Inverse Trigonometric Functions • Topic 2 of 2

Properties of Inverse Trigonometric Functions

A compact set of identities simplifies almost every inverse-trig expression in the syllabus. Each identity holds only on a stated domain, and quoting that domain is part of a correct answer — the most common mistakes in this chapter come from applying an identity outside its valid range.

Self-inverse (cancellation) identities

When the inverse and the direct function meet, they cancel only on the right interval:

  • $\sin(\sin^{-1}x)=x$ and $\cos(\cos^{-1}x)=x$ for all $x\in[-1,1]$; $\tan(\tan^{-1}x)=x$ for all $x\in\mathbb{R}$. (Outer trig of inner inverse — always clean.)
  • $\sin^{-1}(\sin x)=x$ only for $x\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$; $\cos^{-1}(\cos x)=x$ only for $x\in[0,\pi]$; $\tan^{-1}(\tan x)=x$ only for $x\in\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$. (Inverse of inner trig — reduce the angle into the branch first.)

Reciprocal identities

The inverse of a reciprocal turns into the matching inverse function:

$$\csc^{-1}x=\sin^{-1}\frac{1}{x},\qquad \sec^{-1}x=\cos^{-1}\frac{1}{x},\qquad \cot^{-1}x=\tan^{-1}\frac{1}{x}\ (x>0).$$

For $\cot^{-1}$ the last rule needs care: $\cot^{-1}x=\tan^{-1}\tfrac1x$ holds for $x>0$, while for $x<0$, $\cot^{-1}x=\pi+\tan^{-1}\tfrac1x$, because the ranges of the two functions differ.

Negative-argument identities

  • $\sin^{-1}(-x)=-\sin^{-1}x$, $\ \tan^{-1}(-x)=-\tan^{-1}x$, $\ \csc^{-1}(-x)=-\csc^{-1}x$ (odd functions).
  • $\cos^{-1}(-x)=\pi-\cos^{-1}x$, $\ \cot^{-1}(-x)=\pi-\cot^{-1}x$, $\ \sec^{-1}(-x)=\pi-\sec^{-1}x$.

Complementary (co-function) identities

These three sum to a right angle on the stated domains:

$$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}\ (x\in[-1,1]),\quad \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}\ (x\in\mathbb{R}),\quad \sec^{-1}x+\csc^{-1}x=\frac{\pi}{2}\ (|x|\ge1).$$

Sum and difference of two arctangents

The workhorse identity, with its mandatory quadrant correction:

$$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\!\left(\frac{x+y}{1-xy}\right),\qquad xy<1.$$

If $xy>1$ with $x,y>0$, add $\pi$; if $xy>1$ with $x,y<0$, subtract $\pi$. The difference form is

$$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\!\left(\frac{x-y}{1+xy}\right),\qquad xy>-1.$$

Sum and difference for arcsine and arccosine

$$\sin^{-1}x+\sin^{-1}y=\sin^{-1}\!\left(x\sqrt{1-y^2}+y\sqrt{1-x^2}\right),$$

valid when $x^2+y^2\le1$ (or when $xy<0$); otherwise a $\pm\pi$ adjustment is needed. Likewise

$$\cos^{-1}x+\cos^{-1}y=\cos^{-1}\!\left(xy-\sqrt{1-x^2}\sqrt{1-y^2}\right),$$

valid for $x+y\ge0$ (otherwise replace by $2\pi$ minus the right side).

The $2\tan^{-1}x$ family

A single $\tan^{-1}$ can be doubled into $\sin^{-1}$, $\cos^{-1}$ or $\tan^{-1}$ form — the key to simplifying rational arguments:

$$2\tan^{-1}x=\tan^{-1}\frac{2x}{1-x^2}=\sin^{-1}\frac{2x}{1+x^2}=\cos^{-1}\frac{1-x^2}{1+x^2}.$$

Domains: the $\tan^{-1}$ form needs $-1

Simplifying by substitution — the standard trick

To simplify a messy algebraic argument, substitute a trig value. For $\tan^{-1}\dfrac{\sqrt{1+x^2}-1}{x}$ put $x=\tan\theta$; for arguments with $\sqrt{1-x^2}$ put $x=\sin\theta$ or $x=\cos\theta$; for $\sqrt{1+x^2}$ put $x=\tan\theta$. The radical collapses, the expression becomes a clean function of $\theta$, and you read off the answer as a fraction of the original inverse.

Graphs of y equals sin inverse x and y equals cos inverse x over minus 1 to 1 showing their sum is the constant pi over 2 sin−¹x + cos−¹x = π/2 x y −1 1 π π/2 −π/2 y = sin−¹x y = cos−¹x sum stays at π/2 Table of domains and principal value ranges for sin inverse cos inverse and tan inverse Principal value ranges Function Domain Range (branch) sin−¹x [−1, 1] [−π/2, π/2] cos−¹x [−1, 1] [0, π] tan−¹x all real x (−π/2, π/2)
Solved Examples
1
Worked Example
Evaluate $\sin^{-1}\!\left(\tfrac{1}{2}\right)+\cos^{-1}\!\left(\tfrac{1}{2}\right)$.
Solution

By the complementary identity $\sin^{-1}x+\cos^{-1}x=\tfrac{\pi}{2}$ for every $x\in[-1,1]$. Here $x=\tfrac12$, so the sum is $\dfrac{\pi}{2}$ — no need to compute each term.

Answer: $\dfrac{\pi}{2}$

2
Worked Example
Simplify $\cos^{-1}\!\left(-\tfrac{1}{2}\right)$ using a negative-argument identity.
Solution

$\cos^{-1}(-x)=\pi-\cos^{-1}x$. With $x=\tfrac12$: $\cos^{-1}\!\left(-\tfrac12\right)=\pi-\cos^{-1}\tfrac12=\pi-\tfrac{\pi}{3}=\dfrac{2\pi}{3}$.

Answer: $\dfrac{2\pi}{3}$

3
Worked Example
Find $\tan^{-1}1+\tan^{-1}2+\tan^{-1}3$.
Solution

First $\tan^{-1}2+\tan^{-1}3$: here $xy=6>1$ with both positive, so we add $\pi$: $\tan^{-1}2+\tan^{-1}3=\pi+\tan^{-1}\dfrac{2+3}{1-6}=\pi+\tan^{-1}(-1)=\pi-\tfrac{\pi}{4}=\tfrac{3\pi}{4}$. Adding $\tan^{-1}1=\tfrac{\pi}{4}$ gives $\tfrac{3\pi}{4}+\tfrac{\pi}{4}=\pi$.

Answer: $\pi$

4
Worked Example
Write $2\tan^{-1}\!\left(\tfrac{1}{3}\right)$ as a single arctangent.
Solution

$2\tan^{-1}x=\tan^{-1}\dfrac{2x}{1-x^2}$ for $|x|<1$. With $x=\tfrac13$: $\dfrac{2\cdot\tfrac13}{1-\tfrac19}=\dfrac{\tfrac23}{\tfrac89}=\dfrac{2}{3}\cdot\dfrac{9}{8}=\dfrac{3}{4}$. So $2\tan^{-1}\tfrac13=\tan^{-1}\dfrac{3}{4}$.

Answer: $\tan^{-1}\dfrac{3}{4}$

5
Worked Example
Simplify $\tan^{-1}\dfrac{1}{2}+\tan^{-1}\dfrac{1}{3}$.
Solution

Here $xy=\tfrac12\cdot\tfrac13=\tfrac16<1$, so the plain identity applies. $\dfrac{x+y}{1-xy}=\dfrac{\tfrac12+\tfrac13}{1-\tfrac16}=\dfrac{\tfrac56}{\tfrac56}=1$. Hence the sum $=\tan^{-1}1=\dfrac{\pi}{4}$.

Answer: $\dfrac{\pi}{4}$

6
Worked Example
Prove that $2\tan^{-1}\dfrac{1}{5}+\tan^{-1}\dfrac{1}{8}=\tan^{-1}\dfrac{4}{7}$.
Solution

First $2\tan^{-1}\tfrac15=\tan^{-1}\dfrac{2\cdot\tfrac15}{1-\tfrac{1}{25}}=\tan^{-1}\dfrac{\tfrac25}{\tfrac{24}{25}}=\tan^{-1}\dfrac{10}{24}=\tan^{-1}\dfrac{5}{12}$. Now add $\tan^{-1}\tfrac18$ (here $xy=\tfrac{5}{96}<1$): $\dfrac{\tfrac{5}{12}+\tfrac18}{1-\tfrac{5}{96}}=\dfrac{\tfrac{10+3}{24}}{\tfrac{91}{96}}=\dfrac{\tfrac{13}{24}}{\tfrac{91}{96}}=\dfrac{13}{24}\cdot\dfrac{96}{91}=\dfrac{13\cdot4}{91}=\dfrac{52}{91}=\dfrac{4}{7}$. Hence the sum is $\tan^{-1}\dfrac47$.

Answer: $\tan^{-1}\dfrac47$

7
Worked Example
Evaluate $\tan^{-1}\!\left(\tan\tfrac{3\pi}{4}\right)+\cos^{-1}\!\left(\cos\tfrac{3\pi}{4}\right)$.
Solution

For $\tan^{-1}$ reduce into $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$: $\tan\tfrac{3\pi}{4}=\tan\!\left(\tfrac{3\pi}{4}-\pi\right)=\tan\!\left(-\tfrac{\pi}{4}\right)$, so the first term is $-\tfrac{\pi}{4}$. For $\cos^{-1}$ the angle $\tfrac{3\pi}{4}$ already lies in $[0,\pi]$, so the second term is $\tfrac{3\pi}{4}$. Sum $=-\tfrac{\pi}{4}+\tfrac{3\pi}{4}=\dfrac{\pi}{2}$.

Answer: $\dfrac{\pi}{2}$

8
Worked Example
Simplify $\tan^{-1}\!\left(\dfrac{\sqrt{1+x^2}-1}{x}\right)$, $x\ne0$.
Solution

Put $x=\tan\theta$, so $\theta=\tan^{-1}x$ and $\sqrt{1+x^2}=\sec\theta$. Then the argument is $\dfrac{\sec\theta-1}{\tan\theta}=\dfrac{1-\cos\theta}{\sin\theta}=\tan\dfrac{\theta}{2}$ (half-angle identity). Hence the expression equals $\tan^{-1}\!\left(\tan\tfrac{\theta}{2}\right)=\dfrac{\theta}{2}=\dfrac{1}{2}\tan^{-1}x$.

Answer: $\dfrac{1}{2}\tan^{-1}x$

9
Worked Example
Show that $\sin^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{8}{17}=\cos^{-1}\dfrac{36}{85}$.
Solution

Let $\alpha=\sin^{-1}\tfrac35$ and $\beta=\sin^{-1}\tfrac{8}{17}$. Then $\cos\alpha=\tfrac45$, $\cos\beta=\tfrac{15}{17}$. So $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\tfrac45\cdot\tfrac{15}{17}-\tfrac35\cdot\tfrac{8}{17}=\dfrac{60-24}{85}=\dfrac{36}{85}$. Since $\alpha+\beta$ lies in $[0,\pi]$, $\alpha+\beta=\cos^{-1}\dfrac{36}{85}$, as required.

Answer: $\sin^{-1}\dfrac{3}{5}+\sin^{-1}\dfrac{8}{17}=\cos^{-1}\dfrac{36}{85}$ (proved).

10
Worked Example
Solve $\tan^{-1}(2x)+\tan^{-1}(3x)=\dfrac{\pi}{4}$.
Solution

Combine the left side: $\tan^{-1}\dfrac{2x+3x}{1-6x^2}=\tfrac{\pi}{4}$, so $\dfrac{5x}{1-6x^2}=\tan\tfrac{\pi}{4}=1$. Thus $5x=1-6x^2$, i.e. $6x^2+5x-1=0$, which factorises as $(6x-1)(x+1)=0$, giving $x=\tfrac16$ or $x=-1$. Check the domain $xy<1$: for $x=\tfrac16$, $6x^2=\tfrac16<1$, valid. For $x=-1$ the left side becomes $\tan^{-1}(-2)+\tan^{-1}(-3)$, which is negative, so it cannot equal $\tfrac{\pi}{4}$ — reject it. Hence $x=\dfrac{1}{6}$.

Answer: $x=\dfrac{1}{6}$

11
Worked Example
Prove that $2\tan^{-1}\dfrac{1}{3}+\tan^{-1}\dfrac{1}{7}=\dfrac{\pi}{4}$.
Solution

First convert the doubled term: $2\tan^{-1}\tfrac13=\tan^{-1}\dfrac{2\cdot\tfrac13}{1-\tfrac19}=\tan^{-1}\dfrac{\tfrac23}{\tfrac89}=\tan^{-1}\dfrac34$. Now add $\tan^{-1}\tfrac17$; here $xy=\tfrac34\cdot\tfrac17=\tfrac{3}{28}<1$, so the plain sum rule applies: $\dfrac{\tfrac34+\tfrac17}{1-\tfrac{3}{28}}=\dfrac{\tfrac{21+4}{28}}{\tfrac{25}{28}}=\dfrac{25}{25}=1$. Hence the total is $\tan^{-1}1=\dfrac{\pi}{4}$.

Answer: $2\tan^{-1}\dfrac{1}{3}+\tan^{-1}\dfrac{1}{7}=\dfrac{\pi}{4}$ (proved).

12
Worked Example
Find the value of $\cos\!\left(\sin^{-1}\dfrac{3}{5}+\cos^{-1}\dfrac{4}{5}\right)$.
Solution

Let $\alpha=\sin^{-1}\tfrac35$, so $\sin\alpha=\tfrac35,\cos\alpha=\tfrac45$; let $\beta=\cos^{-1}\tfrac45$, so $\cos\beta=\tfrac45,\sin\beta=\tfrac35$. Then $\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta=\tfrac45\cdot\tfrac45-\tfrac35\cdot\tfrac35=\dfrac{16-9}{25}=\dfrac{7}{25}$.

Answer: $\dfrac{7}{25}$

Key Points

  • Cancellation: outer trig of inner inverse is always clean ($\sin(\sin^{-1}x)=x$); inverse of inner trig works only inside the branch.
  • Reciprocal rules: $\csc^{-1}x=\sin^{-1}\tfrac1x$, $\sec^{-1}x=\cos^{-1}\tfrac1x$, $\cot^{-1}x=\tan^{-1}\tfrac1x$ for $x>0$ (add $\pi$ for $x<0$).
  • $\sin^{-1},\tan^{-1},\csc^{-1}$ are odd; $\cos^{-1}(-x)=\pi-\cos^{-1}x$ and likewise for $\cot^{-1},\sec^{-1}$.
  • Complementary: $\sin^{-1}x+\cos^{-1}x=\tan^{-1}x+\cot^{-1}x=\sec^{-1}x+\csc^{-1}x=\tfrac{\pi}{2}$ on their domains.
  • $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\tfrac{x+y}{1-xy}$ only when $xy<1$; otherwise add/subtract $\pi$. Difference form needs $xy>-1$.
  • $2\tan^{-1}x=\tan^{-1}\tfrac{2x}{1-x^2}=\sin^{-1}\tfrac{2x}{1+x^2}=\cos^{-1}\tfrac{1-x^2}{1+x^2}$ on the stated domains.
  • To simplify radical arguments, substitute $x=\tan\theta$ (for $\sqrt{1+x^2}$) or $x=\sin\theta/\cos\theta$ (for $\sqrt{1-x^2}$); the radical collapses.
  • When solving inverse-trig equations, always verify each root satisfies the domain/branch restriction and reject extraneous solutions.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.$\sin^{-1}x+\cos^{-1}x$ equals (for $x\in[-1,1]$):
Explanation: It is the constant $\tfrac{\pi}{2}$ for every valid $x$.
Q2.$\cos^{-1}(-x)=$
Explanation: $\cos^{-1}$ is not odd; $\cos^{-1}(-x)=\pi-\cos^{-1}x$.
Q3.$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\tfrac{x+y}{1-xy}$ holds provided:
Explanation: The plain identity requires $xy<1$; otherwise a $\pm\pi$ correction is needed.
Q4.$2\tan^{-1}\tfrac{1}{2}$ written as a single arctangent is:
Explanation: $\tfrac{2x}{1-x^2}=\tfrac{1}{1-1/4}=\tfrac{1}{3/4}=\tfrac{4}{3}$.
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