We have an answer to "I have to prove the following result :tan⁡θ1−cot⁡θ+coθ1−tan⁡θ=1+sec⁡θ⋅csc⁡θ"

Rex Maxwell

Answered question

2022-03-30

I have to prove the following result :

\frac{\tan θ}{1 - \cot θ} + \frac{c o θ}{1 - \tan θ} = 1 + \sec θ \cdot \csc θ

Answer & Explanation

davane6a7a

Beginner2022-03-31Added 8 answers

Writing $\tan θ$ as $\frac{\sin θ}{\cos θ}$ and $\cot θ$ as $\frac{\cos θ}{\sin θ}$ , we get
$\frac{\frac{\sin θ}{\cos θ}}{1 - \frac{\cos θ}{\sin θ}} + \frac{\frac{\cos θ}{\sin θ}}{1 - \frac{\sin θ}{\cos θ}}$
$= \frac{\sin^{2} θ}{\cos θ \cdot (\sin θ - \cos θ)} + \frac{\cos^{2} θ}{\sin θ \cdot (\cos θ - \sin θ)}$
$= \frac{\sin^{2} θ}{\cos θ \cdot (\sin θ - \cos θ)} - \frac{\cos^{2} θ}{\sin θ \cdot (\sin θ - \cos θ)}$
$= \frac{1}{(\sin θ - \cos θ)} (\frac{\sin^{2} θ}{\cos θ} - \frac{\cos^{2} θ}{\sin θ})$
$= 1 \cdot \frac{1 + \sin θ \cdot \cos θ}{\sin θ \cos θ}$
which is
$1 + \sec θ \cdot \csc θ$

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