RD Sharma Class 12 Exercise 3.8 Inverse Trigonometry functions Solutions Maths - Download PDF Free Online

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RD Sharma Class 12 Solutions Chapter3 Inverse Trigonometric Functions-Other Exercise

• Chapter 3 - Inverse Trigonometric Functions Ex 3.1
• Chapter 3 - Inverse Trigonometric Functions Ex 3.2
• Chapter 3 - Inverse Trigonometric Functions Ex 3.3
• Chapter 3 - Inverse Trigonometric Functions Ex 3.4
• Chapter 3 - Inverse Trigonometric Functions Ex 3.5
• Chapter 3 - Inverse Trigonometric Functions Ex 3.6
• Chapter 3 - Inverse Trigonometric Functions Ex 3.7
• Chapter 3 - Inverse Trigonometric Functions Ex 3.9
• Chapter 3 - Inverse Trigonometric Functions Ex 3.10
• Chapter 3 - Inverse Trigonometric Functions Ex 3.11
• Chapter 3 - Inverse Trigonometric Functions Ex 3.12
• Chapter 3 - Inverse Trigonometric Functions Ex 3.13
• Chapter 3 - Inverse Trigonometric Functions Ex 3.14
• Chapter 3 - Inverse Trigonometric Ex VSA

Inverse trigromtery functions exercise 3.8 question 1 (II)

Answer:

$\frac{12}{13}$
Hint:
As we know that the value of $\sin ({\sin }^{-1}x)\text{ is }x\text{ for }x\in [-1,1]$
Given:
We have,
$\sin ({\cos }^{-1}\frac{5}{13})$
Solution:
Here in the place of ${\sin }^{-1}x$ there is ${\cos }^{-1}x$.
So we convert ${\cos }^{-1}x$ into ${\sin }^{-1}x$.
Let’s suppose that
$\begin{array}{rl}& {\cos }^{-1}\frac{5}{13}=\alpha \\ & \cos \alpha =\frac{5}{13}\\ & \sin ({\cos }^{-1}\frac{5}{13})=\sin \alpha \end{array}$
Let, in$\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B
$\cos \alpha =\frac{5}{13}=\frac{B}{H}$

$\begin{array}{rl}& (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (13{)}^2=5^2+(BC{)}^2\\ & (BC{)}^2=169-25\end{array}$
$BC=\pm 12$ [we will ignore the -ve sign because BC is a length and it can’t be -ve.]
$BC=12$
$\sin =\frac{12}{13}$
So, we required $\alpha$ and that is $\frac{12}{13}$
$\sin ({\cos }^{-1}\frac{5}{13})=\sin \alpha =\frac{12}{13}$

Inverse trigromtery functions exercise 3.8 question 1 (III)

Answer:

$\frac{24}{25}$
Hint:
As we know that the value of $({\sin }^{-1}x)\text{ is }x\text{ for }x\in [-1,1]$
Given:
We have $\sin ({\tan }^{-1}\left(\frac{24}{7}\right))$
Solution:
Here in the place of ${\sin }^{-1}x$we have ${\tan }^{-1}x$.
So, we convert ${\tan }^{-1}x$ to ${\sin }^{-1}x$.
Let’s suppose that,
$\begin{array}{rl}& {\tan }^{-1}\left(\frac{24}{7}\right)=\alpha \\ & \tan \alpha =\frac{24}{7}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}\tan \alpha & =\frac{P}{B}=\frac{24}{7}\\ (AC{)}^2& =(AB{)}^2+(BC{)}^2\\ & =7^2+(24{)}^2\\ & =49+576\\ & =625\end{array}$
$=+25$ [we will ignore the -ve sign because AC is a length and it can’t be -ve.]
$AC=25$
$\begin{array}{rl}& \sin \alpha =\frac{24}{25}\\ & \sin ({\tan }^{-1}\left(\frac{24}{7}\right))=\sin \alpha =\frac{24}{25}\end{array}$ [since, ${\tan }^{-1}\left(\frac{24}{7}\right)=\alpha$]

Inverse trigonometric functions exercise 3.8 question 1 (iv)

Answer:

$\frac{15}{17}$

Hint:
As we know that the value of $({\sin }^{-1}x)\text{ is }x\text{ for }x\in [-1,1]$
Given:
We have
$\sin ({\sec }^{-1}\left(\frac{17}{8}\right))$
Solution:
So here in place of ${\sin }^{-1}x$ there is ${\sec }^{-1}x$.
So, we convert ${\sec }^{-1}x$ into ${\sin }^{-1}x$
Let’s we suppose that,
$\begin{array}{rl}& {\sec }^{-1}\frac{17}{8}=\alpha \\ & \sec \alpha =\frac{17}{8}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& \sec \alpha =\frac{H}{B}=\frac{17}{8}\\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (17{)}^2=8^2+(BC{)}^2\\ & (BC{)}^2=289-64\end{array}$
$BC=\pm 15$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC=15$
$\begin{array}{rl}& \sin \alpha =\frac{P}{H}=\frac{15}{17}\\ & \sin ({\sec }^{-1}\left(\frac{17}{8}\right))=\frac{15}{17}\end{array}$

Inverse trigonometric functions exercise 3.8 question 1 (v)

Answer:

$\frac{5}{4}$
Hint:
As we know that the value of $\cos ec(\cos e{c}^{-1}x)\text{ is }x\text{ for }x\in (-\mathrm{\infty },-1)\cup (1,\mathrm{\infty })$
Given:
We have
$\mathrm{cosec}({\cos }^{-1}\frac{3}{5})$
Solution:
In place of $\cos e{c}^{-1}x$ here is ${\cos }^{-1}x$
So, we convert ${\cos }^{-1}x$ into $\cos e{c}^{-1}x$.
Let’s suppose that,
$\begin{array}{rl}& {\cos }^{-1}\frac{3}{5}=\alpha \\ & \cos \alpha =\frac{3}{5}=\frac{B}{H}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & 5^2=3^2+(BC{)}^2\\ & 25-9=(BC{)}^2\end{array}$
$BC=\pm 4$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC=4$
$\begin{array}{rl}& \cos ec\alpha =\frac{H}{P}=\frac{5}{4}\\ & \cos ec({\cos }^{-1}\frac{3}{5})=\frac{5}{4}\end{array}$

Inverse trigonometric functions exercise 3.8 question 1 (vi)

Answer:

$\frac{13}{5}$

Hint:
As we know that the range of $\sec ({\sec }^{-1}x)\text{ is }x\text{ for }x\in (-\mathrm{\infty },-1)\cup (1,\mathrm{\infty })$
Given:
We have
$\sec ({\sin }^{-1}\left(\frac{12}{13}\right))$
Solution:
In place of ${\sec }^{-1}x$ there is ${\sin }^{-1}x$
So, we convert ${\sin }^{-1}x$ into ${\sec }^{-1}x$ .
Let’s suppose that,

$\begin{array}{rl}& {\sin }^{-1}\frac{12}{13}=\alpha \\ & \sin \alpha =\frac{12}{13}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (13{)}^2=(AB{)}^2+(12{)}^2\\ & (AB{)}^2=25\end{array}$
$AB=\pm 5$ [we will ignore the -ve sign because AB is a length and it can’t be -ve]
$AB=5$
$\begin{array}{rl}& \sec \alpha =\frac{13}{5}\\ & \sec ({\sin }^{-1}\left(\frac{12}{13}\right))=\sec \alpha =\frac{13}{5}\\ & \sec ({\sin }^{-1}\left(\frac{12}{13}\right))=\frac{13}{5}\end{array}$

Inverse Trigonometric functions exercise 3.8 question 1 (vii)

Answer:

$\frac{15}{8}$

Hint:
As we know that the value of $\tan ({\tan }^{-1}x)\text{ is }x\text{ for }x\in \mathbb{R}\text{ ( real number) }$
Given:
We have
$\tan ({\cos }^{-1}\left(\frac{8}{17}\right))$
Solution:
In place of ${\tan }^{-1}x$ here is ${\cos }^{-1}x$
So, we convert ${\cos }^{-1}x$ into ${\tan }^{-1}x$.
Let’s suppose that,
$\begin{array}{rl}& {\cos }^{-1}\left(\frac{8}{17}\right)=\alpha \\ & \cos \alpha =\frac{8}{17}=\frac{B}{H}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.


$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (17{)}^2=8^2+(BC{)}^2\\ & (BC{)}^2=289-64\\ & (BC{)}^2=25\end{array}$
$BC=\pm 15$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC=15$
$\begin{array}{rl}& \tan ({\cos }^{-1}\left(\frac{8}{17}\right))=\tan \alpha \\ & \tan \alpha =\frac{15}{8}\\ & \tan ({\cos }^{-1}\left(\frac{8}{17}\right))=\frac{15}{8}\end{array}$

Inverse Trigonometric functions exercise 3.8 question 1 (viii)

Answer:

$\frac{3}{4}$

Hint:
As we know that the range of $\cot ({\cot }^{-1}x)\text{ is }x\text{ for }x\in \mathbb{R}\text{ ( real number) }$
Given:
$\cot ({\cos }^{-1}\left(\frac{3}{5}\right))$
Solution:
In place of ${\cot }^{-1}x$ here is ${\cos }^{-1}x$
So, we convert ${\cos }^{-1}x$ into ${\cot }^{-1}x$.
Let’s suppose that,
$\begin{array}{rl}& {\cos }^{-1}\left(\frac{3}{5}\right)=\alpha \\ & \cos \alpha =\frac{3}{5}=\frac{B}{H}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (5{)}^2=3^2+(BC{)}^2\\ & 25=9+(BC{)}^2\\ & (BC{)}^2=16\end{array}$
$BC=\pm 4$ [we will ignore the -ve sign because BC is a length and it can’t be -ve]
$BC=4$$\begin{array}{rl}& \cot ({\cos }^{-1}\left(\frac{3}{5}\right))=\cot c\\ & \cot \alpha =\frac{3}{4}\\ & \cot ({\cos }^{-1}\left(\frac{3}{5}\right))=\frac{3}{4}\end{array}$

Inverse Trigonometric functions exercise 3.8 question 1 (ix)

Answer:

$\frac{7}{25}$
Hint:
As we know that the range of $({\cos }^{-1}x)\text{ is }x\text{ for }x\in [-1,1]$
Given:
We have
$\cos ({\tan }^{-1}\left(\frac{24}{7}\right))$
Solution:
In place of ${\cos }^{-1}x$ here is ${\tan }^{-1}x$
So, we convert ${\tan }^{-1}x$ into ${\cos }^{-1}x$ .
Let’s suppose that,
$\begin{array}{rl}& {\tan }^{-1}\left(\frac{24}{7}\right)=\alpha \\ & \tan \alpha =\frac{24}{7}\end{array}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (AC{)}^2=7^2+(24{)}^2\\ & (AC{)}^2=625\end{array}$
$AC=\pm 25$ [we will ignore the -ve sign because AB is a length and it can’t be -ve]
$AC=25$
$\begin{array}{rl}& \cos ({\tan }^{-1}\left(\frac{24}{7}\right))=\cos \alpha \\ & \cos \alpha =\frac{7}{25}\\ & \cos ({\tan }^{-1}\left(\frac{24}{7}\right))=\frac{7}{25}\end{array}$

Inverse Trigonometric functions exercise 3.8 question 2 (i)

Hint:
We convert in the form of $\tan (\alpha +\beta )$ so we know the formula of it.
Given:
We have to prove that $\tan ({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3})=\frac{17}{6}$
Solution:
LHS= $\tan ({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3})$
Let suppose that ${\cos }^{-1}\frac{4}{5}=\alpha$ and ${\tan }^{-1}\frac{2}{3}=\beta$,
$\begin{array}{rl}& \cos \alpha =\frac{4}{5}\text{ and }\tan \beta =\frac{2}{3}\\ & \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\end{array}$
But we have not $\tan \alpha$. We have only $\tan \beta$. So, we convert $\cos \alpha$ into $\tan \alpha$.
$\cos \alpha =\frac{4}{5}=\frac{B}{H}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.


$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (5{)}^2=4^2+(BC{)}^2\\ & (BC{)}^2=25-16\end{array}$
$BC=\pm 9$
$BC=3$
$\tan \alpha =\frac{3}{4}$

$[\because \tan \alpha =\frac{3}{4}and\tan \beta =\frac{2}{3}]$
$\begin{array}{c}\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ =\frac{\frac{3}{4}+\frac{2}{3}}{1-\frac{3}{4}\times \frac{2}{3}}\\ =\frac{\frac{9+8}{12-6}}{12}=\frac{17}{6}=\mathrm{RHS}\\ \tan ({\cos }^{-1}\frac{4}{5}+{\tan }^{-1}\frac{2}{3})=\frac{17}{6}\end{array}$ [ we take LCM of 3,4]
Hence proved.

Inverse trigonometric functions exercise 3.8 question 2 (ii)

Hint:
We convert in the form of $\cos (\alpha +\beta )$and we know the formula of it.
Given:
We have to prove $\cos ({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})=\frac{6}{5\sqrt{13}}$
Solution:
LHS = $\cos ({\sin }^{-1}\frac{3}{5}+{\cot }^{-1}\frac{3}{2})$
Let’s suppose that ${\sin }^{-1}\frac{3}{5}=\alpha$and ${\cot }^{-1}\frac{3}{2}=\beta$.
$\sin \alpha =\frac{3}{5}and\cot \beta =\frac{3}{2}$
$\cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta$
Here, we only have $\sin \alpha$ and $\cos \beta$.So we have to find $\cos \alpha$and $\sin \beta$.
$\sin \alpha =\frac{3}{5}=\frac{B}{H}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (5{)}^2=(AB{)}^2+3^2\\ & (AB{)}^2=25-9\end{array}$
$AB=\pm 4$
$\cos \alpha =\frac{4}{5}$
Let, in $\mathrm{\Delta }PQR$, angle $RPQ=\beta$ and right angle at B.
$\cot \beta =\frac{3}{2}=\frac{B}{P}$

$\begin{array}{rl}& \\ & (PR{)}^2=(QR{)}^2+(PQ{)}^2\\ & (PR{)}^2=2^2+3^2\\ & (PR{)}^2=4+9\end{array}$
$PR=\pm \sqrt{13}$ [we will ignore the -ve sign because PR is a length and it can’t be -ve]
$PR=\sqrt{13}$
$\begin{array}{rl}& PR=\sqrt{13}\\ & \sin \beta =\frac{2}{\sqrt{13}},\cos \beta =\frac{3}{\sqrt{13}}\\ & \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & =\frac{4}{5}\times \frac{3}{\sqrt{13}}-\frac{2}{\sqrt{13}}\times \frac{3}{5}\\ & =\frac{12}{5\sqrt{13}}-\frac{6}{5\sqrt{13}}\\ & \cos (\alpha +\beta )=\frac{6}{5\sqrt{13}}=\mathrm{RHS}\end{array}$

Inverse trigonometric functions exercise 3.8 question 2 (iii)

Hint:
We convert expression in the form of $\tan (\alpha +\beta )$ so we know the formula of it.
Given:
We have to prove $\tan ({\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5})=\frac{63}{16}$
Solution:
LHS = $\tan ({\sin }^{-1}\frac{5}{13}+{\cos }^{-1}\frac{3}{5})$
${\sin }^{-1}\frac{5}{13}=\alpha and{\cos }^{-1}\frac{3}{5}=\beta$
$\sin \alpha =\frac{5}{13}and\cos \beta =\frac{3}{5}$
$\tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ …(i)
But we have not $\tan \alpha$ and $\tan \beta$. So, we find out first then we put in equation (i)
$\sin \alpha =\frac{5}{13}=\frac{P}{H}$
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.

$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (13{)}^2=A{B}^2+5^2\\ & (AB{)}^2=144\end{array}$
$AB=\pm 12$ [we will ignore the -ve sign because AB is a length and it can’t be -ve]
$AB=12$
$\therefore \tan \alpha =\frac{5}{12}$

Let, in $\mathrm{\Delta }PQR$, angle $RPQ=\beta$ and right angle at B.
$\cot \beta =\frac{3}{5}=\frac{B}{P}$

$\begin{array}{rl}& \\ & (PR{)}^2=(QR{)}^2+(PQ{)}^2\\ & (5{)}^2=3^2+(QR{)}^2\\ & (QR{)}^2=16\end{array}$
$QR=\pm 4$
$QR=4$
$\therefore \tan \beta =\frac{4}{3}$
Now, $\begin{array}{rl}\tan (\alpha +\beta )=& \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }\\ & =\frac{\frac{5}{12}+\frac{4}{3}}{1-\frac{5}{12}\times \frac{4}{3}}\end{array}$ [ put, $\tan \alpha =\frac{5}{12}$ and $\tan \beta =\frac{4}{3}$]
[ we take LCM of 3 and 12]
$\begin{array}{rl}& =\frac{\frac{5+16}{12}}{\frac{36-20}{36}}\\ & =\frac{\frac{21}{12}}{\frac{16}{36}}\\ & =\frac{21}{12}\times \frac{36}{16}\\ & =\frac{21}{1}\times \frac{3}{16}\\ \tan (\alpha +\beta )& =\frac{63}{16}=\mathrm{RHS}\end{array}$
Hence proved.

Inverse trigonometric functions exercise 3.8 question 2 (iv)

Answer:

Hint:
We convert in the form of $\sin (\alpha +\beta )$and we know the formula of it.
Given:
We have to prove $\sin ({\cos }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13})=\frac{63}{65}$
Solution:
LHS = $\sin ({\cos }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13})$
Let’s suppose that ${\cos }^{-1}\frac{3}{5}=\alpha and{\sin }^{-1}\frac{5}{13}=\beta$
$\cos \alpha =\frac{3}{5}and\sin \beta \frac{5}{13}$ and
$\begin{array}{rl}& \Rightarrow \sin ({\cos }^{-1}\frac{3}{5}+{\sin }^{-1}\frac{5}{13})\\ & \Rightarrow \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta \end{array}$
…(i)
Here we don’t have $\sin \alpha .\cos \beta$. So, we find out it and then put all the values in equation (i)
Let, in $\mathrm{\Delta }ABC$, angle $CAB=\alpha$ and right angle at B.
$\cos \alpha =\frac{3}{5}=\frac{B}{H}$

$\begin{array}{rl}& \\ & (AC{)}^2=(AB{)}^2+(BC{)}^2\\ & (5{)}^2=(3{)}^2+(BC{)}^2\\ & {25}^2=9^2+(BC{)}^2\\ & (BC{)}^2=16\end{array}$
$BC=\pm 4$
$BC=4$
$\therefore \sin \alpha =\frac{4}{5}$
Let, in $\mathrm{\Delta }PQR$, angle $RPQ=\beta$ and right angle at B.
$\sin \beta =\frac{5}{13}=\frac{P}{H}$

$\begin{array}{rl}& \\ & (PR{)}^2=(PQ{)}^2+(QR{)}^2\\ & (13{)}^2=(PQ{)}^2+5^2\\ & 169=(PQ{)}^2+25\\ & (PQ{)}^2=144\end{array}$
$PQ=\pm 12$ [we will ignore the -ve sign because PQ is a length and it can’t be -ve]
$PQ=12$
$\therefore \cos \beta =\frac{12}{13}$
Now, $\begin{array}{rl}\sin (\alpha +\beta )& =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & =\frac{4}{5}\times \frac{12}{13}+\frac{3}{5}\times \frac{5}{13}\\ & =\frac{48}{65}+\frac{15}{65}\\ \sin (\alpha +\beta )& =\frac{63}{65}=\mathrm{RHS}\end{array}$
Hence proved.

The 3rd chapter in mathematics, Inverse Trigonometry for Class 12, is not easy for many students to understand. Especially in the eighth exercise, which is Exercise 3.8, the students must solve the functions as learned from the previous exercises. The concepts given here are regarding solving sine, cosine, tangent, cotangent, secant, and cosecant functions. There are about fifteen questions, including the subparts in this exercise 3.8. If they find it hard, the RD Sharma Class 12 Chapter 3 Exercise 3.8 solutions will lend them a helping hand.

The solutions in this book are created by experts and verified according to the updated syllabus. You can use this guide to complete your homework and make your assignments too. As it follows the NCERT pattern, it is beneficial for the students who pursue their education at CBSE institutions. If you see yourself scoring low marks in the chapter Inverse Trigonometry, use the solutions given in the Class 12 RD Sharma Chapter 3 Exercise 3.8 Solution and understand the concepts. The different ways to solve a single question are given in the material. Soon, you will find yourself crossing your benchmark score.

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Chapter-wise RD Sharma Class 12 Solutions

• Chapter 1 - Relations
• Chapter 2 - Functions
• Chapter 3 - Inverse Trigonometric Functions
• Chapter 4 - Algebra of Matrices
• Chapter 5 - Determinants
• Chapter 6 - Adjoint and Inverse of a Matrix
• Chapter 7 - Solution of Simultaneous Linear Equations
• Chapter 8 - Continuity
• Chapter 9 - Differentiability
• Chapter 10 - Differentiation
• Chapter 11 - Higher Order Derivatives
• Chapter 12 - Derivative as a Rate Measurer
• Chapter 13 - Differentials, Errors and Approximations
• Chapter 14 - Mean Value Theorems
• Chapter 15 - Tangents and Normals
• Chapter 16 - Increasing and Decreasing Functions
• Chapter 17 - Maxima and Minima
• Chapter 18 - Indefinite Integrals
• Chapter 19 - Definite Integrals
• Chapter 20 - Areas of Bounded Regions
• Chapter 21 - Differential Equations
• Chapter 22 - Algebra of Vectors
• Chapter 23 - Scalar Or Dot Product
• Chapter 24 - Vector or Cross Product
• Chapter 25 - Scalar Triple Product
• Chapter 26 - Direction Cosines and Direction Ratios
• Chapter 27 - Straight Line in Space
• Chapter 28 - The Plane
• Chapter 29 - Linear programming
• Chapter 30- Probability
• Chapter 31 - Mean and Variance of a Random Variable