Mark the tick against the correct answer in the following:
$\cos \left(2 \tan ^{-1} \frac{1}{2}\right)=?$
A. $\frac{3}{5}$
B. $\frac{4}{5}$
C. $\frac{7}{8}$
D. none of these
To Find: The value of $\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$
Let, $x=\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$
$\Rightarrow x=\cos \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}\right)$
Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\Rightarrow \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{\frac{1}{2}+\frac{1}{2}}{\left.1-\frac{1}{2} \times \frac{1}{2}\right)}\right)=\tan ^{-1} \frac{4}{3}$
$\Rightarrow x=\cos \left(\tan ^{-1} \frac{4}{3}\right)$
Now, let $y=\tan ^{-1} \frac{4}{3}$
$\Rightarrow \tan \mathrm{y}=\frac{4}{3}=\frac{\text { opposite side }}{\text { adjacent side }}$
By pythagorus theroem,
(Hypotenuse ) $^{2}=$ (opposite side $)^{2}+(\text { adjacent side })^{2}$
Therefore, Hypotenuse $=5$
$\Rightarrow \cos \left(\tan ^{-1} \frac{4}{3}\right)=\cos y=\frac{3}{5}$
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