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

Home » Mark the tick against the correct answer in the following: A. B. C. D. none of these

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

CBSE | Class 12 | Inverse Trigonometric Functions | Maths | RS Aggarwal

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

Solution:

Option(A) is correct.
To Find: The value of $\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$
Let, $x=\cos \left(2 \tan ^{-1} \frac{1}{2}\right)$
$\Rightarrow \mathrm{x}=\cos \left(\tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}\right)$
Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{array}{l} \Rightarrow \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{2}=\tan ^{-1}\left(\frac{\frac{2}{2}+\frac{2}{2}}{1-\left(\frac{1}{2} x-\frac{1}{2}\right)}\right)=\tan ^{-1} \frac{4}{3} \\ \Rightarrow x=\cos \left(\tan ^{-1} \frac{4}{3}\right) \end{array}$
Now, let $y=\tan ^{-1} \frac{4}{3}$
$\Rightarrow \tan y=\frac{4}{3}=\frac{\text { opposite side }}{\text { adjacent side }}$
By pythagorus theroem,
$(\text { Hypotenuse })^{2}=(\text { opposite side })^{2}+(\text { adjacent side })^{2}$
Therefore, Hypotenuse $=5$
$\Rightarrow \cos \left(\tan ^{-1} \frac{4}{3}\right)=\cos y=\frac{3}{5}$