12) PT (1/Tan 3A+Tan A) - (1/Cot 3A+Cot A)=Cot 4A

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Solution

$LHS = \frac{1}{\tan 3 A + \tan A} - \frac{1}{\cot 3 A + \cot A} = \frac{1}{\frac{1}{\cot 3 A} + \frac{1}{\cot A}} - \frac{1}{\cot 3 A + \cot A} = \frac{\cot 3 A \cot A}{\cot A + \cot 3 A} - \frac{1}{\cot 3 A + \cot A} = \frac{\cot 3 A \cot A - 1}{\cot 3 A + \cot A} = \cot (3 A + A) = \cot 4 A = RHS Hence, proved .$

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