Mark the tick against the correct answer in the following: $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=?$ A. $\frac{13}{6}$ B. $\frac{17}{6}$ C. $\frac{19}{6}$ D. $\frac{23}{6}$

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Mark the tick against the correct answer in the following: $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=?$
A. $\frac{13}{6}$
B. $\frac{17}{6}$
C. $\frac{19}{6}$
D. $\frac{23}{6}$

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

Mark the tick against the correct answer in the following: $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=?$
A. $\frac{13}{6}$
B. $\frac{17}{6}$
C. $\frac{19}{6}$
D. $\frac{23}{6}$

Solution:

Option(B) is correct.
To Find: The value of $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}$
Let $x=\cos ^{-1} \frac{4}{5}$
$\Rightarrow \cos x=\frac{4}{5}=\frac{\text { ad facent stde }}{\text { hypotenuse }}$
By pythagorus theroem,
(Hypotenuse ) $^{2}=$ (opposite side $)^{2}+$ (adjacent side ) $^{2}$
Therefore, opposite side $=3$
$\begin{array}{l} \Rightarrow \tan x=\frac{\text { oppositeside }}{\text { adjacent stde }}=\frac{3}{4} \\ \Rightarrow x=\tan ^{-1} \frac{3}{4} \end{array}$
Now $\tan \left\{\cos ^{-1} \frac{4}{5}+\tan ^{-1} \frac{2}{3}\right\}=\tan \left\{\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right\}$
Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\begin{array}{l} \tan \left\{\tan ^{-1} \frac{3}{4}+\tan ^{-1} \frac{2}{3}\right\}=\tan \left(\tan ^{-1}\left(\frac{\frac{3}{4}+\frac{2}{3}}{1-\left(\frac{3}{4} x^{3}\right)}\right)\right) \\ =\tan \left(\tan ^{-1}\left(\frac{17}{6}\right)\right) \\ =\frac{17}{6} \end{array}$