Question

Prove ${\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}={\tan }^{-1}\frac{63}{16}$

Answer 1

Let $a={\sin }^{-1}\frac{5}{13},b={\cos }^{-1}\frac{3}{5}$

$\Rightarrow \sin a=\frac{5}{13},\cos b=\frac{3}{5}$

We know that ${\cos }^{2}a=1-{\sin }^{2}a$

$\Rightarrow \cos a=\sqrt{1-{\sin }^{2}a}$

$=\sqrt{1-{\left(\frac{5}{13}\right)}^2}$

$=\sqrt{1-\frac{25}{169}}$

$=\sqrt{\frac{169-25}{169}}$

$=\sqrt{\frac{144}{169}}$

$=\frac{12}{13}$

We know that ${\sin }^{2}b=1-{\cos }^{2}b$

$\Rightarrow \sin b=\sqrt{1-{\cos }^{2}b}$

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

$=\sqrt{1-\frac{9}{25}}$

$=\sqrt{\frac{25-9}{25}}$

$=\sqrt{\frac{16}{25}}$

$=\frac{4}{5}$

Let $\tan a=\frac{\sin a}{\cos a}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{13}\times \frac{13}{12}=\frac{5}{12}$

Let $\tan b=\frac{\sin b}{\cos b}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{5}\times \frac{5}{3}=\frac{4}{3}$

Now, we know that $\tan (a+b)=\frac{\tan a+\tan b}{1-\tan a\tan b}$

$=\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times \frac{4}{3}}$

$=\frac{\frac{15+48}{36}}{1-\frac{20}{36}}$

$=\frac{\frac{63}{36}}{\frac{36-20}{36}}$

$=\frac{\frac{63}{36}}{\frac{16}{36}}$

$=\frac{63}{16}$

Hence,$\tan (a+b)=\frac{63}{16}$

$\Rightarrow a+b={\tan }^{-1}\left(\frac{63}{16}\right)$

Putting the values of $a$ and $b$

${\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5}={\tan }^{-1}\left(\frac{63}{16}\right)$

Hence L.H.S$=$R.H.S

Hence proved.