JEE MAIN - Mathematics (2022 - 29th June Morning Shift - No. 19)
$$50\tan \left( {3{{\tan }^{ - 1}}\left( {{1 \over 2}} \right) + 2{{\cos }^{ - 1}}\left( {{1 \over {\sqrt 5 }}} \right)} \right) + 4\sqrt 2 \tan \left( {{1 \over 2}{{\tan }^{ - 1}}(2\sqrt 2 )} \right)$$ is equal to ____________.
Answer
29
Explanation
$50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right)$
$$
+4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right)
$$
$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$
$\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha$
Where $2 \alpha=\tan ^{-1} 2 \sqrt{2}$
$\Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=2 \sqrt{2} \quad$.. (i)
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+2 \tan \alpha-2 \sqrt{2}=0$
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+4 \tan \alpha-2 \tan \alpha-2 \sqrt{2}=0$
$\Rightarrow(2 \sqrt{2} \tan \alpha-2)(\tan \alpha-\sqrt{2})=0$
$\Rightarrow \tan \alpha=\sqrt{2}$ or $\frac{1}{\sqrt{2}}$
$\Rightarrow \tan \alpha=\frac{1}{\sqrt{2}}$
$(\tan \alpha=\sqrt{2}$ doesn't satisfy (i))
$\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29$
$\Rightarrow 50 \tan \left(\pi+\tan ^{-1}\left(\frac{1}{2}\right)\right)+4 \sqrt{2} \tan \left(\frac{1}{2} \tan ^{-1} 2 \sqrt{2}\right)$
$\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha$
Where $2 \alpha=\tan ^{-1} 2 \sqrt{2}$
$\Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=2 \sqrt{2} \quad$.. (i)
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+2 \tan \alpha-2 \sqrt{2}=0$
$\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+4 \tan \alpha-2 \tan \alpha-2 \sqrt{2}=0$
$\Rightarrow(2 \sqrt{2} \tan \alpha-2)(\tan \alpha-\sqrt{2})=0$
$\Rightarrow \tan \alpha=\sqrt{2}$ or $\frac{1}{\sqrt{2}}$
$\Rightarrow \tan \alpha=\frac{1}{\sqrt{2}}$
$(\tan \alpha=\sqrt{2}$ doesn't satisfy (i))
$\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29$
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