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