Question

Prove: \(cos^{-1} \frac45 + cos^{-1} \frac {12}{13} = cos^{-1} \frac {33}{65} \)

Answer 1

Let cos^-1\(\frac 45\) = x. Then cosx = \(\frac 45\) = sin x = \(\sqrt {1-(\frac45)^2}\) = \(\frac 35\). 
tanx = \(\frac 34\). \(\implies\)x = tan^-1\(\frac 34\)
Therefore cos^-1\(\frac 45\) = tan^-1\(\frac 34\) …..... (1)  
Now, let cos^-1\(\frac {12}{13}\) = y.\(\implies\)cos y = \(\frac {12}{13}\) = sin y = \(\frac {5}{13}\). 
therefore tan y = \(\frac {5}{12}\)= tan^-1 \(\frac {5}{12}\).  
therefore cos^-1\(\frac {12}{13}\) = tan^-1 \(\frac {5}{12}\)  …….…. (2)  
Let cos^-1\(\frac {33}{65}\) = z. cosz = \(\frac {33}{65}\). \(\implies\)sin z = \(\frac {56}{65}\).  
therefore tan z = \(\frac {56}{65}\) \(\implies\)z = tan^-1\(\frac {56}{65}\)
therefore cos^-1\(\frac {33}{65}\) = tan^-1\(\frac {56}{65}\) …….... (3) 
Now, we will prove that:
LHS = cos^-1\(\frac 45\) + cos^-1 \(\frac {12}{13}\) 
        =tan^-1\(\frac 34\) + tan^-1\(\frac {5}{12}\)          [using(1) and (2)] 
        =tan^-1\(\frac {\frac 34 + \frac {5}{12}}{1- \frac 34 \times \frac {5}{12}}\)    [tan^-1x + tan^-1y
        =tan-¹\(\frac {x+y}{1-xy}\)
        =tan^-1\(\frac {36+20}{48-15}\)
        =tan^-1\(\frac {56}{33}\)     [by(3)] 
        =RHS