if tan 2 a = 1 – e 2 then show that sec a + tan 3 a cosec a = (2-e 2)3/2

tan2a=1- e2 i.e. sec2a =2-e2seca + tan3a coseca = 1/cosa + sin2a/cos3a = 1/cosa + (1-cos2a)/cos3a= 1/cosa + 1/cos3a – 1/cosa = 1/cos3a = sec3a = (2-e2)3/2

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Grade 12Trigonometry

if tan ²a = 1 – e ²then show that sec a + tan ³a cosec a = (2-e ²)^3/2

Prateek Agrawal

11 Years agoGrade 12

2 Answers

Kaustubh Nayyar

11 Years ago

tan²a=1- e² i.e. sec²a =2-e²

seca + tan³a coseca = 1/cosa + sin²a/cos³a = 1/cosa + (1-cos²a)/cos³a

= 1/cosa + 1/cos³a – 1/cosa = 1/cos³a = sec³a = (2-e²)^3/2

Rishi Sharma

5 Years ago

Dear Student,
Please find below the solution to your problem.

tan²θ=1-a²
LHS
=secθ+tan³θcosecθ
=√(1+tan²θ)+tan²θ×tanθ×√(1+cot²θ)
[∵, sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]
=√1+(1-a²)+(1-a²)×√(1-a²)×√{1+(1/tan²θ)}
=√(2-a²)+(1-a²)×√(1-a²)×√{1+1/(1-a²)}
=√(2-a²)+(1-a²)×√(1-a²)×√{(1-a²+1)/(1-a²)}
=√(2-a²)+(1-a²)×√(2-a²)
=√(2-a²)×(1+1-a²)
=√(2-a²)×(2-a²)
=(2-a²)¹/²⁺¹
=(2-a²)³/²
=RHS (Proved)

Thanks and Regards