1.

If 4a = sec θ and 4/a = tanθ, then \(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right)\) is1). 1/162). 1/83). 1/24). 1/4

Answer»

⇒ 4a = sec θ

⇒ a = sec θ/4----1

⇒ 4/a = tanθ

⇒ 1/a = tanθ/ 4----2

⇒ We NEED to find the value of $(8\left( {{a^2} - \FRAC{1}{{{a^2}}}} \right))$

⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8(a – (1/a)) (a + (1/a))

⇒ Putting the value of a and 1/a in above we get

⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8 ((sec θ/4) - (tanθ/4)) × ((sec θ/4) + (tanθ/4))

⇒ $(8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 8/16 (sec2 θ - tan2 θ)

$(\THEREFORE \;8\left( {{a^2} - \frac{1}{{{a^2}}}} \right))$= 1/2


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