1057last wala term 111121 hoga ki 1121?
\hspace{-16}$Calculate value of expression\\\\\\ $\bf{\sin\left(\tan^{-1}\left(\frac{1}{3}\right)+\tan^{-1}\left(\frac{1}{5}\right)+\tan^{-1}\left(\frac{1}{7}\right)+\tan^{-1}\left(\frac{1}{11}\right)+\tan^{-1}\left(\frac{1}{13}\right)+\tan^{-1}\left(\frac{111}{121}\right)\right)}$
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262last term should have been tan-1 (101121)
because the sum of rest of the terms is tan-1(121101)
thus we get
sin ( tan-1x + tan-11/x) = sin π/2 =1
1708\hspace{-16}$Let $\bf{z=x+iy}$. Then $\bf{\theta = \tan^{-1}\left(\frac{y}{x}\right)\Leftrightarrow \arg(z)=\tan^{-1}\left(\frac{y}{x}\right)}$\\\\\\ Where $\bf{x\;,y>0}$\\\\\\ So Let $\bf{\arg(z_{1})=\tan^{-1}\left(\frac{1}{3}\right)\Leftrightarrow z_{1}=(3+i)}$\\\\\\ Similarly $\bf{z_{2}=(5+i)\;\;, z_{3}=(7+i)\;\;\;,}$\\\\\\ $\bf{z_{4}=(11+i)\;\;,z_{5}=(13+i)}$\\\\\\ So $\bf{\arg\left(z_{1}.z_{2}.z_{3}.z_{4}.z_{5}\right)=(3+i).(5+i).(7+i).(11+i).(13+i)=11100+12100.i}$\\\\\\ So $\bf{\arg(z)=\tan^{-1}\left(\frac{12100}{11100}\right)=\tan^{-1}\left(\frac{121}{111}\right)}$\\\\\\ So $\bf{\sin\left(\tan^{-1}\left(\frac{121}{111}\right)+\tan^{-1}\left(\frac{111}{121}\right)\right)=\sin\left(\frac{\pi}{2}\right)=1}$