Question

${\cos }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}x\Rightarrow x=$

• A. $\frac{3}{65}$
• B. $\frac{-36}{65}$
• C. $\frac{-33}{65}$
• D. $-1$

Answer 1

Answer: (C)

$\frac{-33}{65}$

Given:

${\cos }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right)={\cos }^{-1}\left(x\right)$

Taking cosine on both sides,

$\cos ({\cos }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{3}{5}\right))=\cos \left({\cos }^{-1}\right(x\left)\right)$

Concepts used:

$\cos (A+B)=\cos \left(A\right)\cos \left(B\right)-\sin \left(A\right)\sin \left(B\right)$

${\cos }^{-1}\left(\frac{5}{13}\right)={\sin }^{-1}\left(\frac{12}{13}\right)$ and,

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

$x=\frac{5}{13}\times \frac{3}{5}-\frac{12}{13}\times \frac{4}{5}$

$x=\frac{15-48}{65}$

$x=\frac{-33}{65}$