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