Question #02bba

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Question #02bba

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Answer 1 · 2016-11-02T05:00:11

Given $0 < u < \frac{π}{2}$ $0< u< pi/2$ and $tan u = \frac{5}{3}$ $tanu=5/3$

we get $u = {tan}^{- 1} (\frac{5}{3}) \approx 59^{\circ}$ $u=tan^-1(5/3)~~59^@$

So $2 u = 118^{\circ} \to In 2nd quadrant$ $2u=118^@->"In 2nd quadrant"$

Hence $sin 2 u \to +ve$ $sin2u->"+ve"$

But $tan 2 u and cos 2 u \to -ve$ $tan2u and cos 2u ->"-ve"$

Now

$sin 2 u = \frac{2 tan u}{1 + {tan}^{2} u}$ $sin2u=(2tanu)/(1+tan^2u)$

$\Rightarrow sin 2 u = \frac{2 \times \frac{5}{3}}{1 + {(\frac{5}{3})}^{2}} = \frac{10}{3} \times \frac{9}{34} = \frac{15}{17}$ $=>sin2u=(2xx5/3)/(1+(5/3)^2)=10/3xx9/34=15/17$

$cos 2 u = - \frac{1 - {tan}^{2} u}{1 + {tan}^{2} u}$ $cos2u=-(1-tan^2u)/(1+tan^2u)$

$\Rightarrow cos 2 u = \frac{1 - {(\frac{5}{3})}^{2}}{1 + {(\frac{5}{3})}^{2}}$ $=>cos2u=(1-(5/3)^2)/(1+(5/3)^2$

$\Rightarrow cos 2 u = - \frac{16}{9} \times \frac{9}{34} - \frac{8}{17}$ $=>cos2u=-16/9xx9/34-8/17$

$tan 2 u = \frac{2 tan u}{1 - {tan}^{2} u}$ $tan2u=(2tanu)/(1-tan^2u)$

$\Rightarrow tan 2 u = \frac{2 \times \frac{5}{3}}{1 - {(\frac{5}{3})}^{2}} = - \frac{10}{3} \times \frac{9}{16} = - \frac{15}{8}$ $=>tan2u=(2xx5/3)/(1-(5/3)^2)=-10/3xx9/16=-15/8$

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Question #02bba

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