Question:
Mark the tick against the correct answer in the following:
If $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ then $x=?$
A. $\frac{1}{3}$
B. $\frac{1}{5}$
C. 3
D. 5
Solution:
Given: $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$
To Find: The value of $x$
Here $\tan ^{-1} x+\tan ^{-1} 3=\tan ^{-1} 8$ can be written as
$\tan ^{-1} x=\tan ^{-1} 8-\tan ^{-1} 3$
Since we know that $\tan ^{-1} x$ - $\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$
$\tan ^{-1} x=\tan ^{-1} 8-\tan ^{-1} 3=\tan ^{-1}\left(\frac{8-3}{1+(8 \times 3)}\right)$
$=\tan ^{-1}\left(\frac{5}{25}\right)$
$=\tan ^{-1}\left(\frac{1}{5}\right)$
$\Rightarrow x=\frac{1}{5}$
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