cosecA + secA = cosecB + secB হলে tanA tanB=?

দেওয়া আছে, \(cosec A + sec A = cosec B + sec B\).

সুতরাং, \(\frac{1}{sin A} + \frac{1}{cos A} = \frac{1}{sin B} + \frac{1}{cos B}\)

বা, \(\frac{cos A + sin A}{sin A cos A} = \frac{cos B + sin B}{sin B cos B}\)

বা, \(\frac{cos A + sin A}{sin A cos A} - \frac{cos B + sin B}{sin B cos B} = 0\)

বা, \(\frac{(cos A + sin A)sin B cos B - (cos B + sin B)sin A cos A}{sin A cos A sin B cos B} = 0\)

অতএব, \((cos A + sin A)sin B cos B - (cos B + sin B)sin A cos A = 0\)

বা, \(cos A sin B cos B + sin A sin B cos B - cos B sin A cos A - sin B sin A cos A = 0\)

বা, \(cos A sin B cos B - cos B sin A cos A + sin A sin B cos B - sin B sin A cos A = 0\)

বা, \(cos A cos B (sin B - sin A) + sin A sin B (cos B - cos A) = 0\)

বা, \(cos A cos B (2 cos(\frac{A+B}{2}) sin(\frac{B-A}{2})) + sin A sin B (-2 sin(\frac{A+B}{2}) sin(\frac{B-A}{2})) = 0\)

বা, \(2 sin(\frac{B-A}{2}) [cos A cos B cos(\frac{A+B}{2}) - sin A sin B sin(\frac{A+B}{2})] = 0\)

যদি \(sin(\frac{B-A}{2}) = 0\) হয়, তবে \(\frac{B-A}{2} = 0\) অথবা \(B = A\). সেক্ষেত্রে tanA tanB = tan²A.

যদি \(sin(\frac{B-A}{2}) \neq 0\) হয়, তবে \(cos A cos B cos(\frac{A+B}{2}) - sin A sin B sin(\frac{A+B}{2}) = 0\)

বা, \(cos A cos B cos(\frac{A+B}{2}) = sin A sin B sin(\frac{A+B}{2})\)

বা, \(\frac{sin(\frac{A+B}{2})}{cos(\frac{A+B}{2})} = \frac{cos A cos B}{sin A sin B}\)

বা, \(tan(\frac{A+B}{2}) = \frac{1}{tan A tan B}\)

অতএব, \(tan A tan B = \frac{1}{tan(\frac{A+B}{2})}\)

সুতরাং, \(tan A tan B = cot(\frac{A+B}{2})\). 🎉