Miscellaneous · Q7: $\tan^{-1}\frac{63}{16}=\sin^{-1}\frac{5}{13}+\cos^{-1}\frac35$

Basics you need

Solution

1. Name the two angles and locate them

   Let

   $$A = \sin^{-1}\frac{5}{13}, \qquad B = \cos^{-1}\frac35.$$

   Then $\sin A = \frac{5}{13}$ with $A \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠; the sine is positive and less than 1, so $0 < A < \frac{\pi}{2}$⁠. Also $\cos B = \frac35$ with $B \in [0,\pi]$⁠; the cosine is positive and less than 1, so $0 < B < \frac{\pi}{2}$⁠.

   We must prove $A + B = \tan^{-1}\frac{63}{16}$⁠. (The question writes the equation the other way round; equality reads the same in both directions.)

   The two angles come from different inverse functions, so each is located using its own range. Both are acute, which lets us draw right triangles and gives $0 < A+B < \pi$ for Step 4.
   $y = \sin^{-1}x$⁠: $\sin y = x$⁠, $y \in \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$⁠. $y = \cos^{-1}x$⁠: $\cos y = x$⁠, $y \in [0,\pi]$ (Basic 1).

2. Build the two triangles

   For $A$⁠: opposite 5, hypotenuse 13, adjacent $\sqrt{169 - 25} = \sqrt{144} = 12$⁠. For $B$⁠: adjacent 3, hypotenuse 5, opposite $\sqrt{25 - 9} = 4$⁠. Hence

   $$\begin{aligned} \cos A &= \tfrac{12}{13}, & \tan A &= \tfrac{5}{12}, \\[1mm] \sin B &= \tfrac45, & \tan B &= \tfrac43. \end{aligned}$$
   The other side of the equation is a $\tan^{-1}$⁠, so we want tangents. All values are positive because both angles are acute. Watch the second triangle: $B$ is given by its cosine, so 3 is adjacent and 4 opposite, and $\tan B = \frac43$⁠, not $\frac34$⁠.
   $5^2 + 12^2 = 13^2$ and $3^2 + 4^2 = 5^2$⁠. Lesson: the triangle trick

3. Add the angles: $\tan(A+B) = \frac{63}{16}$

   The pieces:

   $$\begin{aligned} \tan A + \tan B &= \frac{5}{12} + \frac{16}{12} = \frac{21}{12} = \frac74, \\[1mm] \tan A\tan B &= \frac{5}{12}\cdot\frac43 = \frac{20}{36} = \frac59. \end{aligned}$$

   Then

   $$\begin{aligned} \tan(A+B) &= \frac{\tan A + \tan B}{1 - \tan A\tan B} \\[1mm] &= \frac{7/4}{1 - 5/9} = \frac{7/4}{4/9} \\[1mm] &= \frac74\cdot\frac94 = \frac{63}{16}. \end{aligned}$$

   For the next step, also note

   $$\begin{aligned} \cos(A+B) &= \tfrac{12}{13}\cdot\tfrac35 - \tfrac{5}{13}\cdot\tfrac45 \\[1mm] &= \tfrac{36 - 20}{65} = \tfrac{16}{65}, \\[2mm] \sin(A+B) &= \tfrac{5}{13}\cdot\tfrac35 + \tfrac{12}{13}\cdot\tfrac45 \\[1mm] &= \tfrac{15 + 48}{65} = \tfrac{63}{65}. \end{aligned}$$

   So $A+B$ belongs to a 16, 63, 65 triangle (check: $16^2 + 63^2 = 256 + 3969 = 4225 = 65^2$⁠).

   The denominator $1 - \frac59 = \frac49$ is not zero, so $\tan(A+B)$ is defined and the formula applies. Both numerator and denominator are positive, so $\tan(A+B) > 0$⁠, which Step 4 uses.
   $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A\tan B}$⁠, together with the $\cos(A+B)$ and $\sin(A+B)$ formulas (Class XI, Chapter 3).

4. The crucial check: $A+B$ is acute

   Equal tangents do not yet give equal angles: tangent repeats every $\pi$⁠. We need $A+B$ in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠.

   From Step 1, $0 < A+B < \pi$⁠. The value $\frac{\pi}{2}$ is excluded because $\tan(A+B)$ is defined, and $\left(\frac{\pi}{2},\pi\right)$ is excluded because tangent is negative there while $\tan(A+B) = \frac{63}{16} > 0$⁠. So

   $$0 < A + B < \frac{\pi}{2}.$$

   The cosine confirms it: $\cos(A+B) = \frac{16}{65} > 0$⁠. Equivalently, $\tan A\tan B = \frac59 < 1$⁠.

   Of all the angles with tangent $\frac{63}{16}$⁠, namely $\tan^{-1}\frac{63}{16} + k\pi$⁠, only the one with $k = 0$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. Showing $A+B$ is acute is exactly what proves $k = 0$⁠. At about $75.7^\circ$⁠, this sum is the largest in Questions 3 to 7, so it has the smallest margin below $90^\circ$⁠. Had $\tan A\tan B$ exceeded 1, the denominator of the addition formula would have been negative and the sum would have been obtuse.
   For acute $A$⁠, $B$⁠: $\cos(A+B) = \cos A\cos B\,(1 - \tan A\tan B)$⁠, so $A+B$ is acute exactly when $\tan A\tan B < 1$⁠. For a failing case see Miscellaneous Q3, in its final step.

5. Conclude

   $A+B$ lies in $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ (Step 4) and $\tan(A+B) = \frac{63}{16}$ (Step 3). Tangent takes each value only once on that interval, so

   $$\begin{aligned} \tan^{-1}\frac{63}{16} &= \tan^{-1}\bigl(\tan(A+B)\bigr) = A + B \\[1mm] &= \sin^{-1}\frac{5}{13} + \cos^{-1}\frac35. \end{aligned}$$

   This is the required result.

   The same four moves as in Miscellaneous Q4: name the angles, build triangles, apply the addition formula, and place the sum in the range of the inverse function on the other side.

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