Question

please solve this sum ​

Answer 1

$\large\underline{\underline{\tt\purple{Proof:-}}}$

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• Given:-

• $\sf \dfrac{tan^3 \theta - 1}{tan \theta - 1} = sec^2 \theta + tan \theta$

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• Solution:-

✯ $\large\underline{\underline{\bf\blue{L.H.S:-}}}$

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➪ $\sf \dfrac{tan^3 \theta - 1}{tan \theta - 1}$

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We know,

$\pink{\bigstar}$ $\underline{\boxed{\bf\green{a^3 - b^3 = (a-b)(a^2+ab+b^2)}}}$

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• Applying the above identity we get:-

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➪ $\sf \dfrac{(tan \theta - 1)(tan^2 \theta + tan \theta \times 1 + 1^2)}{(tan \theta - 1)}$

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➪ $\sf \dfrac{(tan \theta - 1)(tan^2 \theta + tan \theta + 1)}{(tan \theta - 1)}$

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• Cancelling out the similar terms, we get:-

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➪ $\sf tan^2 \theta + tan \theta + 1$

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We know,

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$\pink{\bigstar}$ $\underline{\boxed{\bf\green{tan^2 \theta + 1 = sec^2 \theta}}}$

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Hence,

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★ $\bf\red{sec^2 \theta + tan \theta}$

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✯ $\large\underline{\underline{\bf\blue{R.H.S:-}}}$

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★ $\bf\red{sec^2 \theta + tan \theta}$

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Since,

L.H.S = R.H.S

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★ HENCE PROVED ★

Answer 2

$\huge\mathrm{Answer}$

To proof :-

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

Prove :-

L.H.S ,

= tan³θ - 1 / tan³θ - 1

= ( tan θ - 1 )( tan² θ + tan θ + 1 ) / tan θ - 1

= tan² θ + tan θ + 1

= sec² θ + tan θ

= R.H.S ( Proved )

Notes ⭐

★ a³ - b³ = ( a - b )(a² + ab + b² )

★ tan² θ + 1 = sec² θ

L.H.S = Left hand side

R.H.S = Right hand side