MCQ
$4tan^{-1} \frac{1}{5} -tan^{-1} \frac{1}{239}$ મેળવો.
- A$\pi$
- B$\frac{\pi}{2}$
- C$\frac{\pi}{3}$
- D$\frac{\pi}{4}$
$=2\left(2 \tan ^{-1} \frac{1}{5}\right)-\tan ^{-1} \frac{1}{239}$
$=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$
$=2 \tan ^{-1} \frac{2 / 5}{24 / 25}-\tan ^{-1} \frac{1}{239}$
$=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}$
$=2 \tan ^{-1} \frac{\frac{2}{5}}{1-\left(\frac{1}{5}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$
$=2 \tan ^{-1} \frac{2 / 5}{24 / 25}-\tan ^{-1} \frac{1}{239}$
$=2 \tan ^{-1} \frac{5}{12}-\tan ^{-1} \frac{1}{239}$
$=\tan ^{-1} \frac{2 \cdot \frac{5}{12}}{1-\left(\frac{5}{12}\right)^{2}}-\tan ^{-1} \frac{1}{239} \quad\left(\because 2 \tan ^{-1} x=\tan ^{-1} \frac{2 x}{1-x^{2}}\right)$
$=\tan ^{-1} \frac{144 \times 5}{119 \times 6}-\tan ^{-1} \frac{1}{239}$
$=\tan ^{-1} \frac{120}{119}-\tan ^{-1} \frac{1}{239}$
$=\tan ^{-1} \frac{\frac{120}{119}-\frac{1}{239}}{1+\frac{120}{119} \cdot \frac{1}{239}} \quad\left(\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1} \frac{x-y}{1+x y}\right)$
$=\tan ^{-1} \frac{120 \times 239-119}{119 \times 239+120}=\tan ^{-1} \frac{28680-119}{28441+120}$
$=\tan ^{-1} \frac{28561}{28561}=\tan ^{-1} 1=\frac{\pi}{4}$
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