Prove that: ^ -1 12 13 = ^ -1 3 5 = ^ -1 3 5 = ^ -1 56 65

Question

Prove that:
$\cos^{-1}\frac{12}{13}=\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{3}{5}=\sin^{-1}\frac{56}{65}$

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Answer

$\text{Let}\ \sin^{-1}\frac{3}{5}=x.\text{Then},\sin x=\frac{3}{5}$ $\Rightarrow\cos x=\sqrt{1-\bigg(\frac{3}{5}\bigg)^2}=\sqrt{\frac{16}{25}}=\frac{4}{5}.$ $\therefore\tan x=\frac{3}{4}\Rightarrow x=\tan^{-1}\frac{3}{4}$ $\therefore\sin^{-1}\frac{3}{5}=\tan^{-1}\frac{3}{4} \dots\dots(1)$ $\text{Now let}\cos^{-1}\frac{12}{13}=y,\cos y=\frac{12}{13}\Rightarrow\sin y=\frac{5}{13}.$ $\therefore\tan y=\frac{5}{12}\Rightarrow y=\tan^{-1}\frac{5}{12}$ $\therefore\cos^{-1}\frac{12}{13}=\tan^{-1}\frac{5}{12} \dots\dots(2)$ $\text{Let}\sin^{-1}\frac{56}{65}=z.\text{Then},\sin z=\frac{56}{65}\Rightarrow\cos z=\frac {33}{65}.$ $\therefore\tan z=\frac{56}{33}\Rightarrow z=\tan^{-1}\frac{56}{33}$ $\therefore\sin^{-1}\frac{56}{65}=\tan^{-1}\frac{56}{33} \dots\dots(3)$ Now, we have: $\text{L.H.S.}=\cos^{-1}\frac{12}{13}+\sin^{-1}\frac{3}{5}$$=\tan^{-1}\frac{5}{12}+\tan^{-1}\frac{3}{4}$ [Using (1) and (2)]
$=\tan^{-1}\frac{\frac{5}{12}+\frac{3}{4}}{1-\frac{5}{12}.\frac{3}{4}}$ $ \bigg[\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}\bigg]$
$=\tan^{-1}\frac{20+36}{48-15}$
$=\tan^{-1}\frac{56}{33}$
$=\sin^{-1}\frac{56}{65}=\text{R.H.S.}$ [Using (3)]

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