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: Converting all to tan⁻¹ form:
\(\sin^{-1}\dfrac{12}{13}=\tan^{-1}\dfrac{12}{5}\) (opposite 12, adjacent 5, hypotenuse 13) and \(\cos^{-1}\dfrac{4}{5}=\tan^{-1}\dfrac{3}{4}\) (adjacent 4, opposite 3, hypotenuse 5), both acute-angle right-triangle ratios.

Step 2: Adding the two tan⁻¹ terms:
\(\tan^{-1}\dfrac{12}{5}+\tan^{-1}\dfrac{3}{4}\): since \(\dfrac{12}{5}\cdot\dfrac34=\dfrac{36}{20}=1.8>1\), the direct addition formula would land outside \((-\pi/2,\pi/2)\), so add \(\pi\) to the raw formula result: \(\tan^{-1}\dfrac{12}{5}+\tan^{-1}\dfrac34=\pi+\tan^{-1}\!\left(\dfrac{\frac{12}{5}+\frac34}{1-\frac{12}{5}\cdot\frac34}\right)=\pi+\tan^{-1}\!\left(-\dfrac{63}{16}\right)=\pi-\tan^{-1}\dfrac{63}{16}\).

Step 3: Adding the third term:
So the full sum is \(\left(\pi-\tan^{-1}\dfrac{63}{16}\right)+\tan^{-1}\dfrac{63}{16}\).

Final Answer:
\[ \boxed{\sin^{-1}\tfrac{12}{13}+\cos^{-1}\tfrac45+\tan^{-1}\tfrac{63}{16}=\pi} \]