Question

${\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{3}{5}\right)=?$

• A. ${\tan }^{-1}\left(\frac{27}{11}\right)$
• B. ${\sin }^{-1}\left(\frac{11}{27}\right)$
• C. ${\cos }^{-1}\left(\frac{11}{27}\right)$
• D. None of these.

Answer 1

The correct option is A

${\tan }^{-1}\left(\frac{27}{11}\right)$

Explanation for correct option:

Simplify the given series using Trigonometric identities

$$\begin{gathered} {\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{3}{5}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{1-{\left({\displaystyle \frac{4}{5}}\right)}^2}}{{\displaystyle \frac{4}{5}}}\right)+{\tan }^{-1}\left(\frac{3}{5}\right);[\because {\cos }^{-1}\left(x\right)={\tan }^{-1}\left(\frac{\sqrt{1-x^2}}{x}\right)] \\ ={\tan }^{-1}\left(\frac{\sqrt{1-{\displaystyle \frac{16}{25}}}}{{\displaystyle \frac{4}{5}}}\right)+{\tan }^{-1}\left(\frac{3}{5}\right) \\ ={\tan }^{-1}\left(\frac{\sqrt{{\displaystyle \frac{9}{25}}}}{{\displaystyle \frac{4}{5}}}\right)+{\tan }^{-1}\left(\frac{3}{5}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}\right)+{\tan }^{-1}\left(\frac{3}{5}\right) \\ ={\tan }^{-1}\left(\frac{3}{5}\times \frac{5}{4}\right)+{\tan }^{-1}\left(\frac{3}{5}\right) \\ ={\tan }^{-1}\left(\frac{3}{4}\right)+{\tan }^{-1}\left(\frac{3}{5}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle \frac{3}{4}}+{\displaystyle \frac{3}{5}}}{1-{\displaystyle \frac{3}{4}}\times {\displaystyle \frac{3}{5}}}\right);[\because {\tan }^{-1}\left(x\right)+{\tan }^{-1}\left(y\right)={\tan }^{-1}\left(\frac{x+y}{1-xy}\right)] \\ ={\tan }^{-1}\left(\frac{{\displaystyle 15+12}}{20-9}\right) \\ ={\tan }^{-1}\left(\frac{{\displaystyle 27}}{11}\right) \end{gathered}$$

Therefore, Option(A) is the correct answer.