Question

The value of ${\sin }^{-1}\left(\frac{12}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)+{\tan }^{-1}\left(\frac{63}{16}\right)$

• A. $\frac{2\pi }{3}$
• B. $\pi$
• C. $2\pi$
• D. $3\pi$

Answer 1

Answer: (B)

$\pi$

${\sin }^{-1}\left[\frac{12}{13}\right]+{\cos }^{-1}\left[\frac{4}{5}\right]+{\tan }^{-1}\left[\frac{63}{16}\right]$

Let ${\sin }^{-1}\left(\frac{12}{13}\right)=\theta$

$\sin \theta =\left(\frac{12}{13}\right)$

$\cos \theta =\sqrt{1-{\left(\frac{12}{13}\right)}^2}=\sqrt{\frac{169-144}{169}}=\frac{5}{13}$

$\tan \theta =\frac{12}{5}$

Let $\cos \beta =\frac{4}{5}$

$\sin \beta =\sqrt{1-\frac{16}{25}}=\frac{3}{5}$

$\tan \beta =\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$

Now, ${\tan }^{-1}\frac{12}{5}+{\tan }^{-1}\frac{3}{4}+{\tan }^{-1}\frac{63}{16}$

${\tan }^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1-\frac{12}{5}\times \frac{3}{4}}\right)+{\tan }^{-1}\frac{63}{16}$

${\tan }^{-1}\left[\frac{48+15}{20-36}\right]+{\tan }^{-1}\left(\frac{63}{16}\right)$

$-{\tan }^{-1}\frac{63}{16}+{\tan }^{-1}\frac{63}{16}=0=\tan \pi$.