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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 |
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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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