(tan³ θ - 1) / (tan θ - 1) = sec² θ + tan θ

As, a 3 - b 3 = (a - b)(a 2 + ab + b 2 ) ∴ tan 3 θ - (1) 3 = (tan θ - 1)(tan 2 θ + tan θ + 1) The L.H.S. of the equation can be written as, Since, L.H.S. = R.H.S. hence, proved that = sec 2 θ + tan θ.

Mathematics

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
$\dfrac{\text{tan}^3 \text{ θ} - 1}{\text{tan θ} - 1} = \text{sec}^2 \text{ θ} + \text{tan θ}.$

Trigonometric Identities

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Answer

As, a³ - b³ = (a - b)(a² + ab + b²)

∴ tan³ θ - (1)³ = (tan θ - 1)(tan² θ + tan θ + 1)

The L.H.S. of the equation can be written as,

$\Rightarrow \dfrac{\text{(tan θ - 1)}(\text{tan}^2 \text{ θ} + \text{tan θ + 1})}{\text{tan θ - 1}} \\[1em] \Rightarrow \text{tan}^2 \text{ θ} + \text{tan θ + 1} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} - 1 + 1 + \text{tan θ} \\[1em] \Rightarrow \text{sec}^2 \text{ θ} + \text{tan θ}$

Since, L.H.S. = R.H.S. hence, proved that $\dfrac{\text{tan}^3 \text{ θ} - 1}{\text{tan θ - 1}}$ = sec² θ + tan θ.

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Mathematics

Prove the following identities, where the angles involved are acute angles for which the trigonometric ratios are defined:
$\dfrac{\text{tan}^3 \text{ θ} - 1}{\text{tan θ} - 1} = \text{sec}^2 \text{ θ} + \text{tan θ}.$

Trigonometric Identities

Answer

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Trigonometric Identities

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