Question

Solve ${\cos }^{-1}\left(\frac{4}{5}\right)+{\cos }^{-1}\left(\frac{63}{65}\right)=$

• A. ${\cos }^{-1}\left(\frac{204}{325}\right)$
• B. ${\cos }^{-1}\left(\frac{300}{325}\right)$
• C. ${\cos }^{-1}\left(\frac{201}{300}\right)$
• D. ${\sin }^{-1}\left(\frac{204}{325}\right)$

Answer 1

Answer: (A)

${\cos }^{-1}\left(\frac{204}{325}\right)$

We have

$={\cos }^{-1}\left(\frac{4}{5}\right)+{\cos }^{-1}\left(\frac{63}{65}\right)$

We know that

${\cos }^{1}x+{\cos }^{-1}y={\cos }^{-1}[xy-\sqrt{1-x^2}\sqrt{1-y^2}]$

Therefore,

$={\cos }^{-1}(\frac{4}{5}\times \frac{63}{65}-\sqrt{1-{\left(\frac{4}{5}\right)}^2}\sqrt{1-{\left(\frac{63}{65}\right)}^2})$

$={\cos }^{-1}(\frac{252}{325}-\sqrt{1-\left(\frac{16}{25}\right)}\sqrt{1-\left(\frac{3969}{4225}\right)})$

$={\cos }^{-1}(\frac{252}{325}-\sqrt{\frac{9}{25}}\sqrt{\frac{256}{4225}})$

$={\cos }^{-1}(\frac{252}{325}-\frac{3}{5}\times \frac{16}{65})$

$={\cos }^{-1}(\frac{252}{325}-\frac{48}{325})$

$={\cos }^{-1}\left(\frac{204}{325}\right)$

Hence, this is the answer.