Question

If ${cos}^{-1}\frac{3}{5}-{sin}^{-1}\frac{4}{5}={cos}^{-1}x$, then x is equal to -

• A. 0
• B. 1
• C. -1
• D. None of these

Answer 1

Answer: (D)

None of these

Given,

${\cos }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{4}{5}={\cos }^{-1}x$

apply $\cos$ on both sides, we get,

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

$Usethefollowingidentity:\cos (s+t)=\cos \left(s\right)\cos \left(t\right)-\sin \left(s\right)\sin \left(t\right)$

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

$\frac{3}{5}\left(\sqrt{1-{\left(\frac{4}{5}\right)}^2}\right)-\frac{4}{5}\left(\sqrt{1-{\left(\frac{3}{5}\right)}^2}\right)=x$

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

$\frac{9}{25}-\frac{16}{25}=x$

$\therefore x=-\frac{7}{25}$