RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA
RD Sharma Solutions Class 12 Mathematics Chapter 3 VSA
Updated on Jan 21, 2022 12:37 PM IST
RD Sharma has been the choice of many students and teachers all over the country. Their simple easy to understand solutions make it very beneficial for students to learn the concepts quickly. Many teachers use these materials for lectures and to set up question papers. As these contain numerous examples, students can be thorough with their preparation and leave no problem unsolved.
Answer: Statement of Rolle’s Theorem Hint: You must know about Rolle’s Theorem. Given: Rolle’s Theorem Solution: Rolle’s Theorem states that if $f\left ( x \right )$ is continuous in the closed interval $\left [ a,b \right ]$ and differentiable on open interval $\left ( a,b \right )$ such that $f\left ( a \right )=f\left ( b \right )$ then ${f}'\left ( x \right )=0$ for some $a< x< b$.
Answer: Statement of mean value theorem. Hint: You must know about mean value theorem. Given: Lagrange’s mean value theorem. Solution: Let be a function that satisfies the following hypothesis.
$f\left ( x \right )$ is continuous on the closed interval $\left [ a,b \right ]$.
$f\left ( x \right )$ is differentiable on the open interval $\left ( a,b \right )$.
Then there is a value in $\left ( a,b \right )$ such that ${f}'\left ( c \right )=\frac{f\left ( b \right )-f\left ( a \right )}{b-a}$ Or equivalently $f\left ( b \right )-f\left ( a \right )={f}'\left ( c \right )\left ( b-a \right )$
Answer: $\pi$ Given: $\sin ^{-1} x, x \in(-1,1)$ Hint: The maximum value of $\sin ^{-1} x$ in $x \in (-1,1)$ is at 1 . Solution: $\sin ^{-1}(1)$ maximum value is, $=\sin ^{-1}\left(\sin \frac{\pi}{2}\right)$ $=\frac{\pi}{2}$ Again the minimum value is at -1 So, $\begin{aligned} \sin ^{-1}(-1) &=-\sin ^{-1}(1) \\\\ &=-\sin ^{-1}\left(\frac{\pi}{2}\right) \\\\ &=-\frac{\pi}{2} \end{aligned}$ The difference between the minimum value, maximum value is $\left(\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)\right)=\pi$
Answer: Given: $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2}, x+y+z=?$ Hint: The maximum value in the range of $\sin ^{-1}$ is $\frac{\pi}{2}$ Solution: $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}$ Here sum of three inverse of $\sin x$ is terms $\frac{\pi}{2}$, i.e. every $\sin$ inverse function is equal to $\frac{\pi}{2}$. $\begin{array}{l} \sin ^{-1} x=\frac{\pi}{2}, \sin ^{-1} y=\frac{\pi}{2}, \sin ^{-1} z=\frac{\pi}{2} \\\\ x=\frac{\sin \pi}{2}, y=\sin \frac{\pi}{2}, z=\sin \frac{\pi}{2} \\\\ x=1, y=1, z=1 \\\\ x+y+z=1+1+1 \\\\ \quad=3 \end{array}$
Answer: $\frac{2 \pi}{3}$ Given: $\cos ^{-1}\left(\frac{1}{2}\right)+2 \sin ^{-1}\left(\frac{1}{2}\right)$ Hint: The range of $\sin is \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right), \ \ range \ \ of \ \ \cos is \ \ \((0, \pi)$ Solution:
Answer:$\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ Given: Range of $\tan ^{-1} x$ Hint: Range of $\tan$ function should be all real number. Solution: $\tan ^{-1} x=\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$
Answer:$\frac{-\pi}{3}$ Given: $\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ Hint: $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ is the range of the principal value branch of inverse sine function. Solution: $Let \ \ y=\sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)$ Then, $\begin{array}{c} \sin y=-\frac{\sqrt{3}}{2}=\sin \left(-\frac{\pi}{3}\right) \\\\ \end{array}$ $\begin{array}{c} y=-\frac{\pi}{3} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \\\\ \end{array}$ $\begin{array}{c} \therefore \sin ^{-1}\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3} \end{array}$
Answer: $\frac{-\pi}{6}$ Given: $\sin ^{-1}\left(-\frac{1}{2}\right)$ Hint: $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ is the range of the principal value branch of inverse sine function. Solution: $\begin{aligned} \text { Let } y=& \sin ^{-1}\left(-\frac{1}{2}\right) \\\\ & \sin y=-\frac{1}{2}=\sin \left(-\frac{\pi}{6}\right) \\\\ & y=-\frac{\pi}{6} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \end{aligned}$ Here, $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is range of principal value of sin $\sin ^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}$
Answer: $\pi$ Given: $\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right)$ Hint: $\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)$ is the range of the principal value branch of sine range of $\cos (0, \pi)$ Solution: We have, $\begin{aligned} \cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{2 \pi}{3}\right) &=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \\ &=\cos ^{-1}\left(\cos \frac{2 \pi}{3}\right)+\sin ^{-1}\left(\sin \frac{\pi}{3}\right) \\ &=\frac{2 \pi}{3}+\frac{\pi}{3} \\ &= \pi \end{aligned}$
Answer: $\frac{\pi}{2}$ Given :$\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}$ Hint: $\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}$ Solution: Using the hint we are able to solve this, $\tan ^{-1} \sqrt{3}+\cot ^{-1} \sqrt{3}=\frac{\pi}{2}$
Answer: $\phi$ Given: $\sec ^{-1}\left(\frac{1}{2}\right)$ Hint: The range of $\sec$ is (-1, 1) . Solution: The value of $\sec ^{-1}\left(\frac{1}{2}\right)$ is undefined as it is outside the range, R-(-1, 1)
Answer: $\emptyset$ Given:$\operatorname{cosec}^{-1}\left(\frac{\sqrt{3}}{2}\right)$ Hint: Range of principal value of $\operatorname{cosec} \ is \ (-1,1)$ Solution: Range of principal value of is undefined as it’s outside the range $(i.e., \(R-(-1,1)\rangle$
Answer: $\pi-\cot ^{-1} x$ Given: $\cot ^{-1}(x) \ \ for \ \ all \ \ x \in R$ in terms of $\cot ^{-1}x$ Hint: $\cot ^{-1}(-x)=\pi-\cot ^{-1} x$ Solution: The value of $\cot ^{-1}(-x)=\pi-\cot ^{-1} x\) \ for \ all \ x \in \mathrm{R}$in terms of $\cot ^{-1} (x)$ is, $=\pi-\cot ^{-1}(x)$
Answer: $\frac{2}{5}$ Given: $\cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0$ Hint: Try to convert the separate $\sin$ , $\cos$ function. So, that the variable get free. Solution: $\begin{array}{l} \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=0 \\\\ \cos \left(\sin ^{-1} \frac{2}{5}+\cos ^{-1} x\right)=\cos \left(\frac{\pi}{2}\right) \\\\ \sin ^{-1} \frac{2}{5}+\cos ^{-1} x=\frac{\pi}{2} \\\\ x=\frac{2}{5} c g f y \end{array}$
Answer: $\frac{\pi}{8}$ Given $\tan ^{-1}\left(\tan \frac{9 \pi}{8}\right)$ Hint: To solve $\tan$ function we can get the solution. Solution: $\begin{aligned} \tan ^{-1}\left(\tan \frac{9 \pi}{8}\right) &=\tan ^{-1}\left(\tan \left(\pi+\frac{\pi}{8}\right)\right) \\ &=\tan ^{-1}\left(\tan \left(\frac{\pi}{8}\right)\right) \\ &=\frac{\pi}{8} \end{aligned}$
Answer: $\frac{3 \pi}{4}$ Given: $\cot ^{-1} 2\ \ and \ \ \cot ^{-1} 3$ , then third angle. Hint: The sum of tree angles of a triangle is $\pi$ Solution: $\begin{array}{l} \tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{2}+\tan ^{-1} x=\pi \\\\ \tan ^{-1} \frac{\left(\frac{1}{3}+\frac{1}{2}\right)}{1-\frac{1}{6}}+\tan ^{-1} x=\pi \end{array}$
RD Sharma Class 12th VSAQ is based on Inverse Trigonometric Functions. To understand this chapter, students need to have a good idea about trigonometry. However, even if you don't have enough experience with trigonometry, you can still learn this chapter as the solution given by Career360 is for students to understand the chapter from scratch. RD Sharma Class 12th VSAQ can be considered small basic questions that the students can quickly go through.
In RD Sharma Class 12th Chapter 3 VSAQ, you will first evaluate the inverse functions and solve basic sums. This can be solved solely through fundamental operations. The higher-level questions which carry more marks are continued in the next section of this chapter.
There are a total of 62 VSAQs in this chapter which might seem quite overwhelming to students. Still, it is nothing to worry about as Class 12 RD Sharma Chapter 3 VSAQ Solution by Career360 has provided the best solutions which you can refer to and efficiently complete this section.VSAQs, although they carry fewer marks in exams but are essential where the weightage is concerned. It is through these VSAQs that you can learn crucial concepts for lengthy and complex problems. This is why you should always practice every one of them.
Practice, Planning, and Execution are the three factors that define an ideal Math student. As trigonometry consists of hundreds of sums, it is best to plan and divide those questions into parts. This method will be easier as you have to solve several sums per day while learning their concepts. Dividing up your work is very beneficial as it lessens the burden and gives the best output.
As Maths is such a vast subject getting a hold of all the concepts is pretty hard. Teachers can’t explain all the sums from the book, so you need to study these concepts beforehand to reduce the work needed afterward.
As far as trigonometry and inverse trigonometry are concerned, they contain hundreds of sums that are quite a lot for students to practice. This is where RD Sharma Class 12th Chapter 3 VSAQ shines. It includes all essential concepts with solutions for Inverse Trigonometry that can be accessible by everyone for free.
A majority of students have joined the team and started learning from Career360's material. You can also avail of this opportunity and download the material for free. This has excellent solutions which will help you score good marks in exams. So stay on top of your class and excel in trigonometric concepts with the help of RD Sharma Class 12 Chapter 3 VSAQ.
This ebook serves as a valuable study guide for NEET exams, specifically designed to assist students in light of recent changes and the removal of certain topics from the NEET exam.
Yes, Career360 offers all the chapters on their website, available for free and accessible by students. You can learn from these solutions to score good marks in exams. For example, to get more insight on Inverse Trigonometry, you can download Class 12 RD Sharma Chapter 3 VSAQ Solution.
2.What are the applications of trigonometry?
These are some of the applications of trigonometry:
Oceanography - To calculate tide heights.
Calculus
Physics - Sine and Cos functions are used as wave representations.
Height calculation - To Measure the height of buildings or mountains.
3.Who invented Inverse Trigonometry?
Daniel Bernoulli introduced inverse trigonometry. This concept was used to determine the angle if two sides of a triangle are given. For more information on Inverse Trigonometry, check out RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ
4.Are NCERT and RD Sharma books similar?
NCERT books for maths are good for general preparation, but RD Sharma books are relatively better to dive deep into the concepts. They are widely considered by teachers and students all over the country and are excellent means of studying for exams. To find the RD Sharma material for Inverse Trigonometry, you can study from RD Sharma Class 12 Solutions Chapter 3 VSAQ.
5.Where is inverse trigonometry used in real life?
nverse trigonometry is a concept through which we can determine the third side of a triangle if two are known. This has its applications across engineering physics, geometry, and navigation on land and sea. To get more information about Inverse Trigonometry, you can refer to RD Sharma Class 12 Solutions Inverse Trigonometric Functions VSAQ.