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STD 12 - 2. inverse trigonometric function — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. inverse trigonometric function4 Marks
Question
$50 \tan \left(3 \tan ^{-1}\left(\frac{1}{2}\right)+2 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1}(2 \sqrt{2})\right)$ is equal to
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Answer

a
$50 \tan \left(3 \tan ^{-1} \frac{1}{2}+2 \cos ^{-1} \frac{1}{\sqrt{5}}\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$

$=50 \tan \left(\tan ^{-1} \frac{1}{2}+2\left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} 2\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$

$\left.=50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \cdot \frac{\pi}{2}\right)\right)+4 \sqrt{2} \times \frac{1}{\sqrt{2}}$

$=50\left(\tan \tan ^{-1} \frac{1}{2}\right)+4$

$=25+4=29$

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