Question
Prove that $\frac{\tan\text{A}}{(1-\cot\text{A})}+\frac{\cot\text{A}}{(1-\tan\text{A})}=(1+\tan\text{A}+\cot\text{A}).$
✓
Answer
$\text{LHS}=\frac{\tan\text{A}}{(1-\cot\text{A})}+\frac{\cot\text{A}}{(1-\tan\text{A})}$$=\frac{\tan\text{A}}{(1-\cot\text{A})}+\frac{\cot^2\text{A}}{(\cot\text{A}-1)}$ $\Big[\because\tan\text{A}=\frac{1}{\cot\text{A}}\Big]$
$=\frac{\tan\text{A}}{(1-\cot\text{A})}-\frac{\cot^2\text{A}}{(\cot\text{A}-1)}$
$=\frac{\tan\text{A}-\cot^2\text{A}}{(1-\cot\text{A})}$
$=\frac{\big(\frac{1}{\cot\text{A}}\big)-\cot^2\text{A}}{(1-\cot\text{A})}$
$=\frac{1-\cot^3\text{A}}{\cot\text{A}(1-\cot\text{A})}$
$=\frac{(1-\cot\text{A})(1+\cot\text{A}+\cot^2\text{A})}{\cot\text{A}(1-\cot\text{A})}$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\big]$
$=\frac{1}{\cot\text{A}}+\frac{\cot^2\text{A}}{\cot\text{A}}+\frac{\cot\text{A}}{\cot\text{A}}$
$=1+\tan\text{A}+\cot\text{A}$
$=\text{RHS}$
Hence proved.
$=\frac{\tan\text{A}}{(1-\cot\text{A})}-\frac{\cot^2\text{A}}{(\cot\text{A}-1)}$
$=\frac{\tan\text{A}-\cot^2\text{A}}{(1-\cot\text{A})}$
$=\frac{\big(\frac{1}{\cot\text{A}}\big)-\cot^2\text{A}}{(1-\cot\text{A})}$
$=\frac{1-\cot^3\text{A}}{\cot\text{A}(1-\cot\text{A})}$
$=\frac{(1-\cot\text{A})(1+\cot\text{A}+\cot^2\text{A})}{\cot\text{A}(1-\cot\text{A})}$ $\big[\because\text{a}^3-\text{b}^3=(\text{a}-\text{b})\big(\text{a}^2+\text{ab}+\text{b}^2\big)\big]$
$=\frac{1}{\cot\text{A}}+\frac{\cot^2\text{A}}{\cot\text{A}}+\frac{\cot\text{A}}{\cot\text{A}}$
$=1+\tan\text{A}+\cot\text{A}$
$=\text{RHS}$
Hence proved.
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