 )^2 + 1^2 + 2𝑝^2 − 4𝑝^2)/(𝑝^2 + 1)^2 sin2 θ = ((𝑝^2 )^2 + 1^2 − 2𝑝^2)/(𝑝^2 + 1)^2 Using (a – b)2 = a2 + b2 – 2ab where a = p2 , b = 1 sin2 θ = (𝑝^2 − 1)^2/(𝑝^2 + 1)^2 sin2 θ = ((𝑝^2 − 1)/(𝑝^2 + 1))^2 Cancelling squares sin θ = ± (𝑝^2 − 1)/(𝑝^2 + 1) For + sin θ = (𝑝^2 − 1)/(𝑝^2 + 1) Therefore, cosec θ = 1/sin𝜃 = (𝒑^𝟐 + 𝟏)/(𝒑^𝟐 − 𝟏) For – sin θ = (−(𝑝^2 − 1))/(𝑝^2 + 1) Therefore, cosec θ = 1/sin𝜃 = (𝒑^𝟐 + 𝟏)/(𝒑^𝟐 − 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

Class 10

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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.