Question

Show that \(\sin^{-1}\dfrac{12}{13}+\cos^{-1}\dfrac{4}{5}+\tan^{-1}\dfrac{63}{16}=\pi\).

Hint:

Convert to acute-angle triangles, add the first two via tan(A+B), then match with π − tan⁻¹(63/16).

Answer 1

Step 1: Naming the first two angles:
Let \(A=\sin^{-1}\dfrac{12}{13}\) so \(\sin A=\dfrac{12}{13}\), \(\cos A=\dfrac{5}{13}\), \(\tan A=\dfrac{12}{5}\). Let \(B=\cos^{-1}\dfrac{4}{5}\) so \(\cos B=\dfrac{4}{5}\), \(\sin B=\dfrac{3}{5}\), \(\tan B=\dfrac{3}{4}\); both \(A,B\) are acute.

Step 2: Computing tan(A+B):
\(\tan(A+B)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}=\dfrac{\frac{12}{5}+\frac34}{1-\frac{12}{5}\cdot\frac34}=\dfrac{\frac{48+15}{20}}{1-\frac{36}{20}}=\dfrac{63/20}{-16/20}=-\dfrac{63}{16}\).

Step 3: Locating A+B:
Since \(A,B\) are both acute, \(0<A+B<\pi\); a negative tangent in this range means \(A+B\) is obtuse, i.e. in the second quadrant.

Step 4: Relating to the given inverse tangent:
For an angle \(\theta\) in \((\pi/2,\pi)\) with \(\tan\theta=-\dfrac{63}{16}\), we have \(\theta=\pi-\tan^{-1}\dfrac{63}{16}\), i.e. \(A+B=\pi-\tan^{-1}\dfrac{63}{16}\).

Final Answer:
So \(A+B+\tan^{-1}\dfrac{63}{16}=\pi\), i.e. \(\sin^{-1}\dfrac{12}{13}+\cos^{-1}\dfrac{4}{5}+\tan^{-1}\dfrac{63}{16}=\pi\).\[ \boxed{=\pi} \]