The equation having roots $tan(A+B)$ and $tan(A+B)$ is given by:
- ${x^2}\cos 2A - x\left( {2\sin 2A} \right)\cos 2B = 0$
- ${x^2}\left( {\cos 2A + \cos 2B} \right) - x\left( {2\sin 2A} \right) + \cos 2B - \cos 2A = 0$
- ${x^2}\left( {\cos 2A + \cos 2B} \right) - x\sin 2A + \cos 2A - \cos 2B = 0$
-
None of these
The question asks for the quadratic equation with roots tan(A+B) and tan(A-B) (the duplicate in the prompt is a typo in the source). Using sum and product of roots where sum = 2*sin(2A)/(cos(2A)+cos(2B)) and product format leads to option B.
The question asks for the quadratic equation with roots tan(A+B) and tan(A-B), assuming a typo in the provided text which lists both roots as tan(A+B). The required quadratic equation is formed using the sum and product of these roots, x^2 - (sum)x + (product) = 0. Using the tangent addition formulas, the sum tan(A+B) + tan(A-B) simplifies to sin(2A) / (cos(2A) + cos(2B)) and the product simplifies to (cos(2B) - cos(2A)) / (cos(2A) + cos(2B)). Substituting these into the quadratic form and clearing the denominator produces x^2(cos(2A) + cos(2B)) - x(2sin(2A)) + cos(2B) - cos(2A) = 0.