# Prove that (tanA/(1-cotA))+(cotA/(1-tanA))=secA.cosecA+1?

Feb 25, 2018

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#### Explanation:

$\left(\frac{\tan A}{1 - \cot A}\right) + \left(\frac{\cot A}{1 - \tan A}\right)$

$= \left(\frac{\sin \frac{A}{\cos} A}{1 - \left(\cos \frac{A}{\sin} A\right)}\right) + \left(\frac{\cos \frac{A}{\sin} A}{1 - \left(\sin \frac{A}{\cos} A\right)}\right)$

$= \left(\frac{\sin \frac{A}{\cos} A}{\frac{\sin A - \cos A}{\sin} A} + \frac{\cos \frac{A}{\sin} A}{\frac{\cos A - \sin A}{\cos} A}\right)$

$= {\left(\sin A\right)}^{2} / \left(\cos A \left(\sin A - \cos A\right)\right) - {\left(\cos A\right)}^{2} / \left(\sin A \left(\sin A - \cos A\right)\right)$

$= \frac{1}{\sin A - \cos A} \cdot \left[{\left(\sin A\right)}^{2} / \cos A - {\left(\cos A\right)}^{2} / \sin A\right]$

$= \frac{1}{\sin A - \cos A} \cdot \left[\frac{{\left(\sin A\right)}^{3} - {\left(\cos A\right)}^{3}}{\sin A \cdot \cos A}\right]$

$= \frac{1}{\cancel{\left(\sin A - \cos A\right)}} \cdot \left[\frac{\cancel{\left(\sin A - \cos A\right)} \left({\left(\sin A\right)}^{2} + {\left(\cos A\right)}^{2} + \sin A \cdot \cos A\right)}{\sin A \cos A}\right]$

Replace ${\sin}^{2} A + {\cos}^{2} A = 1$

$= \frac{1 + \sin A \cdot \cos A}{\sin A \cdot \cos A}$

$= \cos e c A \cdot \sec A + 1$

hope you can get it!!

Feb 25, 2018

$L H S = \left(\tan \frac{A}{1 - \cot A}\right) + \left(\cot \frac{A}{1 - \tan A}\right)$

$= \left(\tan \frac{A}{1 - \cot A}\right) - \left({\cot}^{2} \frac{A}{- \cot A + \cot A \tan A}\right)$

$= \left(\tan \frac{A}{1 - \cot A}\right) - \left({\cot}^{2} \frac{A}{1 - \cot A}\right)$

$= \frac{1}{1 - \cot A} \cdot \left(\frac{1}{\cot} A - {\cot}^{2} A\right)$

$= \frac{\left(1 - \cot A\right) \left(1 + \cot A + {\cot}^{2} A\right)}{\left(1 - \cot A\right) \cot A}$

$= \frac{{\csc}^{2} A + \cot A}{\cot} A$

$= {\csc}^{2} \frac{A}{\cot} A + \cot \frac{A}{\cot} A$

$= \frac{1}{\sin} ^ 2 A \times \sin \frac{A}{\cos} A + 1$

$= \frac{1}{\sin A \times \cos A} + 1$

$= \sec A \cdot \cos e c A + 1 = R H S$

Feb 25, 2018

See the proof below

#### Explanation:

We need

$\tan A = \sin \frac{A}{\cos} A$

$\cot A = \cos \frac{A}{\sin} A$

${a}^{3} - {b}^{3} = \left(a - b\right) \left({a}^{2} + a b + {b}^{2}\right)$

$\sec A = \frac{1}{\cos} A$

$\csc A = \frac{1}{\sin} A$

Therefore,

$L H S = \tan \frac{A}{1 - \cot A} + \cot \frac{A}{1 - \tan A}$

$= \left(\frac{\sin \frac{A}{\cos} A}{1 - \cos \frac{A}{\sin} A}\right) + \left(\frac{\cos \frac{A}{\sin} A}{1 - \sin \frac{A}{\cos} A}\right)$

$= \left(\frac{\sin \frac{A}{\cos} A}{\frac{\sin A - \cos A}{\sin} A}\right) - \left(\frac{\cos \frac{A}{\sin} A}{\frac{\sin A - \cos A}{\cos} A}\right)$

$= \left(\frac{{\sin}^{2} A}{\cos A \left(\sin A - \cos A\right)}\right) - \left(\frac{{\cos}^{2} A}{\sin A \left(\sin A - \cos A\right)}\right)$

$= \frac{{\sin}^{3} A - {\cos}^{3} A}{\cos A \sin A \left(\sin A - \cos A\right)}$

$= \frac{\cancel{\sin A - \cos A} \left({\sin}^{2} A + \sin A \cos A + {\cos}^{2} A\right)}{\left(\cos A \sin A\right) \cancel{\sin A - \cos A}}$

$= \frac{1 + \sin A \cos A}{\cos A \sin A}$

$= \frac{1}{\cos A \sin A} + 1$

$= \sec A \csc A + 1$

$= R H S$

$Q E D$