## If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$

[question] Question. If $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$, then $x=$ (a) 5 (b) 1/5 (c) 5/14 (d) 14/5 [/question] [solution] Solution: (b) $\frac{1}{5}$ We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1} \frac{x+y}{1-x y}$. Now, $\tan ^{-1} 3+\tan ^{-1} x=\tan ^{-1} 8$ $\Rightarrow \tan ^{-1}\left(\frac{3+x}{1-3 x}\right)=\tan ^{-1} 8$ $\Rightarrow \frac{3+x}{1-3 x}=8$ $\Rightarrow 3+x=8-24 x$ $\Rightarrow 3-8=-24 x-x$ $\Rightarrow-5=-25 x$ $\Rightarrow x=\frac{5}{25}=\frac{1}{5}$ [/soluti...

## $2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to

[question] Question. $2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}$ is equal to (a) $\cot ^{-1} x$ (b) $\cot ^{-1} \frac{1}{x}$ (c) $\tan ^{-1} x$ (d) none of these [/question] [solution] Solution: (c) $\tan ^{-1} x$ Let $\tan ^{-1} x=y$ So, $x=\tan y$ \$\therefore 2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \left(\cot ^{-1} x\right)\right\}=2 \tan ^{-1}\left\{\operatorname{cosec}\left(\tan ^{-1} x\right)-\tan \l...

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