Question

Prove the following result:

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

Answer 1

Consider $\tan ({\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5})$

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

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

$\Rightarrow {\sin }^{2}a=\frac{25}{169},{\cos }^{2}b=\frac{9}{25}$

$\Rightarrow {\cos }^{2}a=1-{\sin }^{2}a=1-\frac{25}{169},{\sin }^{2}b=1-{\cos }^{2}b=1-\frac{9}{25}$

$\Rightarrow {\cos }^{2}a=\frac{169-25}{169},{\sin }^{2}b=\frac{25-9}{25}$

$\Rightarrow {\cos }^{2}a=\frac{144}{169},{\sin }^{2}b=\frac{16}{25}$

$\Rightarrow \cos a=\frac{12}{13},\sin b=\frac{4}{5}$

$\therefore \tan a=\frac{\sin a}{\cos a}=\frac{\frac{5}{13}}{\frac{12}{13}}=\frac{5}{12},\tan b=\frac{\sin b}{\cos b}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}$

Consider $\tan ({\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5})$

L.H.S$=\tan ({\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5})$

$=\tan (a+b)$ where $a={\sin }^{-1}\frac{5}{13}$ and $b={\cos }^{-1}\frac{3}{5}$

$=\frac{\tan a+\tan b}{1-\tan a\tan b}$

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

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

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

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

$=\frac{63}{16}=$R.H.S

Hence proved.