Question:
If tan^2Ï´=1-e^2 then show that secÏ´ tan^3Ï´.cosecÏ´ = (2-e^2)^3
Answer:
Given, tan2 θ = 1 - e2
=> tan θ = √(1 - e2 )
From the figure and Pythagorus theorem,
AC2 = AB2 + BC2
=> AC2 = {√(1 - e2 )}2 + 12
=> AC2 = 1 - e2 + 1
=> AC2 = 2 - e2
=> AC = √(2 - e2 )
Now, sec θ = √(2 - e2 )
cosec θ = √(2 - e2 )/√(1 - e2 )
and tan θ = √(1 - e2 )
Given, sec θ + tan3 θ * cosec θ
= √(2 - e2 ) + {(1 - e2 )3/2 * √(2 - e2 )/√(1 - e2 )}
= √(2 - e2 ) + {(1 - e2 ) * (1 - e2 ) * √(2 - e2 )/√(1 - e2 )}
= √(2 - e2 ) + (1 - e2 ) * √(2 - e2 )
= √(2 - e2 )*(1 + 1 - e2 )
= √(2 - e2 )*(2 - e2 )
= (2 - e2 )3/2
So, sec θ + tan3 θ * cosec θ = (2 - e2 )3/2
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