Question

Si tgA + ctgA=m. Hallar el valor de sen 2A:

Answer 1

$m=\tan A+\cot A\\ \\ m= \dfrac{\sin A}{\cos A}+\dfrac{\cos A}{\sin A}\\ \\ m=\dfrac{\sin^2A+\cos^2A}{\sin A\cos A}\\ \\ m=\dfrac{1}{\sin A\cos A}\\ \\ m=\dfrac{2}{2\sin A\cos A}\\ \\ m=\dfrac{2}{\sin 2A}\\ \\ \\ \boxed{\boxed{\sin 2A =\dfrac{2}{m}}}$

Answer 2

$Hola~~ Cath \\ \\ Veamos: \\ \\ Si~~ tgA+ctgA=m \\ Resolucio'n \\ \frac{senA}{cosA} + \frac{cosA}{senA} =m~~~---\ \textgreater \ por~propiedad[ \frac{a}{b} + \frac{c}{d} = \frac{ad+cb}{bd} ] \\ \\ \frac{sen ^{2} A+cos ^{2} A}{cosA.senA} =m~~~----\ \textgreater \ se~sabe~que[ sen^{2}a+ cos^{2}a=1] \\ \\ \frac{1}{cosA.senA} =m$

$cosA.senA= \frac{1}{m} \\ Se~sabe~que~\boxed{\{cosB.senC= \frac{1}{2} [sen(B+C)-sen(B-C)]\}} \\ Con~esa~condicio'n~resolvemos: \\ \\ \frac{1}{2} [sen(A+A)-sen(A-A)]= \frac{1}{m} \\ \\ \frac{1}{2} [sen2A-sen0]= \frac{1}{m} ~~~---\ \textgreater \ se~sabe~que[sen0=0] \\ \\ \frac{1}{2} sen2A= \frac{1}{m} \\ \\ \boxed{\boxed{sen2A= \frac{2}{m} }}----\ \textgreater \ respuesta$

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