Inverse Circular Functions | NEB Class 11 Mathematics Notes

Chapter Overview

Welcome to one of the most scoring chapters in NEB Class 11 Mathematics! Think of Inverse Circular Functions as the 'undo' button for trigonometry. If $\sin \theta$ gives you a ratio, the inverse function helps you find the angle $\theta$ back. This chapter is the critical bridge to Calculus—without it, complex integration and differentiation are impossible. We'll explore how to restrict domains to make these functions work and how to solve equations involving them.

What this chapter covers

Invertible Functions and One-to-One correspondencePrincipal Values and Range RestrictionsThe Golden Table (Domain and Range)Fundamental Properties of Inverse FunctionsComplementary Angle IdentitiesAddition and Subtraction Formulas ($\tan^{-1}x \pm \tan^{-1}y$)Converting Inverse Functions using Right-Angled TrianglesSolving Inverse Trigonometric Equations

Learning Objectives

1

Understand why we must restrict the domains of trigonometric functions (Sine, Cosine, etc.) to make them invertible.

2

Identify the Principal Value Branch and calculate principal values for any given ratio.

3

Memorize the 'Golden Table' of Domains and Ranges for all six inverse functions.

4

Apply addition and subtraction theorems (like $\tan^{-1}x + \tan^{-1}y$) to simplify complex expressions.

5

Use the Right-Angled Triangle method to convert one inverse function into another (e.g., $\sin^{-1}x$ to $\sec^{-1}x$).

6

Solve trigonometric equations involving inverse functions by taking trig functions of both sides.

The 'Undo' Button

Key Insight

Inverse functions are all about reversibility, but they require a strict rule to work.

In basic trig, $\sin 30^\circ = 1/2$. The inverse, $\sin^{-1}(1/2)$, asks: 'Which angle has a sine of 1/2?'

However, there are infinite angles with that ratio ($30^\circ, 150^\circ, 390^\circ...$). To be a function, we can only have one answer. That is why we restrict the 'Principal Range'.

Example: For $\sin^{-1}x$, we only look at angles from $-90^\circ$ to $90^\circ$ ($-\pi/2$ to $\pi/2$). This ensures every input gives a unique output.

Key Definitions

Invertible Function

A function that is both one-to-one (every input has a unique output) and onto (every possible output is mapped). Only one-to-one functions have inverses.

Inverse Circular Function

If $\sin \theta = x$, then $\theta = \sin^{-1}x$ (or $\text{arc sine } x$). It represents the angle whose sine is $x$.

Principal Value

The numerical value of the angle that lies within the specific restricted range (principal value branch) of the inverse function. It is the 'least' or 'primary' angle satisfying the equation.

The Golden Table (Domain & Range)

You must memorize this table. The 'Domain' is the allowed input (x), and the 'Range' is the allowed output angle (y).

Inverse Function

$$Notation$$

$y = \sin^{-1}x$

$y = \cos^{-1}x$

$y = \tan^{-1}x$

$y = \cot^{-1}x$

$y = \csc^{-1}x$

$y = \sec^{-1}x$

VS

Domain & Range

$$Allowed Values$$

Domain: $[-1, 1]$
Range: $[-\frac{\pi}{2}, \frac{\pi}{2}]$

Domain: $[-1, 1]$
Range: $[0, \pi]$

Domain: $R$ (All Real)
Range: $(-\frac{\pi}{2}, \frac{\pi}{2})$

Domain: $R$ (All Real)
Range: $(0, \pi)$

Domain: $R - (-1, 1)$
Range: $[-\frac{\pi}{2}, \frac{\pi}{2}] - \{0\}$

Domain: $R - (-1, 1)$
Range: $[0, \pi] - \{\frac{\pi}{2}\}$

Memory Tip: The 'Co' functions ($cos, cot, csc$) have ranges generally in the positive quadrants ($0$ to $\pi$), while the non-Co functions range from negative to positive ($-\pi/2$ to $\pi/2$).

Essential Formulas & Derivations

Fundamental Properties

\sin(\sin^{-1}x) = x, \quad \sin^{-1}(\sin \theta) = \theta

Valid only if $\theta$ is in the principal range of $\sin^{-1}$.

1By definition, if $y = \sin^{-1}x$, then $x = \sin y$.

2Substituting $\sin^{-1}x$ for $y$ in the sine function:

3$$ \sin(\sin^{-1}x) = \sin(y) = x $$

4The reverse, $\sin^{-1}(\sin \theta) = \theta$, is only true if $\theta$ is between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$. If $\theta$ is outside this range, you must find an equivalent angle inside the range.

Complementary Identities

\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}

The sum of inverse sine and inverse cosine of the same value is always $90^\circ$ ($\pi/2$). Also applies to $\tan^{-1}x + \cot^{-1}x$.

1Let $\sin^{-1}x = \theta$. This implies $\sin \theta = x$.

2We know from basic trigonometry that $\sin \theta = \cos(\frac{\pi}{2} - \theta)$.

3Therefore, $x = \cos(\frac{\pi}{2} - \theta)$.

4Taking the inverse cosine of both sides: $\cos^{-1}x = \frac{\pi}{2} - \theta$.

5Substitute $\theta = \sin^{-1}x$ back into the equation:

6$$ \cos^{-1}x = \frac{\pi}{2} - \sin^{-1}x \implies \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} $$

Addition Formula (Tangent)

\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)

This is true only if $xy < 1$. If $xy > 1$, the answer is $\tan^{-1}\left(\frac{x+y}{1-xy}\right) + \pi$.

1Let $\tan^{-1}x = A$ and $\tan^{-1}y = B$. Then, $\tan A = x$ and $\tan B = y$.

2We need to find $A + B$. We know $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$.

3Substitute $x$ and $y$: $\tan(A+B) = \frac{x+y}{1-xy}$.

4Taking the inverse tangent of both sides gives the result.

5$$ A + B = \tan^{-1}\left(\frac{x+y}{1-xy}\right) $$

Question Recognition Strategy

When you see a question in the exam, use this flow to decide your method.

Is the question asking for the 'Principal Value'?

Yes

Check the Domain (input validity) and find the angle within the specific Range (e.g., $[-\pi/2, \pi/2]$ for sine).

No

Continue to next step ↓

Is the question asking to 'Prove the Identity' (LHS = RHS)?

Yes

Use the Addition/Subtraction formulas. If the functions are different (e.g., sin and tan), use the Right-Angled Triangle method to convert them.

No

Continue to next step ↓

Is the question asking to 'Solve for x'?

Yes

Apply formulas to combine the inverse terms into a single term (e.g., $\tan^{-1}A$), then take the trig function (tan) of both sides to remove the inverse and solve the algebraic equation.

No

undefined

The 'Co' Rule & Negation

Trick: How to instantly remember the Ranges and Negative inputs.

Steps to Remember:

For Principal Ranges: Functions starting with 'Co' ($\cos^{-1}, \cot^{-1}, \sec^{-1}$) live in the upper quadrants ($0$ to $\pi$). All others live around the x-axis ($-\pi/2$ to $\pi/2$).
For Negative Inputs ($-x$):
1. Odd functions ($\sin, \tan, \csc$) just take the negative sign out: $\sin^{-1}(-x) = -\sin^{-1}x$.
2. Even-like 'Co' functions use $\pi$: $\cos^{-1}(-x) = \pi - \cos^{-1}x$.

Mnemonic: 'Co' stands for 'Complete' (0 to pi). 'Non-Co' stands for 'Negative allowed' (-pi/2 to pi/2).

Common Mistakes in NEB Exams

Avoid these traps to secure your marks!

The Range Trap: Writing $\sin^{-1}(\sin \frac{2\pi}{3}) = \frac{2\pi}{3}$. This is WRONG because $2\pi/3$ is not in the range $[-\pi/2, \pi/2]$. You must adjust the angle.
The $\tan^{-1}$ Trap ($xy > 1$): When using $\tan^{-1}x + \tan^{-1}y$, if $xy > 1$, you must add $\pi$ to your answer because the angle shifts quadrants.
Formula Misuse: Using the wrong addition formula for sine and cosine. Remember, sine addition involves square roots ($\sqrt{1-x^2}$), while tangent does not.
Forgetting the Triangle: Trying to memorize every conversion ($\sin^{-1}$ to $\tan^{-1}$). It is safer and faster to draw a simple right-angled triangle (p, b, h) and find the missing sides.

Solved Examples (Easy → Hard)

1

Find the principal value of $\sin^{-1}(-\frac{1}{2})$.

Easy

Solution

Let's solve this step by step.
1. Let $\sin^{-1}(-\frac{1}{2}) = y$.
2. This implies $\sin y = -\frac{1}{2}$.
3. We know $\sin(\frac{\pi}{6}) = \frac{1}{2}$, so $\sin(-\frac{\pi}{6}) = -\frac{1}{2}$.
4. Check the range for $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $-\frac{\pi}{6}$ falls in this interval, it is the principal value.

∴ $-\frac{\pi}{6}$

2

Evaluate $\cos(\tan^{-1}\frac{3}{4})$.

Medium

Solution

We can use the Right-Angled Triangle method here.
1. Let $\tan^{-1}\frac{3}{4} = \theta$. This means $\tan \theta = \frac{3}{4} = \frac{\text{Perpendicular}}{\text{Base}}$.
2. Draw a triangle with Perpendicular $p=3$ and Base $b=4$.
3. Find Hypotenuse $h$ using Pythagoras: $h = \sqrt{3^2 + 4^2} = 5$.
4. We need $\cos \theta$. By definition, $\cos \theta = \frac{b}{h} = \frac{4}{5}$.

∴ $\frac{4}{5}$

3

Solve $\tan^{-1}x + \tan^{-1}2x = \frac{\pi}{4}$.

Hard

Solution

This involves an equation with inverse tangents.
1. Apply the addition formula: $\tan^{-1}\left(\frac{x + 2x}{1 - x(2x)}\right) = \frac{\pi}{4}$
2. Simplify the argument: $\tan^{-1}\left(\frac{3x}{1 - 2x^2}\right) = \frac{\pi}{4}$.
3. Take the tangent of both sides to remove the inverse: $\frac{3x}{1 - 2x^2} = \tan(\frac{\pi}{4}) = 1$.
4. Cross-multiply: $3x = 1 - 2x^2 \implies 2x^2 + 3x - 1 = 0$.
5. Use the quadratic formula to find $x$. (Note: You must check if the solution satisfies $x(2x) < 1$ or adjust with $\pi$ if not).

∴ $x = \frac{-3 \pm \sqrt{17}}{4}$

MCQ Practice (Original Variations)

Mixed Level — MCQ

1

The principal value of $\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ is:

Correct!

Incorrect. Correct: $-\frac{\pi}{3}$

Solution

We know the range for $\sin^{-1}x$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Since $\sin(\pi/3) = \sqrt{3}/2$, then $\sin(-\pi/3) = -\sqrt{3}/2$. This value falls within the required range.

∴ $-\frac{\pi}{3}$

2

If $\cos^{-1}x = \frac{\pi}{3}$, then the value of $\sin^{-1}x$ is:

Correct!

Incorrect. Correct: $\frac{\pi}{6}$

Solution

We use the identity $\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2}$. So, $\sin^{-1}x = \frac{\pi}{2} - \frac{\pi}{3} = \frac{3\pi - 2\pi}{6} = \frac{\pi}{6}$.

∴ $\frac{\pi}{6}$

3

The value of $\tan\left[\tan^{-1}(4) + \tan^{-1}(1)\right]$ is:

Correct!

Incorrect. Correct: $-5/3$

Solution

Using $\tan(A+B)$ logic or the inverse formula: $\tan^{-1}\left(\frac{4+1}{1-4(1)}\right) = \tan^{-1}\left(\frac{5}{-3}\right)$. Taking the tangent of this gives $-5/3$.

∴ $-\frac{5}{3}$

4

What is the domain of the function $y = \cos^{-1}x$?

Correct!

Incorrect. Correct: $[-1, 1]$

Solution

The domain refers to the input $x$. For both sine and cosine inverse, the input must be between -1 and 1 inclusive.

∴ $[-1, 1]$

5

The value of $\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)$ is:

Correct!

Incorrect. Correct: $\frac{\pi}{6}$

Solution

$\sec^{-1}(x) = \cos^{-1}(1/x)$. So, $\cos^{-1}(\frac{\sqrt{3}}{2}) = \frac{\pi}{6}$ because $\cos(30^\circ) = \sqrt{3}/2$.

∴ $\frac{\pi}{6}$

6

If $\tan^{-1}x + \cot^{-1}(3) = \frac{\pi}{2}$, then $x$ is:

Correct!

Incorrect. Correct: $3$

Solution

Since $\tan^{-1}x + \cot^{-1}x = \frac{\pi}{2}$, for the equation to hold true, $x$ must be equal to the argument of the cotangent inverse, which is 3.

∴ $3$

7

The principal value of $\cos^{-1}\left(\cos \frac{7\pi}{6}\right)$ is:

Correct!

Incorrect. Correct: $\frac{5\pi}{6}$

Solution

The range of $\cos^{-1}$ is $[0, \pi]$. $\frac{7\pi}{6}$ is outside this. We rewrite: $\cos(2\pi - \frac{5\pi}{6}) = \cos(\frac{5\pi}{6})$. Now $5\pi/6$ is in range.

∴ $\frac{5\pi}{6}$

8

The value of $2\tan^{-1}(1/3)$ in terms of $\tan^{-1}$ is:

Correct!

Incorrect. Correct: $\tan^{-1}(3/4)

Solution

Use formula $2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right)$. Here $x=1/3$, so $\frac{2/3}{1-1/9} = \frac{2/3}{8/9} = \frac{2}{3} \cdot \frac{9}{8} = \frac{3}{4}$.

∴ $\tan^{-1}\left(\frac{3}{4}\right)$

9

$\sin^{-1}x$ is equal to:

Correct!

Incorrect. Correct: $\csc^{-1}(1/x)$

Solution

This is a reciprocal property: $\sin^{-1}(1/x) = \csc^{-1}x$, so $\sin^{-1}x = \csc^{-1}(1/x)$.

∴ $\csc^{-1}(1/x)$

10

If $xy > 1$, then $\tan^{-1}x + \tan^{-1}y$ is equal to:

Correct!

Incorrect. Correct: $\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$

Solution

This is a special restriction. When the product $xy$ exceeds 1, the angle moves into the next quadrant, requiring the addition of $\pi$.

∴ $\pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right)$

Short Questions (2 Marks Each)

1

Evaluate: $\sin(\cos^{-1} \frac{3}{5})$

[2 marks]

Solution

Let $\cos^{-1}\frac{3}{5} = \theta$. This means $\cos \theta = \frac{3}{5}$.
We know $\sin \theta = \sqrt{1 - \cos^2 \theta}$.
$\sin \theta = \sqrt{1 - (\frac{3}{5})^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$.

∴ $\frac{4}{5}$

2

Prove that $\tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \tan^{-1}\frac{1}{2}$.

[2 marks]

Solution

Use formula: $\tan^{-1}x + \tan^{-1}y = \tan^{-1}\left(\frac{x+y}{1-xy}\right)$.
LHS = $\tan^{-1}\left(\frac{1/4 + 2/9}{1 - (1/4)(2/9)}\right)$.
Simplify numerator: $\frac{9+8}{36} = \frac{17}{36}$.
Simplify denominator: $1 - \frac{2}{36} = \frac{34}{36}$.
LHS = $\tan^{-1}\left(\frac{17/36}{34/36}\right) = \tan^{-1}(\frac{17}{34}) = \tan^{-1}(1/2) = \text{RHS}$.

∴ Proved as $\tan^{-1}(1/2)$

3

Find the principal value of $\cos^{-1}(-1/\sqrt{2})$.

[2 marks]

Solution

Let $\cos^{-1}(-1/\sqrt{2}) = y$. So, $\cos y = -1/\sqrt{2}$.
We know $\cos(\pi/4) = 1/\sqrt{2}$.
For negative values in $\cos^{-1}$, we use the range $[0, \pi]$, so $y = \pi - \pi/4$.
$y = 3\pi/4$.

∴ $\frac{3\pi}{4}$

4

Evaluate: $\tan(2\tan^{-1} \frac{1}{5})$.

[2 marks]

Solution

First, find $2\tan^{-1}(1/5)$. Using $\tan^{-1}(\frac{2x}{1-x^2})$:
$2\tan^{-1}(1/5) = \tan^{-1}(\frac{2/5}{1-1/25}) = \tan^{-1}(\frac{2/5}{24/25}) = \tan^{-1}(\frac{5}{12})$.
Now, $\tan(\tan^{-1} \frac{5}{12}) = 5/12$.

∴ $\frac{5}{12}$

5

Prove that $\sin^{-1}x = \cos^{-1}\sqrt{1-x^2}$.

[2 marks]

Solution

Let $\sin^{-1}x = \theta$. Then $\sin \theta = x$.
We know $\cos^2 \theta = 1 - \sin^2 \theta = 1 - x^2$.
Taking square root: $\cos \theta = \sqrt{1-x^2}$.
Therefore, $\theta = \cos^{-1}\sqrt{1-x^2}$.

∴ Proved

6

Solve for $x$: $\tan^{-1}x = \cot^{-1}(4)$.

[2 marks]

Solution

We know $\cot^{-1}x = \tan^{-1}(1/x)$.
So, $\cot^{-1}(4) = \tan^{-1}(1/4)$.
The equation becomes $\tan^{-1}x = \tan^{-1}(1/4)$.
Comparing both sides, $x = 1/4$.

∴ $x = \frac{1}{4}$

7

Evaluate: $\sin(2\sin^{-1} 0.6)$.

[2 marks]

Solution

Let $\sin^{-1} 0.6 = \theta$. Then $\sin \theta = 0.6 = 3/5$.
We need to find $\sin 2\theta$. Formula: $\sin 2\theta = 2\sin \theta\cos \theta$.
Find $\cos \theta$: $\sqrt{1 - (3/5)^2} = 4/5$.
$\sin 2\theta = 2 \cdot (3/5) \cdot (4/5) = 24/25 = 0.96$.

∴ $0.96$

8

If $\sin^{-1}x = \pi/5$, find $\cos^{-1}x$.

[2 marks]

Solution

Identity: $\sin^{-1}x + \cos^{-1}x = \pi/2$.
$\pi/5 + \cos^{-1}x = \pi/2$.
$\cos^{-1}x = \pi/2 - \pi/5 = \frac{5\pi - 2\pi}{10} = 3\pi/10$.

∴ $\frac{3\pi}{10}$

9

Evaluate: $\cos(\tan^{-1} 1 + \tan^{-1} 0)$.

[2 marks]

Solution

$\tan^{-1} 1 = \pi/4$ and $\tan^{-1} 0 = 0$.
The expression is $\cos(\pi/4 + 0) = \cos(\pi/4)$.
Result: $1/\sqrt{2}$.

∴ $\frac{1}{\sqrt{2}}$

10

Express $\tan(\text{arc}\sin x)$ in terms of $x$.

[2 marks]

Solution

Let $\text{arc}\sin x = \theta$. Then $\sin \theta = x = p/h$.
Consider a triangle: $p=x, h=1$. Then base $b = \sqrt{1-x^2}$.
$\tan \theta = p/b = \frac{x}{\sqrt{1-x^2}}$.

∴ $\frac{x}{\sqrt{1-x^2}}$

Long Questions (3-4 Marks Each)

1

If $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \frac{\pi}{2}$, prove that $xy + yz + zx = 1$.

[4 marks]

Solution

Given: $\tan^{-1}x + \tan^{-1}y = \frac{\pi}{2} - \tan^{-1}z$.
We know $\frac{\pi}{2} - \tan^{-1}z = \cot^{-1}z$.
Use addition formula: $\tan^{-1}\left(\frac{x+y}{1-xy}\right) = \cot^{-1}z$.
Convert $\cot^{-1}z$ to $\tan^{-1}(1/z)$: $\tan^{-1}\left(\frac{x+y}{1-xy}\right) = \tan^{-1}(1/z)$.
Equate arguments: $\frac{x+y}{1-xy} = \frac{1}{z}$.
Cross multiply: $z(x+y) = 1-xy \Rightarrow zx + zy = 1 - xy$.
Final Form: $xy + yz + zx = 1$. Proved.

∴ Proved as $xy + yz + zx = 1$

2

Solve for $x$: $\tan^{-1}(x-1) + \tan^{-1}x + \tan^{-1}(x+1) = \tan^{-1}3x$.

[4 marks]

Solution

Rearrange: $\tan^{-1}(x-1) + \tan^{-1}(x+1) = \tan^{-1}3x - \tan^{-1}x$.
Apply formula to both sides: $\tan^{-1}\left(\frac{(x-1)+(x+1)}{1-(x-1)(x+1)}\right) = \tan^{-1}\left(\frac{3x-x}{1+3x^2}\right)$.
$\tan^{-1}\left(\frac{2x}{1-(x^2-1)}\right) = \tan^{-1}\left(\frac{2x}{1+3x^2}\right)$.
$\frac{2x}{2-x^2} = \frac{2x}{1+3x^2}$.
Case 1: $2x = 0 \Rightarrow x = 0$.
Case 2: $2-x^2 = 1+3x^2 \Rightarrow 1 = 4x^2 \Rightarrow x = \pm 1/2$.
Solutions: $x = 0, \pm 1/2$.

∴ $x = 0, \pm \frac{1}{2}$

3

Prove that: $\sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} + \sin^{-1}\frac{16}{65} = \frac{\pi}{2}$.

[4 marks]

Solution

Let's combine the first two using $\sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2})$.
$\sin^{-1}\left(\frac{4}{5}\sqrt{1-(\frac{5}{13})^2} + \frac{5}{13}\sqrt{1-(\frac{4}{5})^2}\right)$.
$\sin^{-1}\left(\frac{4}{5} \cdot \frac{12}{13} + \frac{5}{13} \cdot \frac{3}{5}\right) = \sin^{-1}(\frac{48+15}{65}) = \sin^{-1}(\frac{63}{65})$.
Now LHS becomes: $\sin^{-1}\frac{63}{65} + \sin^{-1}\frac{16}{65}$.
Since $\sin^{-1}x = \cos^{-1}\sqrt{1-x^2}$, $\sin^{-1}\frac{16}{65} = \cos^{-1}(\frac{63}{65})$.
LHS: $\sin^{-1}\frac{63}{65} + \cos^{-1}\frac{63}{65} = \frac{\pi}{2}$. Proved.

∴ Proved

4

If $\cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi$, prove $x^2 + y^2 + z^2 + 2xyz = 1$.

[4 marks]

Solution

$\cos^{-1}x + \cos^{-1}y = \pi - \cos^{-1}z$.
$\cos^{-1}(xy - \sqrt{1-x^2}\sqrt{1-y^2}) = \pi - \cos^{-1}z$.
Take $\cos$ of both sides: $xy - \sqrt{1-x^2}\sqrt{1-y^2} = \cos(\pi - \cos^{-1}z)$.
$xy - \sqrt{(1-x^2)(1-y^2)} = -z$ (since $\cos(\pi - \theta) = -\cos \theta$).
$xy + z = \sqrt{(1-x^2)(1-y^2)}$.
Squaring: $(xy+z)^2 = (1-x^2)(1-y^2)$.
$x^2y^2 + 2xyz + z^2 = 1 - x^2 - y^2 + x^2y^2$.
Cancelling $x^2y^2$ and rearranging: $x^2 + y^2 + z^2 + 2xyz = 1$.

∴ Proved

5

Prove that $2\tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{7} = \frac{\pi}{4}$.

[4 marks]

Solution

$2\tan^{-1}(1/3) = \tan^{-1}\left(\frac{2/3}{1-1/9}\right) = \tan^{-1}\left(\frac{2/3}{8/9}\right) = \tan^{-1}(3/4)$.
Now, $\tan^{-1}(3/4) + \tan^{-1}(1/7) = \tan^{-1}\left(\frac{3/4 + 1/7}{1 - (3/4)(1/7)}\right)$.
Numerator: $\frac{21+4}{28} = \frac{25}{28}$.
Denominator: $1 - \frac{3}{28} = \frac{25}{28}$.
$\tan^{-1}(\frac{25/28}{25/28}) = \tan^{-1}(1) = \frac{\pi}{4}$. Proved.

∴ Proved

6

Solve: $\sin^{-1}\frac{2a}{1+a^2} + \sin^{-1}\frac{2b}{1+b^2} = 2\tan^{-1}x$.

[4 marks]

Solution

We know $2\tan^{-1}a = \sin^{-1}\frac{2a}{1+a^2}$.
So, $2\tan^{-1}a + 2\tan^{-1}b = 2\tan^{-1}x$.
$\tan^{-1}a + \tan^{-1}b = \tan^{-1}x$.
$\tan^{-1}\left(\frac{a+b}{1-ab}\right) = \tan^{-1}x$.
$x = \frac{a+b}{1-ab}$.

∴ $x = \frac{a+b}{1-ab}$

7

If $\sin^{-1}x + \sin^{-1}y + \sin^{-1}z = \pi$, prove that $x\sqrt{1-x^2} + y\sqrt{1-y^2} + z\sqrt{1-z^2} = 2xyz$.

[4 marks]

Solution

Let $\sin^{-1}x = A, \sin^{-1}y = B, \sin^{-1}z = C$. Then $A+B+C = \pi$.
Also $x = \sin A, y = \sin B, z = \sin C$.
Expression: $\sin A \cos A + \sin B \cos B + \sin C \cos C$.
Multiply by 2: $\sin 2A + \sin 2B + \sin 2C$.
From trigonometry identities, if $A+B+C = \pi$, then $\sin 2A + \sin 2B + \sin 2C = 4\sin A \sin B \sin C$.
Dividing back by 2: $2\sin A \sin B \sin C = 2xyz$. Proved.

∴ Proved

8

Prove: $\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)$.

[3 marks]

Solution

Let $\tan^{-1}\sqrt{x} = \theta$. Then $\tan \theta = \sqrt{x} \Rightarrow \tan^2 \theta = x$.
We know $\cos 2\theta = \frac{1-\tan^2 \theta}{1+\tan^2 \theta} = \frac{1-x}{1+x}$.
$2\theta = \cos^{-1}\left(\frac{1-x}{1+x}\right)$.
$\theta = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)$.
Substitute $\theta$: $\tan^{-1}\sqrt{x} = \frac{1}{2}\cos^{-1}\left(\frac{1-x}{1+x}\right)$.

∴ Proved

9

Solve for $x$: $\sin^{-1}x + \sin^{-1}2x = \frac{\pi}{3}$.

[4 marks]

Solution

$\sin^{-1}2x = \frac{\pi}{3} - \sin^{-1}x$.
Take sine of both sides: $2x = \sin(\frac{\pi}{3} - \sin^{-1}x)$.
$2x = \sin\frac{\pi}{3}\cos(\sin^{-1}x) - \cos\frac{\pi}{3}\sin(\sin^{-1}x)$.
$2x = \frac{\sqrt{3}}{2}\sqrt{1-x^2} - \frac{1}{2}x$.
$2.5x = \frac{\sqrt{3}}{2}\sqrt{1-x^2} \Rightarrow 5x = \sqrt{3}\sqrt{1-x^2}$.
Square: $25x^2 = 3(1-x^2) = 3 - 3x^2 \Rightarrow 28x^2 = 3$.
$x = \sqrt{3/28} = \frac{\sqrt{3}}{2\sqrt{7}}$.

∴ $x = \frac{\sqrt{3}}{2\sqrt{7}}$

10

Prove: $3\tan^{-1}x = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$.

[3 marks]

Solution

Let $\tan^{-1}x = \theta$. Then $x = \tan \theta$.
We know $\tan 3\theta = \frac{3\tan \theta - \tan^3 \theta}{1 - 3\tan^2 \theta}$.
Substitute $x$: $\tan 3\theta = \frac{3x - x^3}{1 - 3x^2}$.
$3\theta = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$.
Substitute $\theta$ back: $3\tan^{-1}x = \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$.

∴ Proved

One Page Revision Cheat Sheet

1

Definition Check: Invertible = One-to-One + Onto. Many-to-one functions (like basic sin/cos) need domain restriction.

2

The Golden Table: Memorize Domains and Ranges. Sine/Tan range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. Cos/Cot range is $[0, \pi]$.

3

Key Formula: $\tan^{-1}x \pm \tan^{-1}y = \tan^{-1}\left(\frac{x \pm y}{1 \mp xy}\right)$. Watch out for the $xy > 1$ case!

4

Complementary: Sum of $\sin^{-1}$ and $\cos^{-1}$ of same $x$ is always $\frac{\pi}{2}$ (90 degrees).

5

Negation Rule: For 'Co' functions, $\cos^{-1}(-x) = \pi - \cos^{-1}x$. For others, $\sin^{-1}(-x) = -\sin^{-1}x$.

6

Mistake to avoid: $\sin^{-1}(\sin \theta)$ is not always $\theta$. It is only $\theta$ if $\theta$ is in the principal range.

7

Conversion Trick: Use Right-Angled Triangles (p, b, h) to convert $\sin^{-1}$ to $\tan^{-1}$ or vice versa.

8

Double Angle: $\sin(2\sin^{-1}x) = 2x\sqrt{1-x^2}$. Useful for simplification.

9

Sum of Series: If sum of angles is $\pi$ (e.g., $\tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi$), then sum of angles formula for tan applies resulting in sum = 0.