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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Trignometry
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If $A$ lies in the third quadrant and $3 \tan A-4=0$ then $5\sin 2A+3\sin A+4\cos A$ is equal to


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$3\tan A-4=0$
$3\tan A=4$
$\tan A=\large\frac{4}{3}$
$\cos A=-\large\frac{3}{5}$
$\sin A=-\large\frac{4}{5}$
$\sin 2A=\large\frac{2\tan A}{1+\tan^2A}$
$\therefore 5\sin 2A+3\sin A+4\cos A=5\times\large\frac{24}{25}$$+3\times\large\frac{-4}{5}+$$4\times\large\frac{-3}{5}$
Hence (a) is the correct answer.
answered Oct 8, 2013 by sreemathi.v

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