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Parametric Differentiation

Prove that: ...

Question

Prove that: ${s e c}^{6} θ = {t a n}^{6} θ + {3 t a n}^{2} {θ s e c}^{2} θ + 1$

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Solution

We have, ${sec}^{6} θ = ({sec}^{2} θ)^{3} = (1 + {tan}^{2} θ)^{3}$

$= (1)^{3} + ({tan}^{2} θ)^{3} + 3 \times 1 \times {tan}^{2} θ (1 + {tan}^{2} θ)$

$= 1 + {tan}^{6} θ + 3 {tan}^{2} θ {sec}^{2} θ$

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