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QJEEM22Ma318c: JEE Main 2022 Mathematics - Class 12 JEEM22Ma318c Solution

50tan⁡(3tan⁡−1(12)+2cos⁡−1(15))+42tan⁡(12tan⁡−1(22))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 ____________.

Step-by-Step Verified Solution

50tan⁡(tan⁡−112+2tan⁡−1(12)+2tan⁡−1(2))50 \tan \left(\tan ^{-1} \frac{1}{2}+2 \tan ^{-1}\left(\frac{1}{2}\right)+2 \tan ^{-1}(2)\right) +42tan⁡(tan⁡−12(22))+4 \sqrt{2} \tan \left(\frac{\tan ^{-1}}{2}(2 \sqrt{2})\right)

⇒50tan⁡(π+tan⁡−1(12))+42tan⁡(12tan⁡−122)\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)

⇒50(12)+42tan⁡α\Rightarrow \quad 50\left(\frac{1}{2}\right)+4 \sqrt{2} \tan \alpha

Where 2α=tan⁡−1222 \alpha=\tan ^{-1} 2 \sqrt{2}

⇒2tan⁡α1−tan⁡2α=22\Rightarrow \frac{2 \tan \alpha}{1-\tan ^{2} \alpha}=2 \sqrt{2} \quad .. (i)

⇒22tan⁡2α+2tan⁡α−22=0\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+2 \tan \alpha-2 \sqrt{2}=0

⇒22tan⁡2α+4tan⁡α−2tan⁡α−22=0\Rightarrow \quad 2 \sqrt{2} \tan ^{2} \alpha+4 \tan \alpha-2 \tan \alpha-2 \sqrt{2}=0

⇒(22tan⁡α−2)(tan⁡α−2)=0\Rightarrow(2 \sqrt{2} \tan \alpha-2)(\tan \alpha-\sqrt{2})=0

⇒tan⁡α=2\Rightarrow \tan \alpha=\sqrt{2} or 12\frac{1}{\sqrt{2}}

⇒tan⁡α=12\Rightarrow \tan \alpha=\frac{1}{\sqrt{2}}

(tan⁡α=2(\tan \alpha=\sqrt{2} doesn't satisfy (i))

⇒25+4212=29\Rightarrow \quad 25+4 \sqrt{2} \frac{1}{\sqrt{2}}=29

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