DavidCopperfield
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Solve this identity, where the angles involved are acute angles.

[tex] \dfrac{\tan\theta}{1 -\cot\theta } + \dfrac{\cot\theta}{1 -\tan \theta } = 1 + \sec \theta \cosec\theta[/tex]

Respuesta :

SmaгtieGurL

[tex]\therefore \sf{ LHS = \dfrac{tan \theta}{1 – cot \theta } + \dfrac{cot \theta}{1 – tan \theta}}[/tex]

[tex]\\[/tex]

[tex]\therefore\sf{ LHS = \dfrac{tan \theta}{1 – \dfrac{1}{tan \theta}} + \dfrac{ \dfrac{1}{tan \theta}}{1 – tan \theta} }[/tex]

[tex]\\[/tex]

[tex]\therefore\sf{ LHS = \dfrac{ tan^{2} \theta }{tan \theta – 1} + \dfrac{1}{tan \theta (1–tan \theta) } }[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \dfrac{tan^{2} \theta}{tan \theta–1} – \dfrac{1}{tan \theta(tan \theta – 1 )}}[/tex]

[tex]\\[/tex]

[tex]\therefore \sf{LHS = \dfrac{tan^{3}\theta–1}{tan \theta(tan \theta –1)}}[/tex]

[tex]\\[/tex]

[tex]\therefore\sf{LHS = \dfrac{tan^{3}\theta–(1)^{3}}{tan\theta(tan\theta–1)}}[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \dfrac{ (tan \theta -1)(tan^{2}\theta+tan\theta .1 + (1)^{2})} {tan \theta (tan\theta(tan \theta –1 )}}[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \dfrac{(tan^{2} \theta×tan\theta .1+1)}{tan \theta}}[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \dfrac{(sec^{2}\theta+tan\theta)}{tan\theta}}[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \dfrac{sec^{2}\theta}{tan\theta} + \dfrac{tan \theta}{tan\theta}}[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = sec^{2}\theta \times cot \theta + 1 }[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \dfrac{1}{cos^{2}\theta} \times \dfrac{cos \theta }{sin\theta}+1}[/tex]

[tex]\\[/tex]

[tex] \therefore\sf{ LHS = \left( \dfrac{ 1}{cos \theta} \times \dfrac{1}{sin \theta} \right) + 1}[/tex]

∴ LHS = 1 + secθ.cosecθ = RHS

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We have proven that the trigonometric identity [(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] equals 1 + (secθ * cosec θ)

How to solve Trigonometric Identities?

We want to prove the trigonometric identity;

[(tan θ)/(1 - cot θ)] + [(cot θ)/(1 - tan θ)] = 1 + sec θ

The left hand side can be expressed as;

[(tan θ)/(1 - (1/tan θ)] + [(1/tan θ)/(1 - tan θ)]

⇒ [tan²θ/(tanθ - 1)] - [1/(tan θ(tanθ - 1)]

Taking the LCM and multiplying gives;

(tan³θ - 1)/(tanθ(tanθ - 1))

This can also be expressed as;

(tan³θ - 1³)/(tanθ(tanθ - 1))

By expansion of algebra this gives;

[(tanθ - 1)(tan²θ + tanθ.1 + 1²)]/[tanθ(tanθ(tanθ - 1))]

Solving Further gives;

(sec²θ + tanθ)/tanθ

⇒ sec²θ * cotθ + 1

⇒ (1/cos²θ * cos θ/sin θ) + 1

⇒ (1/cos θ * 1/sin θ) + 1

⇒ 1 + (secθ * cosec θ)

Read more about Trigonometric Identities at; https://brainly.com/question/7331447

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