# Inverse Trigonometric Ratios

Inverse Trigonometric Ratios
Go back to  'Trigonometry Basics'

We have heard about function $$f$$, and the inverse of a function $$f^{-1}$$ exist if $$f$$ is a one-one function. But on observation, trigonometric functions are not one-one function.

Further limiting these trigonometric functions to only its principal values, we have inverse trigonometric functions. In this mini-lesson, we shall explore the world of inverse trigonometric ratios by finding answers to questions like what are inverse trigonometric ratios, and what are the inverse trigonometric ratios of arbitrary values.

## Lesson Plan

 1 What Do You Mean by Inverse Trigonometric Ratios? 2 Tips and Tricks 3 Challenging Questions 4 Solved Examples on Inverse Trigonometric Ratios 5 Interactive Questions on Inverse Trigonometric Ratios

## What Do You Mean by Inverse Trigonometric Ratios?

Inverse trigonometric functions are the inverse ratio of the basic trigonometric ratios.

\begin{align} \text{Sin } \theta &= x \\ \theta &= \text{Sin }^{-1} x \end{align}

Here x can have values in whole numbers, decimals, fractions, or exponents.

For $$\theta = 30 ^{\circ}$$ we have $$\theta = Sin^{-1} \frac{1}{2}$$

## What Are Inverse Trigonometric Ratios for Arbitrary Values?

The inverse trigonometric ratio formula for arbitrary values are as follows.

• $$\text{Sin }^{-1}(-x) = -\text{Sin }^{-1}(x)$$
• $$\text{Tan }^{-1}(-x) = -\text{Tan }^{-1}(x)$$
• $$\text{Cosec }^{-1}(-x) = -\text{Cosec }^{-1}(x)$$
• $$\text{Cos }^{-1}(-x) =\pi -\text{Cos }^{-1}(x)$$
• $$\text{Sec }^{-1}(-x) = \pi -\text{Sec }^{-1}(x)$$
• $$\text{Cot }^{-1}(-x) = \pi -\text{Cot }^{-1}(x)$$

### Reciprocal Functions

The inverse trigonometric formula of inverse sine, inverse cosine, and inverse tangent can also be expressed in the following forms.

• $$\text{Sin }^{-1}(x) = \text{Cosec }^{-1}\dfrac{1}{x}$$
• $$\text{Cos }^{-1}(x) = \text{Sec }^{-1}\dfrac{1}{x}$$
• $$\text{Tan }^{-1}(x) = \text{Cot }^{-1}\dfrac{1}{x}$$ Tips and Tricks
1. $$Sin^{-1}(Sinx )= Sin( Sin^{-1}x )= x$$
2. $$Sin^{-1} x$$ is different from $$(Sin x)^{-1}$$. Also $$(Sinx)^{-1} = \dfrac{1}{Sinx}$$
3. $$Sin^{-1} x = \theta$$ and $$\theta$$ refer to the angle, which is the principal value of this inverse trigonometric function.

### Complementary Functions

• $$\text{Sin }^{-1}(x) + \text{Cos }^{-1}(x) = \dfrac{\pi}{2}$$
• $$\text{Tan }^{-1}(x) + \text{Cot }^{-1}(x) = \dfrac{\pi}{2}$$
• $$\text{Sec }^{-1}(x) + \text{Cosec }^{-1}(x) = \dfrac{\pi}{2}$$

### Sum and Difference of Functions

• $$Sin ^{-1}x + Sin ^{-1}y = Sin ^{-1}(x \cdot \sqrt{(1 - y^2)} + y \cdot \sqrt{(1 - x^2)} )$$
• $$Sin ^{-1}x - Sin ^{-1}y = Sin ^{-1}(x \cdot \sqrt{(1 - y^2)} - y \cdot \sqrt{(1 - x^2)} )$$
• $$Cos ^{-1}x + Cos ^{-1}y = Cos ^{-1}(x\cdot y -\sqrt{(1 - x^2)} \cdot\sqrt{(1 - y^2)} )$$
• $$Cos ^{-1}x - Cos ^{-1}y = Cos ^{-1}(x\cdot y +\sqrt{(1 - x^2)} \cdot\sqrt{(1 - y^2)} )$$
• $$Tan ^{-1}x + Tan ^{-1}y = Tan ^{-1}\left(\dfrac{x + y}{1 - xy}\right)$$
• $$Tan ^{-1}x - Tan ^{-1}y = Tan ^{-1}\left(\dfrac{x - y}{1 + xy}\right)$$

### Double of the Function

• $$2Sin ^{-1}x = Sin ^{-1}(2x \cdot \sqrt{1 - x^2})$$
• $$2Cos ^{-1}x = Cos ^{-1}(2x ^2 - 1)$$
• $$2Tan ^{-1}x = Tan ^{-1}\left(\dfrac{2x}{1 - x^2}\right)$$

### Triple of the Function

• $$3Sin^{-1}x = Sin^{-1}(3x - 4x^3 )$$
• $$3Cos^{-1}x = Cos^{-1}(4x^3 - 3x )$$
• $$3Tan^{-1}x = Tan^{-1}\left(\frac{3x - x^3 }{1 - 3x^2}\right)$$

## Solved Examples

 Example 1

Find the value of $$Tan^{-1}(1) + Cos^{-1}(\frac{-1}{2}) + Sin^{-1}(\frac {-1}{2})$$.

Solution

\begin{align}Tan^{-1}(1) + Cos^{-1}(\frac{-1}{2}) + Sin^{-1}(\frac {-1}{2}) &=Tan^{-1}(1) +\pi - Cos^{-1}(\frac{1}{2}) - Sin^{-1}(\frac {1}{2})\\&= \frac{\pi}{4} + \pi - \frac{\pi}{3} -\frac{\pi}{6} \\&=\frac{\pi}{4} + \pi - \frac{\pi}{2} \\&=\frac{\pi}{4} + \frac{\pi}{2} \\&= \frac{3\pi}{4}\end{align}

 $$\therefore$$ The answer is $$\frac{3\pi}{4}$$
 Example 2

Solve the expression $$Tan^{-1}\left(\frac{2}{11}\right) + Tan^{-1}\left(\frac{7}{24}\right)$$.

Solution

Here we shall use the formula $$Tan ^{-1}x + Tan ^{-1}y = Tan ^{-1}\left(\dfrac{x + y}{1 - xy}\right)$$\begin{align}Tan^{-1}\left(\frac{2}{11}\right) + Tan^{-1}\left(\frac{7}{24}\right) &=Tan^{-1}\left(\dfrac{\frac{2}{11} + \frac{7}{24}}{1 - \frac{2}{11}\cdot\frac{7}{24}}\right) \\ &=Tan^{-1}\left(\dfrac{\frac{2\times 24 + 7 \times 11}{11 \times 24} }{ \frac{11 \times 24 - 2 \times 7}{11 \times 24}}\right) \\&=Tan^{-1}\left(\frac{\frac{2\times 24 + 7 \times 11}{11 \times 24} }{ \frac{11 \times 24 - 2 \times 7}{11 \times 24}} \right)\\ &= Tan^{-1}\left(\frac{\frac{48 + 77}{264} }{ \frac{264 - 14}{264}}\right)\\&=Tan^{-1}\left(\frac{\frac{125}{264} }{ \frac{250}{264}}\right) \\&=Tan^{-1}\left(\dfrac{125}{250} \right)\\&=Tan^{-1}\left(\frac{1}{2}\right)\end{align}

 $$\therefore$$ The answer is $$Tan^{-1}\left(\frac{1}{2}\right)$$
 Example 3

$$Sin^{-1}\frac{3}{5} + Sin^{-1}\frac{12}{13} = Sin^{-1}x$$.  Find the value of $$x$$.

Solution

Here we shall apply the sine inverse formula $$Sin ^{-1}x + Sin ^{-1}y = Sin ^{-1}(x \cdot \sqrt{(1 - y^2)} + y \cdot \sqrt{(1 - x^2)} )$$
\begin{align}Sin^{-1}x &= Sin^{-1}\left(\frac{3}{5} + Sin^{-1}\frac{12}{13}\right) \\ &= Sin^{-1}\left(\frac{3}{5}\cdot\sqrt{1 -\left(\frac{12}{13}\right)^2} + \frac{12}{13}\cdot\sqrt{1 -\left(\frac{3}{5}\right)^2}\right)\\&=Sin^{-1}\left(\frac{3}{5}\cdot\frac{5}{13} + \frac{12}{13}\cdot\frac{4}{5}\right)\\&=Sin^{-1}\left(\frac{15}{65} + \frac{48}{65}\right)\\&=Sin^{-1}\left(\frac{63}{65}\right )\end{align}

 $$\therefore$$ $$x = \dfrac{63}{65}$$
 Example 4

Find the value of $$Tan^{-1}(\sqrt3) – Cot^{-1}(-\sqrt3 )$$.

Solution

\begin{align}Tan^{-1}(\sqrt3) – Cot^{-1}(-\sqrt3 )&=Tan^{-1}(\sqrt3) –(\pi - Cot^{-1}(\sqrt3 )) \\ &=Tan^{-1}(\sqrt3) –\pi +Cot^{-1}(\sqrt3 ) \\&= \frac{\pi}{3} -\pi + \frac{\pi}{6} \\&=\frac{\pi}{2} - \pi \\&=-\frac{\pi}{2}\end{align}

 $$\therefore$$ The answer is $$-\frac{\pi}{2}$$
 Example 5

Simplify $$Tan^{-1}\sqrt{\frac{1 - Cosx}{1 + Cosx}}$$

Solution

\begin{align}Tan^{-1}\sqrt{\frac{1 - Cosx}{1 + Cosx}} & = Tan^{-1}\sqrt{\frac{2Sin^2\frac{x}{2}}{2Cos^2\frac{x}{2} }}\\&= Tan^{-1}{\frac{Sin\frac{x}{2}}{Cos\frac{x}{2} }}\\ &=Tan^{-1}(Tan\frac{x}{2})\\&=\frac{x}{2}\end{align}

 $$\therefore$$ The answer is $$\frac{x}{2}$$ Challenging Questions

Simplify the following expressions.

1. $$Tan^{-1}\sqrt{\dfrac{Cosx}{1 - Sinx}}$$
2. $$Tan^{-1}\dfrac{\sqrt{1 + x^2} - 1}{x}$$
3. $$Cot^{-1}\left(\dfrac{\sqrt{1 + Sinx} + \sqrt{1 - Sinx}}{\sqrt{1 + Sinx} - \sqrt{1 - Sinx}}\right)$$

## Interactive Questions on Inverse Trigonometric Ratios

Here are a few activities for you to practice. Select/Type your answer and click the "Check Answer" button to see the result.

## Let's Summarize

The mini-lesson targeted the fascinating concept of inverse trigonometric ratios. The math journey around inverse trigonometric ratios starts with what a student already knows, and goes on to creatively crafting a fresh concept in the young minds. Done in a way that is not only relatable and easy to grasp, but will also stay with them forever. Here lies the magic with Cuemath.

At Cuemath, our team of math experts is dedicated to making learning fun for our favorite readers, the students!

Through an interactive and engaging learning-teaching-learning approach, the teachers explore all angles of a topic.

Be it worksheets, online classes, doubt sessions, or any other form of relation, it’s the logical thinking and smart learning approach that we, at Cuemath, believe in.

## FAQs on Inverse Trigonometric Ratios

### 1. How do you find inverse trigonometric ratios?

The inverse trigonometric ratios of invere sine, inverse cosine, or inverse tangent can be found from the basic trigonometric ratios.
\begin{align}Sin \theta &= x \\ \theta &=Sin^{-1}x \end{align}

### 2. What is inverse trigonometry?

Inverse Trigonometry ratio formula  helps to find the angle for the given inverse trigonometric ratio of the sides of a right-angled triangle.

$Sin^{-1}x = \theta$

### 3. What are the six inverse trigonometric ratios?

The six inverse trigonometric ratios are:

$$Sin^{-1}x,~Cos^{-1}x,~Tan^{-1}x,~Cot^{-1}x,~Sec^{-1}x,~and~Cosec^{-1}x$$

### 4. How do you convert sine inverse to tan inverse?

The sine inverse is converted to tan inverse, as per the formula below.
$Sin^{-1}x = Tan^{-1} \frac{x}{\sqrt{1 - x^2}}$

### 5. What is the inverse of sine called?

The inverse sine is equal to cosecant. $Sin \theta = \dfrac{1}{Cosec \theta}$

### 6. What are the 3 basic trigonometric ratios?

The three basic trigonometric ratios are $$Sin \theta,~ Cos \theta,~ Tan \theta$$.

### 7. Which trigonometric ratios are even?

The Trigonometric ratios of cosine and secant are even.

$Cos( - \theta) = Cos \theta$

$Sec(- \theta) = Sec \theta$

### 8. What is arcsine, arccosine, and arctangent?

The terms arcsine, arccosine, and arctangent are the inverse ratio of the trigonometric ratios $$Sin \theta,~ Cos \theta,~ Tan \theta$$.

\begin{align} \theta &= \sin^{-1}x \\ \theta &= \cos^{-1}y\\ \theta &= \tan^{-1}z\end{align}

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