Exercise 2.2 · Q12: $\tan\left(\sin^{-1}\frac35+\cot^{-1}\frac32\right)$

Basics you need

Solution

1. Name the two angles and find where they lie

   Let

   $$\alpha = \sin^{-1}\frac35, \qquad \beta = \cot^{-1}\frac32.$$

   By definition of the principal values,

   • $\sin\alpha = \frac35$ and $-\frac{\pi}{2} \le \alpha \le \frac{\pi}{2}$⁠. Since $\sin\alpha > 0$⁠, $\alpha$ cannot be negative or zero (sine is $\le 0$ on $\left[-\frac{\pi}{2}, 0\right]$⁠). And $\alpha \ne \frac{\pi}{2}$⁠, because $\sin\frac{\pi}{2} = 1 \ne \frac35$⁠. So $0 < \alpha < \frac{\pi}{2}$⁠.
   • $\cot\beta = \frac32$ and $0 < \beta < \pi$⁠. Since $\cot\beta > 0$⁠, $\beta$ is not in the second quadrant (where $\cot$ is negative) and is not $\frac{\pi}{2}$ (where $\cot$ is 0), so $0 < \beta < \frac{\pi}{2}$⁠.

   We want $\tan(\alpha + \beta)$⁠.

   Naming the angles turns an expression full of inverses into ordinary trigonometry about two angles. Locating them is not optional: knowing both are acute is what allows us to draw them in right triangles, and it fixes every sign (all ratios of an acute angle are positive).
   $y = \sin^{-1}x$ means $\sin y = x$ and $y$ is in the principal range. Both halves are used here. Lesson: principal values

2. The triangle for $\alpha$⁠: read off $\tan\alpha$

   $\sin\alpha = \frac35$⁠, and $\sin\alpha = \frac{\text{opposite}}{\text{hypotenuse}}$⁠, so draw a right triangle with opposite side 3 and hypotenuse 5. By Pythagoras the adjacent side is

   $$\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4.$$

   Therefore

   $$\tan\alpha = \frac{\text{opposite}}{\text{adjacent}} = \frac34.$$
   The triangle is a way of finding all the other ratios of $\alpha$ from the one we are given. It is legitimate only because $\alpha$ is acute (Step 1): an angle of a right triangle is always between 0 and $\frac{\pi}{2}$⁠. The same answer comes from the identity $\cos\alpha = \sqrt{1 - \sin^2\alpha} = \frac45$⁠, where the positive root is right because $\cos\alpha > 0$ for acute $\alpha$⁠.
   This is the familiar 3, 4, 5 right triangle: $9 + 16 = 25$⁠. Lesson: the triangle trick

3. The triangle for $\beta$⁠: read off $\tan\beta$

   $\cot\beta = \frac32 = \frac{\text{adjacent}}{\text{opposite}}$⁠, so draw a right triangle with adjacent side 3 and opposite side 2. (The hypotenuse is $\sqrt{3^2 + 2^2} = \sqrt{13}$⁠, though we will not need it.) Then

   $$\tan\beta = \frac{\text{opposite}}{\text{adjacent}} = \frac23.$$

   Equivalently, $\tan\beta = \dfrac{1}{\cot\beta} = \dfrac{1}{3/2} = \dfrac23$⁠, which is allowed because $\cot\beta \ne 0$⁠.

   $\cot$ is "adjacent over opposite", the reciprocal of $\tan$⁠. A frequent slip is to read $\cot\beta = \frac32$ as $\tan\beta = \frac32$⁠; the triangle makes it clear which side is which. Again the triangle is valid because $\beta$ is acute.
   $\tan\theta = \dfrac{1}{\cot\theta}$ whenever $\cot\theta \ne 0$⁠. Basics: tan, cot, sec and cosec

4. Check that the addition formula applies

   The formula $\tan(\alpha+\beta) = \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$ needs $\tan\alpha$⁠, $\tan\beta$ and $\tan(\alpha+\beta)$ to exist. The first two exist (Steps 2 and 3). For the third, we need $\alpha + \beta \ne \frac{\pi}{2}$⁠. Compute the product that appears in the denominator:

   $$\tan\alpha\tan\beta = \frac34\cdot\frac23 = \frac12 < 1.$$

   For two acute angles, $\tan\alpha\tan\beta < 1$ means $\alpha + \beta < \frac{\pi}{2}$⁠. Here is why. $\tan\alpha < \frac{1}{\tan\beta} = \cot\beta = \tan\left(\frac{\pi}{2} - \beta\right)$⁠, and both $\alpha$ and $\frac{\pi}{2} - \beta$ are acute, where $\tan$ is increasing, so $\alpha < \frac{\pi}{2} - \beta$⁠. Hence

   $$0 < \alpha + \beta < \frac{\pi}{2},$$

   so $\tan(\alpha + \beta)$ is defined (and positive), and the denominator $1 - \frac12$ is not zero.

   The addition formula is an identity only where every $\tan$ in it is defined. If $\alpha + \beta$ were exactly $\frac{\pi}{2}$⁠, the left side would not exist and the denominator would be 0. Checking first means the number we compute next really is $\tan(\alpha+\beta)$⁠. Knowing that $\alpha+\beta$ is acute also tells us in advance that the answer must be positive, which is a useful check.
   $\cot\beta = \tan\left(\frac{\pi}{2} - \beta\right)$⁠, the complementary-angle rule. Basics: special angles and reference angles

5. Substitute and simplify

   $$\begin{aligned} \tan(\alpha + \beta) &= \frac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} \\ &= \frac{\frac34 + \frac23}{1 - \frac34\cdot\frac23} \\ &= \frac{\frac{9 + 8}{12}}{1 - \frac12} = \frac{17/12}{1/2} \\ &= \frac{17}{12}\cdot 2 = \frac{17}{6}. \end{aligned}$$

   So $\tan\left(\sin^{-1}\frac35 + \cot^{-1}\frac32\right) = \frac{17}{6}$, which is positive, as Step 4 predicted.

   The numerator adds two fractions over the common denominator 12. The denominator is $1 - \frac12 = \frac12$⁠. Dividing by $\frac12$ is multiplying by 2. Every step is exact arithmetic with fractions, so no decimals are needed until the check.
   In the language of Chapter 2, this says $\tan^{-1}\frac34 + \tan^{-1}\frac23 = \tan^{-1}\frac{17}{6}$⁠. That form is correct here because $\frac34\cdot\frac23 < 1$ and both numbers are positive, so the sum of the two angles stays inside $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$⁠. Lesson: the triangle trick

Every fact used, and where it comes from