Solve the following Equation: $\frac{\tan (A + B)}{\cot (A - B)}$

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\frac{\tan^{2} A - \tan^{2} B}{1 - \tan^{2} A \tan^{2} B}

\frac{A - \tan^{2} B}{1 - \tan^{2} A \tan^{2} B}

\frac{A + \tan^{2} B}{1 + \tan^{2} A \tan^{2} B}

\frac{\tan^{2} A - \tan^{2} B}{1 + \tan^{2} A \tan^{2} B}

answer is A.

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Detailed Solution

Given equation is,

\frac{\tan (A + B)}{\cot (A - B)}

Using trigonometry identities, it can be written as,

\frac{\tan (A + B)}{\cot (A - B)} = \frac{\frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}}{\frac{\cot A \cdot \cot B + 1}{\cot B - \cot A}}

\frac{\tan (A + B)}{\cot (A - B)} = \frac{\frac{\tan A + \tan B}{1 - \tan A \cdot \tan B}}{\frac{1 + \tan A \cdot \tan B}{\tan A - \tan B}}

\frac{\tan (A + B)}{\cot (A - B)} = \frac{(\tan A + \tan B) (\tan A - \tan B)}{(1 - \tan A \cdot \tan B) (1 + \tan A \cdot \tan B)}

Using algebraic formula, it can be written as,

\frac{\tan (A + B)}{\cot (A - B)} = \frac{\tan^{2} A - \tan^{2} B}{1 - \tan^{2} A \cdot \tan^{2} B} .

Option 1 is Correct.

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