"Find the value:tan⁡θ1−cot⁡θ+cot⁡θ1−tan⁡θ" was answered by our community

Aine Sellers

Answered question

2022-03-01

Find the value:

\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ}

ozebnikvpi

Beginner2022-03-02Added 5 answers

\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} = \frac{\tan θ}{1 - \frac{1}{\tan θ}} + \frac{\cot θ}{1 - \tan θ}

= \frac{- \tan^{2} θ}{1 - \tan θ} + \frac{\cot θ}{1 - \tan θ}

= \frac{\frac{1}{\tan θ} - \tan^{2} θ}{1 - \tan θ}

= \frac{1 - \tan^{3} θ}{\tan θ (1 - \tan θ)}

= \frac{(1 - \tan θ) (1 + \tan θ + \tan^{2} θ)}{\tan θ (1 - \tan θ)}

= \frac{1 + \tan θ + \tan^{2} θ}{\tan θ}

= 1 + \cot θ + \tan θ

Chance Mill

Beginner2022-03-03Added 8 answers

I'm not sure where you got stuck after writing the expression in terms of

\sin θ

and

\cos θ

but here's how I would do it:

\frac{\tan θ}{1 - \cot θ} + \frac{\cot θ}{1 - \tan θ} = \frac{\sin θ \tan θ}{\sin θ + \cos θ} + \frac{\cos θ \cot θ}{\cos θ - \sin θ}

= \frac{\sin θ \tan θ - \cos θ \cot θ}{\sin θ - \cos θ}

= \frac{\sin^{3} θ + \sin θ \cos θ + \cos^{2} θ}{\sin θ \cos θ}

= 1 + \tan θ + \cot θ

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