2026–27 Syllabus Focus
The course structure groups marks by unit, but the paper design does not guarantee a fixed allocation for any chapter. These checkpoints follow the current syllabus; they are not a mark prediction.
- Definitions and domains of the inverse trigonometric functions.
- Ranges and principal-value branches.
- Graphs of inverse trigonometric functions.
Identity practice later on this page is optional extension material; the current syllabus lists the topics above.
Quick Review
Domains, Ranges, and Principal Branches
An inverse trigonometric function returns the angle in its chosen principal range. Check that the input is in the function's domain before evaluating it.
Branches Centered at Zero
Functions: $\sin^{-1}x$, $\tan^{-1}x$, and $\csc^{-1}x$.
Sign mnemonic: These branches are odd, so a negative input keeps its minus sign. For $\csc^{-1}x$, this identity applies when $|x|\ge 1$.
$\sin^{-1}(-x)=-\sin^{-1}x$, $\tan^{-1}(-x)=-\tan^{-1}x$, and $\csc^{-1}(-x)=-\csc^{-1}x$.
Branches from Zero to $\pi$
Functions: $\cos^{-1}x$, $\cot^{-1}x$, and $\sec^{-1}x$.
Sign mnemonic: For these branches, a negative input reflects the principal angle using $\pi-\theta$.
$\cos^{-1}(-x)=\pi-\cos^{-1}x$, $\cot^{-1}(-x)=\pi-\cot^{-1}x$, and $\sec^{-1}(-x)=\pi-\sec^{-1}x$.
| Function | Domain | Principal-value range |
|---|---|---|
| $\sin^{-1}x$ | $[-1,1]$ | $[-\frac{\pi}{2},\frac{\pi}{2}]$ |
| $\cos^{-1}x$ | $[-1,1]$ | $[0,\pi]$ |
| $\tan^{-1}x$ | $\mathbb{R}$ | $(-\frac{\pi}{2},\frac{\pi}{2})$ |
| $\cot^{-1}x$ | $\mathbb{R}$ | $(0,\pi)$ |
| $\sec^{-1}x$ | $(-\infty,-1]\cup[1,\infty)$ | $[0,\pi]\setminus\{\frac{\pi}{2}\}$ |
| $\csc^{-1}x$ | $(-\infty,-1]\cup[1,\infty)$ | $[-\frac{\pi}{2},\frac{\pi}{2}]\setminus\{0\}$ |
Quick check: Find $\sin^{-1}(-\frac{1}{2})$.
Identities and Principal-Value Checks
Use an identity only after checking the input domain and the inverse function's principal range.
Odd Inverse Functions
These functions keep the sign of a negative input:
- $\sin^{-1}(-x)=-\sin^{-1}x$, for $x\in[-1,1]$.
- $\tan^{-1}(-x)=-\tan^{-1}x$, for $x\in\mathbb{R}$.
- $\csc^{-1}(-x)=-\csc^{-1}x$, for $|x|\ge1$.
Angles Reflected About $\frac{\pi}{2}$
- $\cos^{-1}(-x)=\pi-\cos^{-1}x$, for $x\in[-1,1]$.
- $\cot^{-1}(-x)=\pi-\cot^{-1}x$, for $x\in\mathbb{R}$.
- $\sec^{-1}(-x)=\pi-\sec^{-1}x$, for $|x|\ge1$.
Choose the Principal Value
- Check the domain. For $\sin^{-1}$ and $\cos^{-1}$ the input must lie in $[-1,1]$; for $\sec^{-1}$ and $\csc^{-1}$ it must satisfy $|x|\ge1$.
- Find a reference angle from the corresponding trigonometric ratio.
- Place the answer in the principal range shown in the table above. Do not return an equivalent angle outside that interval.
| Expression | Principal value | Why |
|---|---|---|
| $\sin^{-1}(\sin\frac{2\pi}{3})$ | $\frac{\pi}{3}$ | $\frac{\pi}{3}$ lies in $[-\frac{\pi}{2},\frac{\pi}{2}]$ and has the same sine. |
| $\cos^{-1}(\cos\frac{7\pi}{6})$ | $\frac{5\pi}{6}$ | $\frac{5\pi}{6}$ lies in $[0,\pi]$ and has the same cosine. |
| $\tan^{-1}(\tan\frac{3\pi}{4})$ | $-\frac{\pi}{4}$ | Tangent has period $\pi$; the result must lie in $(-\frac{\pi}{2},\frac{\pi}{2})$. |
Optional Identity Practice
The following identities extend the current syllabus checklist. Use them when they are included in your class material, and keep each stated domain condition.
Reciprocal and Complementary Identities
$\cos^{-1}(\frac{1}{x})=\sec^{-1}x,\quad |x|\ge1$
$\tan^{-1}(\frac{1}{x})=\cot^{-1}x,\quad x>0$
$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2},\quad x\in[-1,1]$
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2},\quad x\in\mathbb{R}$
$\sec^{-1}x+\csc^{-1}x=\frac{\pi}{2},\quad |x|\ge1$
Inverse-Tangent Sum and Difference
$\tan^{-1}x-\tan^{-1}y=\tan^{-1}(\frac{x-y}{1+xy}),\quad xy>-1$
When these conditions fail, check the angle ranges separately; a $\pi$ correction can be required. At $xy=1$ or $xy=-1$, the corresponding sum or difference is $\pm\frac{\pi}{2}$.
Equivalent Forms of $2\tan^{-1}x$
| Form | Condition |
|---|---|
| $2\tan^{-1}x=\sin^{-1}(\frac{2x}{1+x^2})$ | $|x|\le1$ |
| $2\tan^{-1}x=\cos^{-1}(\frac{1-x^2}{1+x^2})$ | $x\ge0$ |
| $2\tan^{-1}x=\tan^{-1}(\frac{2x}{1-x^2})$ | $-1<x<1$ |
Substitutions for Radicals
For $a>0$, these substitutions can turn a radical expression into a trigonometric ratio. Choose the angle range so the square root stays nonnegative.
- $\sqrt{a^2-x^2}$: set $x=a\sin\theta$ or $x=a\cos\theta$.
- $\sqrt{a^2+x^2}$: set $x=a\tan\theta$ or $x=a\cot\theta$.
- $\sqrt{x^2-a^2}$: set $x=a\sec\theta$ or $x=a\csc\theta$.
- $\sqrt{\frac{a-x}{a+x}}$ or $\sqrt{\frac{a+x}{a-x}}$: set $x=a\cos 2\theta$, with the domain checked before substitution.
Worked Examples
Example 1. Find the principal value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$.
$\tan^{-1}(\sqrt{3}) = \frac{π}{3}$ (since $\tan \frac{π}{3} = \sqrt{3}$)
$\sec^{-1}(-2) = π - \sec^{-1}(2)$ (using property)
$= π - \frac{π}{3} = \frac{2π}{3}$
Expression $= \frac{π}{3} - \frac{2π}{3}$
$= -\frac{π}{3}$
Example 2. Simplify $\tan^{-1}(\frac{\sqrt{1+x^2}-1}{x})$, where $x\ne0$.
Put $x = \tanθ \Rightarrow θ = \tan^{-1}x$.
Expression becomes $\tan^{-1}(\frac{\sqrt{1+\tan^2θ}-1}{\tanθ}) = \tan^{-1}(\frac{\secθ-1}{\tanθ})$.
$= \tan^{-1}(\frac{\frac{1}{\cosθ}-1}{\frac{\sinθ}{\cosθ}}) = \tan^{-1}(\frac{1-\cosθ}{\sinθ})$
Using half-angle formulas: $1-\cosθ = 2\sin^2\frac{θ}{2}$ and $\sinθ = 2\sin\frac{θ}{2}\cos\frac{θ}{2}$.
$= \tan^{-1}(\frac{2\sin^2\frac{θ}{2}}{2\sin\frac{θ}{2}\cos\frac{θ}{2}}) = \tan^{-1}(\tan\frac{θ}{2})$.
Because $\theta=\tan^{-1}x\in(-\frac{\pi}{2},\frac{\pi}{2})$, we have $\frac{\theta}{2}\in(-\frac{\pi}{4},\frac{\pi}{4})$. Thus $\tan^{-1}(\tan\frac{\theta}{2})=\frac{\theta}{2}=\frac{1}{2}\tan^{-1}x$.
Previous Year Questions (PYQs)
(A) $\frac{3π}{5}$ (B) $\frac{-7π}{5}$ (C) $\frac{π}{10}$ (D) $\frac{-π}{10}$
Ans: (D). $\cos\frac{33π}{5} = \cos(6π + \frac{3π}{5}) = \cos\frac{3π}{5} = \sin(\frac{π}{2} - \frac{3π}{5}) = \sin(\frac{-π}{10})$. So $\sin^{-1}(\sin\frac{-π}{10}) = \frac{-π}{10}$.
Ans: Let $\theta=\tan^{-1}x$. Then $\theta\in(-\frac{\pi}{2},\frac{\pi}{2})$, so $\cos\theta>0$. Using $\tan\theta=x$ and $1+\tan^2\theta=\sec^2\theta$:
$\sin\theta=\frac{\tan\theta}{\sec\theta}=\frac{x}{\sqrt{1+x^2}}$. The signed formula also holds for negative $x$.
Solution: The no-correction addition formula applies when $(2x)(3x)=6x^2<1$. It gives $\frac{5x}{1-6x^2}=1$, hence $6x^2+5x-1=0=(6x-1)(x+1)$. The candidates are $x=\frac{1}{6}$ and $x=-1$; only $x=\frac{1}{6}$ satisfies $6x^2<1$. Check: $\tan^{-1}(\frac{1}{3})+\tan^{-1}(\frac{1}{2})=\tan^{-1}(1)=\frac{\pi}{4}$.
Common Errors and Useful Checks
Common Errors
Checks Before You Finish
Practice Problems (Self-Assessment)
Level 1: Basic
Q1. Find the principal value of $\cos^{-1}(-\frac{1}{2})$.
Use $\cos^{-1}(-x)=\pi-\cos^{-1}x$.
$\cos^{-1}(-\frac{1}{2}) = π - \cos^{-1}(\frac{1}{2})$
Which angle has $\cos = 1/2$? It is $60^°$ or $\frac{π}{3}$.
$= π - \frac{π}{3} = \frac{3π - π}{3} = \frac{2π}{3}$.
Q2. Evaluate $\tan^{-1}(1) + \cos^{-1}(-\frac{1}{2}) + \sin^{-1}(-\frac{1}{2})$.
1. $\tan^{-1}(1) = \frac{π}{4}$
2. $\cos^{-1}(-\frac{1}{2}) = π - \frac{π}{3} = \frac{2π}{3}$
3. $\sin^{-1}(-\frac{1}{2}) = -\frac{π}{6}$
$= \frac{π}{4} + \frac{2π}{3} - \frac{π}{6}$
LCM of 4, 3, 6 is 12.
$= \frac{3π + 8π - 2π}{12} = \frac{9π}{12} = \frac{3π}{4}$.
Level 2: Intermediate
Q3. Prove that $3\sin^{-1}x = \sin^{-1}(3x - 4x^3), x \in [-\frac{1}{2}, \frac{1}{2}]$.
Put $x = \sinθ \Rightarrow θ = \sin^{-1}x$.
$3\sin\theta-4\sin^3\theta=\sin3\theta$.
Since $x\in[-\frac{1}{2},\frac{1}{2}]$, $\theta=\sin^{-1}x\in[-\frac{\pi}{6},\frac{\pi}{6}]$. Thus $3\theta$ lies in the principal range $[-\frac{\pi}{2},\frac{\pi}{2}]$.
Therefore $\sin^{-1}(\sin3\theta)=3\theta$.
Substitute back $θ$: $= 3\sin^{-1}x = LHS$.
Hence Proved.
Q4. Express $\tan^{-1}(\frac{\cos x}{1-\sin x})$ in simplest form for $-\frac{\pi}{2}<x<\frac{\pi}{2}$.
$\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}$
$1 - \sin x = (\cos\frac{x}{2} - \sin\frac{x}{2})^2$
Expression $= \tan^{-1}( \frac{(\cos\frac{x}{2}-\sin\frac{x}{2})(\cos\frac{x}{2}+\sin\frac{x}{2})}{(\cos\frac{x}{2}-\sin\frac{x}{2})^2} )$
$= \tan^{-1}( \frac{\cos\frac{x}{2}+\sin\frac{x}{2}}{\cos\frac{x}{2}-\sin\frac{x}{2}} )$
Divide the numerator and denominator by $\cos\frac{x}{2}$:
$= \tan^{-1}( \frac{1+\tan\frac{x}{2}}{1-\tan\frac{x}{2}} )$
$= \tan^{-1}( \tan(\frac{\pi}{4} + \frac{x}{2}) )$
On the stated interval, $\frac{\pi}{4}+\frac{x}{2}\in(0,\frac{\pi}{2})$, which is the principal range of $\tan^{-1}$. Therefore the answer is $\frac{\pi}{4}+\frac{x}{2}$.
Level 3: Challenge
Q5. Solve for $x$: $\tan^{-1}(\frac{x-1}{x-2}) + \tan^{-1}(\frac{x+1}{x+2}) = \frac{π}{4}$.
$\tan^{-1}A + \tan^{-1}B = \tan^{-1}(\frac{A+B}{1-AB})$
$\frac{x-1}{x-2} + \frac{x+1}{x+2} = \frac{(x-1)(x+2) + (x+1)(x-2)}{(x-2)(x+2)}$
$A+B=\frac{2x^2-4}{x^2-4}$.
$1-AB=\frac{(x^2-4)-(x^2-1)}{x^2-4}=\frac{-3}{x^2-4}$, so $\frac{A+B}{1-AB}=\frac{2x^2-4}{-3}$.
$\frac{2x^2 - 4}{-3} = \tan(\frac{π}{4}) = 1$
$2x^2 - 4 = -3 \Rightarrow 2x^2 = 1$
$x^2 = 1/2 \Rightarrow x = \pm \frac{1}{\sqrt{2}}$.
At either candidate, $x\ne\pm2$ and $AB=\frac{x^2-1}{x^2-4}=\frac{1}{7}<1$. The addition formula therefore has no $\pi$ correction, and both values satisfy the original equation.
10-Mark Chapter Check
25 minAttempt all 4 practice questions (10 marks total) before opening the worked solutions. Suggested time: 25 minutes.
(a) $π$ (b) $\frac{2π}{3}$ (c) $\frac{4π}{3}$ (d) $\frac{π}{3}$
$2\sin^{-1}(\frac{1}{2}) = 2(\frac{π}{6}) = \frac{π}{3}$.
Sum $= \frac{2π}{3} + \frac{π}{3} = \frac{3π}{3} = π$.
Correct Option: (a) $π$ [1 Mark]
- $\frac{7π}{6} \notin (-\frac{π}{2}, \frac{π}{2})$. Since $\tan(\frac{7π}{6}) = \tan(π + \frac{π}{6}) = \tan\frac{π}{6}$, we have $\tan^{-1}(\tan \frac{7π}{6}) = \frac{π}{6}$ [1 Mark].
- $\frac{7π}{6} \notin [0, π]$. Since $\cos(\frac{7π}{6}) = \cos(2π - \frac{5π}{6}) = \cos\frac{5π}{6}$, we have $\cos^{-1}(\cos \frac{7π}{6}) = \frac{5π}{6}$ [1 Mark].
Sum $= \frac{π}{6} + \frac{5π}{6} = \frac{6π}{6} = π$.
Answer: $π$ [2 Marks]
$$\frac{\sqrt{1 + x^2} - 1}{x} = \frac{\sqrt{1 + \tan^2θ} - 1}{\tanθ} = \frac{\secθ - 1}{\tanθ} = \frac{\frac{1}{\cosθ} - 1}{\frac{\sinθ}{\cosθ}} = \frac{1 - \cosθ}{\sinθ}$$ $$= \frac{2\sin^2(θ/2)}{2\sin(θ/2)\cos(θ/2)} = \frac{\sin(θ/2)}{\cos(θ/2)} = \tan(\frac{θ}{2}) [1 Mark]$$ Therefore, $\tan^{-1}(\tan\frac{θ}{2}) = \frac{θ}{2} = \frac{1}{2}\tan^{-1}x$ [1 Mark].
Answer: $\frac{1}{2}\tan^{-1}x$ [3 Marks]
Part 1 (2 Marks): Divide numerator and denominator by $\cos x$:
$$\tan^{-1}(\frac{1 - \tan x}{1 + \tan x}) = \tan^{-1}(\frac{\tan\frac{π}{4} - \tan x}{1 + \tan\frac{π}{4}\tan x}) = \tan^{-1}(\tan(\frac{π}{4} - x)) = \frac{π}{4} - x$$ Part 2 (2 Marks):
$\sin^{-1}(-\frac{1}{2}) = -\sin^{-1}(\frac{1}{2}) = -\frac{π}{6}$.
$$\sin(\frac{π}{3} - (-\frac{π}{6})) = \sin(\frac{π}{3} + \frac{π}{6}) = \sin(\frac{3π}{6}) = \sin(\frac{π}{2}) = 1$$ Answer: Part 1: $\frac{π}{4} - x$; Part 2: $1$ [4 Marks]