Addition formulas

When two inverse trig terms appear in a single problem, the goal is usually to combine them into one. The combining is done by the addition formulas: identities expressing $\sin^{-1} x \pm \sin^{-1} y$ and $\tan^{-1} x \pm \tan^{-1} y$ as a single inverse term. Each comes with conditions on the signs and magnitudes that you must check before applying.

Sum of two arctangents

$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \frac{x + y}{1 - xy}, \quad xy < 1.$

If $xy = 1$ then $1 - xy = 0$ and the right side is $\pm \pi/2$ depending on signs. If $xy > 1$ the formula needs a $\pm \pi$ correction:

$\tan^{-1} x + \tan^{-1} y = \pi + \tan^{-1} \frac{x + y}{1 - xy}, \quad x, y > 0, \ xy > 1,$

and similarly with $-\pi$ for both negative.

Derivation. Let $\alpha = \tan^{-1} x, \beta = \tan^{-1} y$. Then $\tan(\alpha + \beta) = \tfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha \tan\beta} = \tfrac{x + y}{1 - xy}$. So $\alpha + \beta \equiv \tan^{-1} \tfrac{x + y}{1 - xy} \pmod{\pi}$. The correction term comes from checking which branch $\alpha + \beta$ falls into.

Difference of two arctangents

$\tan^{-1} x - \tan^{-1} y = \tan^{-1} \frac{x - y}{1 + xy}, \quad xy > -1.$

Same conditions: a $\pm \pi$ correction is needed when $xy < -1$.

Sum of two arcsines

For $x, y \in [-1, 1]$ with $x^2 + y^2 \le 1$ (i.e., both small enough),

$\sin^{-1} x + \sin^{-1} y = \sin^{-1}(x \sqrt{1 - y^2} + y \sqrt{1 - x^2}).$

The right side comes from the identity $\sin(\alpha + \beta) = \sin\alpha \cos\beta + \cos\alpha \sin\beta$. The condition $x^2 + y^2 \le 1$ ensures the sum $\alpha + \beta$ stays in $[-\pi/2, \pi/2]$.

If $xy < 0$ the same formula applies for any $x, y \in [-1, 1]$.

If $x^2 + y^2 > 1$ and $xy > 0$, we have $\sin^{-1} x + \sin^{-1} y = \pi - \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$ when both are positive, and $-\pi - (\ldots)$ when both are negative.

Difference of two arcsines

$\sin^{-1} x - \sin^{-1} y = \sin^{-1}(x \sqrt{1 - y^2} - y \sqrt{1 - x^2}),$ valid when both sides lie in $[-\pi/2, \pi/2]$ (typically when $xy \ge 0$ or $x^2 + y^2 \le 1$).

Sum of two arccosines

$\cos^{-1} x + \cos^{-1} y = \cos^{-1}(xy - \sqrt{1 - x^2}\sqrt{1 - y^2}),$ when $x + y \ge 0$. Otherwise replace by $2\pi -$ that expression.

Worked examples

Example 1. Compute $\tan^{-1}(1/2) + \tan^{-1}(1/3)$.

$xy = 1/6 < 1$, apply directly: $\tan^{-1}\tfrac{1/2 + 1/3}{1 - 1/6} = \tan^{-1}\tfrac{5/6}{5/6} = \tan^{-1} 1 = \pi/4$.

Example 2. Compute $\tan^{-1} 2 + \tan^{-1} 3$.

$xy = 6 > 1$, both positive. So $\tan^{-1} 2 + \tan^{-1} 3 = \pi + \tan^{-1}\tfrac{5}{1 - 6} = \pi + \tan^{-1}(-1) = \pi - \pi/4 = 3\pi/4$.

Example 3. Show $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 3 = \pi$.

$\tan^{-1} 2 + \tan^{-1} 3 = 3\pi/4$ (above), and $\tan^{-1} 1 = \pi/4$. Sum $= \pi$.

Example 4. Compute $\sin^{-1}(3/5) + \sin^{-1}(4/5)$.

Note $(3/5)^2 + (4/5)^2 = 1$, so $x^2 + y^2 = 1$, exactly at the boundary. By formula: $\sin^{-1}(3/5 \cdot 3/5 + 4/5 \cdot 4/5) = \sin^{-1}(9/25 + 16/25) = \sin^{-1}(1) = \pi/2$.

Cross-check: $\sin^{-1}(3/5)$ and $\sin^{-1}(4/5)$ are complementary because $(3/5)^2 + (4/5)^2 = 1$ means they correspond to a right triangle. So they sum to $\pi/2$. $\checkmark$

Example 5. Compute $\sin^{-1}(12/13) + \cos^{-1}(4/5)$.

Convert: $\cos^{-1}(4/5) = \sin^{-1}(3/5)$. So we want $\sin^{-1}(12/13) + \sin^{-1}(3/5)$.

Check $x^2 + y^2 = 144/169 + 9/25$. Use addition: $\sin^{-1}(\tfrac{12}{13}\sqrt{1 - 9/25} + \tfrac{3}{5}\sqrt{1 - 144/169}) = \sin^{-1}(\tfrac{12}{13} \cdot \tfrac{4}{5} + \tfrac{3}{5} \cdot \tfrac{5}{13}) = \sin^{-1}(\tfrac{48 + 15}{65}) = \sin^{-1}(63/65)$.

Example 6. Show $\tan^{-1}\tfrac{1}{5} + \tan^{-1}\tfrac{1}{7} + \tan^{-1}\tfrac{1}{3} + \tan^{-1}\tfrac{1}{8} = \pi/4$.

Pair them: $\tan^{-1}\tfrac{1}{5} + \tan^{-1}\tfrac{1}{7} = \tan^{-1}\tfrac{12/35}{1 - 1/35} = \tan^{-1}\tfrac{12}{34} = \tan^{-1}\tfrac{6}{17}$.

$\tan^{-1}\tfrac{1}{3} + \tan^{-1}\tfrac{1}{8} = \tan^{-1}\tfrac{11/24}{1 - 1/24} = \tan^{-1}\tfrac{11}{23}$.

Now $\tan^{-1}\tfrac{6}{17} + \tan^{-1}\tfrac{11}{23} = \tan^{-1}\tfrac{6/17 + 11/23}{1 - 66/391}$.

Numerator: $\tfrac{138 + 187}{391} = \tfrac{325}{391}$. Denominator: $\tfrac{391 - 66}{391} = \tfrac{325}{391}$. Ratio $= 1$. So the sum is $\tan^{-1} 1 = \pi/4$. $\checkmark$

Try it yourself

1. Compute $\tan^{-1}\tfrac{1}{2} + \tan^{-1}\tfrac{1}{3}$.
2. Compute $\tan^{-1} 3 + \tan^{-1} 4$.
3. Compute $\sin^{-1}\tfrac{1}{2} + \sin^{-1}\tfrac{1}{2}$.
4. Compute $\sin^{-1}\tfrac{3}{5} - \sin^{-1}\tfrac{4}{5}$.
5. Solve for $x$: $\tan^{-1}(x + 1) + \tan^{-1}(x - 1) = \tan^{-1}\tfrac{8}{31}$.
6. Show $\tan^{-1}\tfrac{1}{7} + \tan^{-1}\tfrac{1}{8} = \tan^{-1}\tfrac{3}{11}$.
7. Show $4 \tan^{-1}\tfrac{1}{5} - \tan^{-1}\tfrac{1}{239} = \pi/4$ (Machin's formula).
8. Compute $\cos^{-1}\tfrac{3}{5} + \cos^{-1}\tfrac{4}{5}$.
9. Compute $\sin^{-1}\tfrac{5}{13} + \sin^{-1}\tfrac{12}{13}$.
10. Solve $\sin^{-1} x + \sin^{-1} 2x = \pi/3$ for $x$.
11. Show $\tan^{-1}\tfrac{x + 1}{x + 2} + \tan^{-1}\tfrac{x - 1}{x - 2} = \pi/4$ for $x = \sqrt 2$.
12. Compute $\tan^{-1} 1 + \tan^{-1} 2 + \tan^{-1} 4 + \tan^{-1} \tfrac{1}{8}$ , careful with branches.
13. Show $\cos^{-1} x + \cos^{-1}\sqrt{1 - x^2} = \pi/2$ for $x \in [0, 1]$.
14. If $\tan^{-1} x + \tan^{-1} y = \pi/4$, find a relation between $x$ and $y$.

Pitfalls / Tricks

• Always check the sign condition: $xy < 1$ for the basic arctangent sum formula. Forgetting the $\pm \pi$ correction is the single biggest source of error.
• Sum of arcsines becomes painful when $x^2 + y^2 > 1$; convert to arctangents.
• When you have many terms, pair them strategically: try to make products inside the formula simplify.
• Machin-style identities (problem 7) are a JEE classic , you may need iteration.

Next, the most useful single-variable identity collection: the multi-angle formulas.