Question

Show that ${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\pi$.

Answer 1

To show ${\sin }^{-1}\frac{12}{13}+{\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{63}{16}=\pi$

Let ${\sin }^{-1}\frac{12}{13}=x$, ${\cos }^{-1}\frac{4}{5}=y$ and ${\tan }^{-1}\frac{63}{15}=z$

$\Rightarrow \sin x=\frac{12}{13}\Rightarrow \cos y=\frac{4}{5}\Rightarrow \tan z=\frac{63}{15}\Rightarrow \sin x=\frac{12}{13}\frac{⟶P}{⟶H}\Rightarrow P^2+B^2=H^2\Rightarrow {12}^2+B^2={13}^2\Rightarrow B^2=169-144\Rightarrow B=5\Rightarrow \cos x=\frac{B}{H}=\frac{5}{13}$

Similarly

$\Rightarrow \sin y=\frac{3}{5}\Rightarrow \cos z=\frac{13}{5}\Rightarrow \sin z=\frac{12}{5}$

As we know $x+y+z=\pi$

Taking $\sin$ on both sides,

$\Rightarrow \sin (x+y+z)=\sin \pi \Rightarrow \sin (x+y)\cos z+\cos (x+y)\sin z=0\Rightarrow \sin x\cos y\cos z+\cos x\sin y\cos z+\cos x\cos y\sin z-\sin x\sin y\sin z=0$

Putting the values on LHS,

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

$=\frac{48}{25}+\frac{15}{25}+\frac{48}{65}-\frac{432}{325}=0$

Hence LHS=RHS