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Prove the following:$\frac{tanA}{1-cotA} +\frac{cotA}{1-tanA} =1+tanA+cotA$
Given: $\frac{tanA}{1-cotA} +\frac{cotA}{1-tanA} =1+tanA+cotA$
To do: To prove L.H.S.$=$R.H.S.
Solution:
L.H.S.$=\frac{tanA}{1-cotA} +\frac{cotA}{1-tanA}$
$=\frac{tanA}{1-\frac{1}{tanA}} +\frac{cotA}{1-tanA}$ $( As\ known\ cotA=\frac{1}{tanA})$
$=\frac{-tan^{2} A}{1-tanA} +\frac{cotA}{1-tanA}$
$=\frac{1}{1-tanA}\left( -tan^{2} A+\frac{1}{tanA}\right)$
$=\frac{1-tan^{3} A}{tanA( 1-tanA)}$
$=\frac{( 1-tanA)\left( 1+tan^{2} A+tanA\right)}{tanA( 1-tanA)}$
$=\frac{\left( 1+tan^{2} A+tanA\right)}{tanA}$
$=\frac{1}{tanA} +\frac{tan^{2} A}{tanA} +\frac{tanA}{tanA}$
$=cotA+tanA+1$ $( \because cotA=\frac{1}{tanA} )$
$=$R.H.S.
Hence proved.
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