Question

Evaluate \[ \frac{\cot A+\tan B}{\cot B+\tan A}. \]

• A. \(\dfrac{\cot A}{\cot B}\)
• B. \(\dfrac{\cos A}{\sin B}\)
• C. \(\dfrac{\sin A}{\sin B}\)
• D. \(\cot A\cos B\)

Hint:

Whenever an expression contains both \(\tan\theta\) and \(\cot\theta\), replace \[ \boxed{\tan\theta=\frac1{\cot\theta}} \] or \[ \boxed{\cot\theta=\frac1{\tan\theta}} \] to simplify the expression.

Answer 1

The Correct Option is A

Concept: Use the identity \[ \boxed{\tan\theta=\frac{1}{\cot\theta}}. \]

Step 1: Rewrite the expression.
\[ \frac{\cot A+\tan B}{\cot B+\tan A} = \frac{\cot A+\frac1{\cot B}} {\cot B+\frac1{\cot A}}. \]

Step 2: Multiply numerator and denominator.
Multiply both numerator and denominator by \(\cot A\cot B\): \[ = \frac{\cot^2A\cot B+\cot A} {\cot^2B\cot A+\cot B}. \] Factor the numerator and denominator: \[ = \frac{\cot A(\cot A\cot B+1)} {\cot B(\cot A\cot B+1)}. \] Cancel the common factor \((\cot A\cot B+1)\): \[ = \frac{\cot A}{\cot B}. \]

Step 3: Final conclusion.
Hence, \[ \boxed{ \frac{\cot A+\tan B} {\cot B+\tan A} = \frac{\cot A}{\cot B} } \]