Question

If ${\cos }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}k$, then the value of $k$ is

• A. $\frac{16}{65}$
• B. $\frac{12}{65}$
• C. $\frac{11}{65}$
• D. $\frac{19}{65}$

Answer 1

Answer: (A)

We have,
${\cos }^{-1}\frac{3}{5}+{\cos }^{-1}\frac{12}{13}={\cos }^{-1}k$
$\Rightarrow {\cos }^{-1}\left\{\frac{3}{5}\cdot \frac{12}{13}-\sqrt{1-{\left(\frac{3}{5}\right)}^2\sqrt{1-{\left(\frac{12}{13}\right)}^2}}\right\}={\cos }^{-1}k$
$[\because {\cos }^{-1}x+{\cos }^{-1}y={\cos }^{-1}\left\{xy-\sqrt{1-x^2}\sqrt{1-y^2}\right\}$, if
$-1\le x,y\le 1$ and $x+y\ge 0]$
$\Rightarrow {\cos }^{-1}\left\{\frac{36}{65}-\frac{4}{5}\times \frac{5}{13}\right\}={\cos }^{-1}k$
$\Rightarrow {\cos }^{-1}\left\{\frac{36}{65}-\frac{20}{65}\right\}={\cos }^{-1}k$
$\Rightarrow {\cos }^{-1}\frac{16}{65}={\cos }^{-1}k$
$\Rightarrow k=\frac{16}{65}$