Question
Prove that : $\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}$
✓
Answer
$\begin{array}{l}\text { L.H.S. }=\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13} \\ \text { Take, } \cos ^{-1} \frac{4}{5}=\alpha, \quad \cos ^{-1} \frac{12}{13}=\beta \\ \therefore \quad \cos \alpha=\frac{4}{5} \quad, \quad \cos \beta=\frac{12}{13}\end{array}$
$\begin{array}{l}\therefore \quad \sin \alpha=\frac{3}{5} \quad, \quad \sin \beta=\frac{5}{13} \\ \text { Here, } \cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta\end{array}$
$\begin{array}{l}=\left(\frac{4}{5} \times \frac{12}{13}\right)-\left(\frac{3}{5} \times \frac{5}{13}\right) \\ =\frac{48}{65}-\frac{15}{65}\end{array}$
$\cos (\alpha+\beta)=\frac{33}{65}$
$\begin{array}{cc}\therefore & \alpha+\beta=\cos ^{-1}\left(\frac{33}{65}\right) \\ \therefore & \cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\end{array}$
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