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{\text }{ If } 3 \tan A=2 \tan B  then what is the value of tan (A+B) ? 

Option: 1

\frac{5 \sin 2A}{5 \cos 2A-1}


Option: 2

\frac{ \sin 2A}{ \cos 2A-1}


Option: 3

\frac{5 \sin 2A}{5 \cos 2A-4}


Option: 4

\frac{\frac{5}{2} \sin 2A}{5 \cos 2A-1}


Answers (1)

best_answer

Sum/Difference into Product -

Sum/Difference into Product-

 

\\1.\;\;\sin \alpha+\sin \beta=2 \sin \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)\\\\2.\;\;\sin \alpha-\sin \beta=2 \sin \left(\frac{\alpha-\beta}{2}\right) \cos \left(\frac{\alpha+\beta}{2}\right)\\\\3.\;\;\cos \alpha-\cos \beta=-2 \sin \left(\frac{\alpha+\beta}{2}\right) \sin \left(\frac{\alpha-\beta}{2}\right)\\\\4.\;\;\cos \alpha+\cos \beta=2 \cos \left(\frac{\alpha+\beta}{2}\right) \cos \left(\frac{\alpha-\beta}{2}\right)

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Given\ 3 \tan A = 2\tan B\\ \tan (A+B)= \frac{\tan A+\tan B}{1-\tan A \tan B}\\ = \frac{\tan A+\frac{3}{2} \tan A}{1-\tan A\ \frac{3}{2}\tan A}\\ =\frac{ \frac{5}{2}\ \tan A}{1-\frac{3}{2} \tan^2 A}\\ =\frac{5 \sin A \cos A}{2 \cos^2 A-3\sin^2 A}\\ =\frac{\frac{5}{2} \sin 2A}{(1+ \cos 2A)- \frac{3}{2}(1-cos 2A)} \\ =\frac{5 \sin 2A}{5 \cos 2A-1}

Posted by

Ajit Kumar Dubey

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