Question

Prove that : ${\sin }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)=\frac{1}{2}{\sin }^{-1}\left(\frac{3696}{4225}\right)$

Answer 1

To prove:-

${\sin }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)=\frac{1}{2}{\sin }^{-1}\left(\frac{3696}{4225}\right)$

Proof:-

Let $\alpha ={\sin }^{-1}\frac{5}{13}$

$\Rightarrow \sin \alpha =\frac{5}{13}$

From right angled triangle,

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

Similarly,

Let $\beta ={\cos }^{-1}\frac{4}{5}$

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

From right angled triangle,

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

Therefore,

${\sin }^{-1}\left(\frac{5}{13}\right)+{\cos }^{-1}\left(\frac{4}{5}\right)=\frac{1}{2}{\sin }^{-1}\left(\frac{3696}{4225}\right)$

${\tan }^{-1}\frac{5}{12}+{\tan }^{-1}\frac{3}{4}=\frac{1}{2}{\sin }^{-1}\left(\frac{3696}{4225}\right)$

$2{\tan }^{-1}\frac{5}{12}+2{\tan }^{-1}\frac{3}{4}={\sin }^{-1}\left(\frac{3696}{4225}\right)$

Taking L.H.S.

$2{\tan }^{-1}\frac{5}{12}+2{\tan }^{-1}\frac{3}{4}$

$={\tan }^{-1}\left(\frac{2\times \frac{5}{12}}{1-{\left(\frac{5}{12}\right)}^2}\right)+{\tan }^{-1}\left(\frac{2\times \frac{3}{4}}{1-{\left(\frac{3}{4}\right)}^2}\right)$

$={\tan }^{-1}\frac{120}{119}+{\tan }^{-1}\frac{24}{7}$

$={\tan }^{-1}\left(\frac{\frac{120}{119}+\frac{24}{7}}{1-\left(\frac{120}{119}\right)\left(\frac{24}{7}\right)}\right)$

$={\tan }^{-1}\left(\frac{840+2856}{833-2880}\right)$

$={\tan }^{-1}(-\frac{3696}{2047})$

$={\sin }^{-1}\left(\frac{3696}{4225}\right)$

$=$ R.H.S.

Hence proved.