સાબિત કરો કે $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}$

(A) ધારો કે $\sin ^{-1} \frac{3}{5}=x$. તેથી,$\sin x=\frac{3}{5} \Rightarrow \cos x=\sqrt{1-\left(\frac{3}{5}\right)^{2}}=\sqrt{\frac{16}{25}}=\frac{4}{5}$.
$\therefore \tan x=\frac{3}{4} \Rightarrow x=\tan ^{-1} \frac{3}{4}$.
$\therefore \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{3}{4} \dots(1)$.
હવે,ધારો કે $\cos ^{-1} \frac{12}{13}=y$. તેથી,$\cos y=\frac{12}{13} \Rightarrow \sin y=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{\frac{25}{169}}=\frac{5}{13}$.
$\therefore \tan y=\frac{5}{12} \Rightarrow y=\tan ^{-1} \frac{5}{12}$.
$\therefore \cos ^{-1} \frac{12}{13}=\tan ^{-1} \frac{5}{12} \dots(2)$.
હવે,ડાબી બાજુ ($L$.$H$.$S$) ધ્યાનમાં લો: $\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}$.
સમીકરણ $(1)$ અને $(2)$ નો ઉપયોગ કરતા,આપણને મળે $\tan ^{-1} \frac{5}{12}+\tan ^{-1} \frac{3}{4}$.
સૂત્ર $\tan ^{-1} A + \tan ^{-1} B = \tan ^{-1} \left( \frac{A+B}{1-AB} \right)$ નો ઉપયોગ કરતા:
$= \tan ^{-1} \left( \frac{\frac{5}{12}+\frac{3}{4}}{1-\left(\frac{5}{12} \cdot \frac{3}{4}\right)} \right) = \tan ^{-1} \left( \frac{\frac{20+36}{48}}{\frac{48-15}{48}} \right) = \tan ^{-1} \left( \frac{56}{33} \right)$.
$\tan ^{-1} \frac{56}{33}$ ને $\sin ^{-1}$ માં ફેરવવા માટે,ધારો કે $\tan ^{-1} \frac{56}{33} = z$. તેથી $\tan z = \frac{56}{33}$.
નિત્યસમ $\sin z = \frac{\tan z}{\sqrt{1+\tan^2 z}} = \frac{56/33}{\sqrt{1+(56/33)^2}} = \frac{56/33}{\sqrt{(1089+3136)/1089}} = \frac{56/33}{65/33} = \frac{56}{65}$.
આમ,$\tan ^{-1} \frac{56}{33} = \sin ^{-1} \frac{56}{65} = \text{જમણી બાજુ (R.H.S)}$.