Question

Prove: \(cos^{-1} \frac {12}{13} + sin^{-1} \frac 35 = sin^{-1} \frac {56}{65}\)

Answer 1

Let sin^-1\(\frac 35\) = x. Then, sin x=\(\frac 35\) \(\implies\)cos x=\(\sqrt {1-(\frac 35)^2}\) = \(\frac 45\),
therefore tan x=\(\frac 34\) \(\implies\) x = tan^-1 \(\frac 34\) 
\(\implies\)sin^-1\(\frac 35\) = tan^-1 \(\frac 34\) ……...... (1) 
Now let cos^-1\(\frac {12}{13}\) = y. Then cos y=\(\frac {12}{13}\) \(\implies\)sin y=\(\frac {5}{13}\). 
tan y = \(\frac 5{12}\) \(\implies\) y = tan^-1\(\frac 5{12}\).  
therefore cos^-1 \(\frac {12}{13}\) = tan^-1\(\frac 5{12}\) ………......(2) 
Let sin-¹\(\frac {56}{65}\) = z. Then sin z = \(\frac {56}{65}\). \(\implies\)cos z = \(\frac {33}{65}\). 
tan z = \(\frac {56}{33}\) \(\implies\)z = tan^-1\(\frac {56}{33}\). 
so sin^-1\(\frac {56}{33}\) = tan^-1\(\frac {56}{33}\) ……..... (3)  
Now, we have: 
L.H.S.= cos^-1\(\frac {12}{13}\)+sin^-1\(\frac {3}{5}\)
         = tan^-1\(\frac 5{12}\) + tan^-1\(\frac 34\)       [using(1) and (2)] 
         = tan^-1\(\frac {\frac {5}{12}+ \frac 34}{1- \frac {5}{12}. \frac 34}\) 
         = tan^-1\(\frac {56}{33}\)
         = sin^-1\(\frac {56}{65}\)
         = R.H.S