RD Sharma Class 12 Exercise 3.3 Inverse Trigonometry Function Solutions Maths

RD Sharma Class 12 Exercise 3.3 Inverse Trigonometry Function Solutions Maths - Download PDF Free Online

Updated on Jan 21, 2022 11:59 AM IST

RD Sharma solutions are pretty popular and well-known among students and teachers. They have been involved in making NCERT solutions for a very long time and have already established their books to be the perfect guide for home study. The RD Sharma class 12th exercise 3.3 is one such NCERT solution set that is extremely helpful to students who have opted for the mathematics subject. RD Sharma class 12 chapter 3 exercise 3.3 is a great book section on NCERT solution that focuses on the third chapter of the mathematics book. Chapter 3 of the book deals with Inverse Trigonometric Functions, which is a critical chapter with many complicated theories. The solutions have nine questions, including subparts in level 1, which cover the concept of principal values.

RD Sharma Class 12 Solutions Chapter 3 Inverse Trigonometric Functions - Other Exercise

Inverse Trigonometric Functions Excercise: 3.3

Inverse Trigonometric Functions Exercise 3.3 Question 1 (i)

Answer :$\frac{\pi }{6}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.
$\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Given : $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
Explanation :
Let $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=y$ .....(i)
$\tan y=\frac{1}{\sqrt{3}} \quad\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[\because \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$
$\tan y=\tan \frac{\pi}{6}$
$\begin{aligned} &y=\frac{\pi}{6} \\ &\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \end{aligned}$ [ From equation (i) ]
The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Hence, principal value of $\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)=\frac{\pi}{6}$.

Inverse Trigonometric Functions Exercise 3.3 Question 1 (ii)

Answer :$\frac{-\pi }{6}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan ^{-1}$.
$\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Given :$\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$
Explanation :
Let $y=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$ ....(i)
$\tan y=\frac{-1}{\sqrt{3}}$
$\tan y=-\tan \left(\frac{\pi}{6}\right)$
$\tan y=\tan \left(\frac{-\pi}{6}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad \left [ \because \tan (-\theta)=-\tan \theta \right ]$
$\begin{aligned} &y=\frac{-\pi}{6} \\ &\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \end{aligned}$ [From equation (i)]
The range of principal value branch of $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\because \quad \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Hence, principal value of $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6}$.

Inverse Trigonometric Functions Exercise 3.3 Question 1 (iii)

Answer : 0
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.
$\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Given :$\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$
Explanation :
Let $y=\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$ ....(i)
$y=\tan ^{-1}(0) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{2}=0\right]$
$\tan \; y=0$
$\tan y=\tan 0^{\circ} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan 0^{\circ}=0\right]$
$y=0$
$\tan ^{-1}\left(\cos \frac{\pi}{2}\right)=0$
The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\because \quad \tan ^{-1}\left(\cos \frac{\pi}{2}\right)=0 \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Hence, the principal value of $\tan ^{-1}\left(\cos \frac{\pi}{2}\right)$ is 0.

Inverse Trigonometric Functions Exercise 3.3 Question 1 (iv)

Answer : $\frac{-\pi }{4}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$
Thus, $\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Given : $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$
Explanation :
Let $y=\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$ ....(i)
$y=\tan ^{-1}\left[2\left(\frac{-1}{2}\right)\right]$ $\left [ \because \cos\frac{2\pi }{3}=\frac{-1}{2} \right ]$
$\begin{aligned} &y=\tan ^{-1}(-1) \\ &\tan y=-1 \end{aligned}$
$\tan y=-\tan \left(\frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \left [ \because \tan (-\theta)=-\tan \theta \right ]$
$y=\frac{-\pi}{4}$
$\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=\frac{-\pi}{4}$
The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\because \quad \tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)=\frac{-\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Hence, the principal value of $\tan ^{-1}\left(2 \cos \frac{2 \pi}{3}\right)$ is $\frac{-\pi}{4}$.

Inverse Trigonometric Functions Exercise 3.3 Question 2 (i)

Answer :$\frac{\pi }{2}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.
Thus, $\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
The branch with range $\left [ 0,\pi \right ]$ is called the principal value branch of the function $\cos^{-1}$ and domain of the function$\cos ^{-1} \text { is }[-1,1]$.
Thus, $\cos ^{-1}:[-1,1] \rightarrow[0, \pi]$
Given : $\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
Explanation :
Let us first solve for $\tan ^{-1}(-1)$
Let $y=\tan ^{-1}(-1)$ ....(i)
$\text { tan } y=-1 \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{4}=1\right]$
$\tan y=-\tan \frac{\pi}{4}$
$\tan y=\tan \left(\frac{-\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta]$
$\begin{aligned} &y=\frac{-\pi}{4} \\ &\tan ^{-1}(-1)=\frac{-\pi}{4} \end{aligned}$ [from equation (i)]
The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\begin{aligned} &\tan ^{-1}(-1)=\frac{-\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\\ &\therefore \text { Principal value of }\\ &\tan ^{-1}(-1)=\frac{-\pi}{4} \end{aligned}$ .....(ii)
Let us solve for $\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$
Let $x=\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)$ .....(iii)
$\cos x=\left(\frac{-1}{\sqrt{2}}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{4}=\frac{1}{\sqrt{2}}\right]$
$\cos x=-\cos \frac{\pi}{4}$
$\cos x=\cos \left(\pi-\frac{\pi}{4}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (\pi-\theta)=-\cos \theta]$
$\begin{aligned} &\cos x=\cos \left(\frac{3 \pi}{4}\right) \\ &x=\frac{3 \pi}{4} \\ &\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \end{aligned}$ [From equation (iii)]
The range of principal value branch of $\cos^{-1}$ is $\left [ 0,\pi \right ]$.
$\therefore \quad \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right)=\frac{3 \pi}{4} \in[0, \pi]$
$\therefore \text { Principal value of } \cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) \text { is } \frac{3 \pi}{4} \text { . }$ .....(iv)
Now,
$\begin{aligned} &\tan ^{-1}(-1)+\cos ^{-1}\left(\frac{-1}{\sqrt{2}}\right) \\ &=\frac{-\pi}{4}+\frac{3 \pi}{4} \end{aligned}$ [From equation (ii) and (iv)]
$\begin{aligned} &=\frac{2 \pi}{4} \\ &=\frac{\pi}{2} \end{aligned}$.

Inverse Trigonometric Functions Exercise 3.3 Question 2 (ii)

Answer :$\frac{\pi }{3}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.
Thus, $\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
The branch with range $\left [ 0,\pi \right ]$ is called the principal value branch of the function $\cos^{-1}$ and domain of the function $\cos^{-1}$ is $\left [ -1, 1\right ]$.
Thus, $\cos ^{-1}:[-1,1] \rightarrow[0, \pi]$
Given :
$\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$
Explanation :
Let is first solve for $\cos ^{-1} \frac{\sqrt{3}}{2}$.
Let, $y=\cos ^{-1} \frac{\sqrt{3}}{2}$ .....(i)
$\begin{aligned} &\cos y=\frac{\sqrt{3}}{2} \\ &\cos y=\cos \frac{\pi}{6} \\ &y=\frac{\pi}{6} \\ &\cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6} \end{aligned}$
$\because$ The range of principal value branch of $\cos^{-1}$ is $\left [ 0,\pi \right ]$.
$\therefore \; \; \; \; \; \; \; \; \quad \cos ^{-1} \frac{\sqrt{3}}{2}=\frac{\pi}{6} \in[0, \pi]$
The principal value of $\cos ^{-1} \frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$ .....(ii)
Now,
$\begin{aligned} &\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\} \\ &=\tan ^{-1}\left\{2 \sin \left(4 \times \frac{\pi}{6}\right)\right\} \end{aligned}$ [From equation (ii)]
$=\tan ^{-1}\left\{2 \sin \left(\frac{2 \pi}{3}\right)\right\}$
$=\tan ^{-1}\left(2 \times \frac{\sqrt{3}}{2}\right)$ $\left[\because \sin \frac{2 \pi}{3}=\frac{\sqrt{3}}{2}\right]$
$=\tan ^{-1}(\sqrt{3})$ .....(iii)
Now,
Let $x=\tan ^{-1}(\sqrt{3})$ ......(iv)
$\begin{aligned} &\tan x=\sqrt{3} \\ &\tan x=\tan \frac{\pi}{3} \\ &x=\frac{\pi}{3} \end{aligned}$
$\tan ^{-1}(\sqrt{3})=\frac{\pi}{3}$ [From equation iv]
$\because$ The range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\because \quad \tan ^{-1}(\sqrt{3})=\frac{\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore$ Principal value of ${\tan ^{-1}}(\sqrt{3})$ is $\frac{\pi}{3}$.
Hence principal vaue of $\tan ^{-1}\left\{2 \sin \left(4 \cos ^{-1} \frac{\sqrt{3}}{2}\right)\right\}$is $\frac{\pi}{3}$.

Inverse Trignometric Function Exercise 3.3 Question 3 (i) .

Answer : $\frac{3\pi}{4}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.
Thus, $\tan ^{-1}: R \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
The branch with range $\left [ 0,\pi \right ]$ is called the principal value branch of the function $\cos^{-1}$.
Thus, $\cos ^{-1}:[-1,1] \rightarrow[0, \pi]$
The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch.
Thus, $\sin ^{-1}:[-1,1] \rightarrow\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Domain of functions $\sin^{-1}$ and $\cos^{-1}$ are $\left [ -1,1 \right ]$ and $\left [ -1,1 \right ]$ respectively.
Given : $\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$
Explanation :
Let us first solve for $\tan ^{-1}(1)$
Let
$\begin{aligned} &y=\tan ^{-1}(1) \\ &\tan y=1 \end{aligned}$ ...(i)
$\tan y=\tan \frac{\pi}{4} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{4}=1\right]$
$\begin{aligned} &y=\frac{\pi}{4} \\ &\tan ^{-1}(1)=\frac{\pi}{4} \end{aligned}$ [From equation (i)]
$\because$ Range of principal value branch of $\tan^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore \quad \tan ^{-1}(1)=\frac{\pi}{4} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Therefore principal of value of $\tan^{-1} (1)$ is $\frac{\pi}{4}$ ....(ii)
Now,
Let us solve for $\sin ^{-1}\left(\frac{-1}{2}\right)$
Let $z=\sin ^{-1}\left(\frac{-1}{2}\right)$ ....(iii)
$\begin{array}{ll} \sin z=\frac{-1}{2} &\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {\left[\because \sin \frac{\pi}{6}=\frac{1}{2}\right]} \\ \sin z=-\sin \frac{\pi}{6} & \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; {[\because \sin (-\theta)=-\sin \theta]} \end{array}$
$\begin{aligned} &\sin z=\sin \left(\frac{-\pi}{6}\right) \\ &z=\frac{-\pi}{6} \\ &\sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6} \end{aligned}$ [From equation (iii)]
The range of principal value branch of $\sin^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\because \sin ^{-1}\left(\frac{-1}{2}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
$\therefore$ Principal value of $\sin ^{-1}\left(\frac{-1}{2}\right) \text { is } \frac{-\pi}{6}$ ......(iv)
Now,
Let us solve for $\cos ^{-1}\left(\frac{-1}{2}\right)$
Let $x=\cos ^{-1}\left(\frac{-1}{2}\right)$ ....(v)
$\begin{aligned} &\cos x=\frac{-1}{2} \\ &\cos x=-\cos \frac{\pi}{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}$
$\cos x=\cos \left(\pi-\frac{\pi}{3}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (\pi-\theta)=-\cos \theta]$
$\begin{aligned} &\cos x=\cos \left(\frac{2 \pi}{3}\right) \\ &x=\frac{2 \pi}{3} \\ &\cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3} \end{aligned}$ [From equation (v)]
$\because$ The range of principal value branch of $\cos^{-1}$ is $\left [ 0,\pi \right ]$.
$\because \quad \cos ^{-1}\left(\frac{-1}{2}\right)=\frac{2 \pi}{3} \in[0, \pi]$
$\therefore$ Principal value of $\cos ^{-1}\left(\frac{-1}{2}\right)$ is $\frac{2\pi}{3}$ ...(vi)
Now,
$\tan ^{-1}(1)+\cos ^{-1}\left(\frac{-1}{2}\right)+\sin ^{-1}\left(\frac{-1}{2}\right)$
Putting the values from equations (ii), (iv) and (vi)
$\begin{aligned} &=\frac{\pi}{4}+\frac{2 \pi}{3}+\left(\frac{-\pi}{6}\right) \\ &=\frac{\pi}{4}+\frac{2 \pi}{3}-\frac{\pi}{6} \\ &=\frac{3 \pi+8 \pi-2 \pi}{12} \\ &=\frac{9 \pi}{12} \\ &=\frac{3 \pi}{4} \end{aligned}$

Inverse Trignometric Function Exercise 3.3 Question 3 (ii) .

Answer : $\frac{-3\pi}{4}$
Hint : The branch with range $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$ is called the principal value branch of function $\tan^{-1}$.
Given : $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)$
Explanation :
Let us first solve for $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$
Let $x=\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)$
$\begin{aligned} &\tan x=\frac{-1}{\sqrt{3}} \\ &\tan x=-\tan \frac{\pi}{6} \end{aligned}$ $\left[\because \tan \frac{\pi}{6}=\frac{1}{\sqrt{3}}\right]$
$\tan x=\tan \left(\frac{-\pi}{6}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta]$
$\begin{aligned} &x=\frac{-\pi}{6} \\ &\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \end{aligned}$ [From equation (i) ]
The range of principal value branch of $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\because \; \; \; \; \tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Hence principal value of $\tan ^{-1}\left(\frac{-1}{\sqrt{3}}\right)=\frac{-\pi}{6}$ ... (ii)
Now,
Let us solve for $\tan ^{-1}(-\sqrt{3})$
Let $y=\tan ^{-1}(-\sqrt{3})$ ...(iii)
$\begin{aligned} &\tan y=-\sqrt{3} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan \frac{\pi}{3}=\sqrt{3}\right] \\ &\tan y=-\tan \left(\frac{\pi}{3}\right) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (-\theta)=-\tan \theta] \end{aligned}$
$\begin{aligned} &\tan y=\tan \left(\frac{-\pi}{3}\right) \\ &y=\frac{-\pi}{3} \end{aligned}$ [From equation (iii)]
$\tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}$
The Range of principal value branch of $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\because \tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3} \in\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$
Hence principal value of $\tan ^{-1}(-\sqrt{3})=\frac{-\pi}{3}$ ...(iv)
Now,
Let us solve for $\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)$
$\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\tan ^{-1}(-1) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \sin \frac{\pi}{2}=1\right]$
$\begin{aligned} &=\tan ^{-1}\left(\tan \left(\frac{-\pi}{4}\right)\right) \\ &=\frac{-\pi}{4} \\ \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right) &=\frac{-\pi}{4} \end{aligned}$
The Range of principal value branch of $\tan ^{-1}$ is $\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]$.
$\therefore \tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\frac{-\pi}{4} \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$
Hence principal value $\tan ^{-1}\left(\sin \left(\frac{-\pi}{2}\right)\right)=\frac{-\pi}{4}$ .....(v)
Now consider
$\tan ^{-1}\left(-\frac{1}{\sqrt{3}}\right)+\tan ^{-1}(-\sqrt{3})+\tan ^{-1}\left(\sin \left(-\frac{\pi}{2}\right)\right)$
From equation (ii), (iv) and (v)
$\begin{aligned} &=-\frac{\pi}{6}+\left(-\frac{\pi}{3}\right)+\left(\frac{-\pi}{4}\right) \\ &=-\frac{\pi}{6}-\frac{\pi}{3}-\frac{\pi}{4} \\ &=\frac{-2 \pi-4 \pi-3 \pi}{12} \\ &=\frac{-9 \pi}{12} \\ &=\frac{-3 \pi}{4} \end{aligned}$

Inverse Trignometric Function Exercise 3.3 Question 3 (iii) .

Answer : 0
Given : $\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)$
Explanation :
$\begin{aligned} &\tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right) \\ &=\tan ^{-1}\left\{\tan \left(\pi-\frac{\pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(2 \pi+\frac{\pi}{6}\right)\right\}\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \;\; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \cos (2 \pi+\theta)=\cos \theta] \end{aligned}$
$\begin{aligned} &=\tan ^{-1}\left\{-\left(\tan \frac{\pi}{6}\right)\right\}+\cos ^{-1}\left\{\cos \left(\frac{\pi}{6}\right)\right\} \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad[\because \tan (\pi-\theta)=-\tan \theta] \\ &=\tan ^{-1}\left\{\tan \left(\frac{-\pi}{6}\right)\right\}+\frac{\pi}{6} \\ &=\frac{-\pi}{6}+\frac{\pi}{6} \\ &=0 \end{aligned}$
$\therefore \quad \tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=0$

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There are many NCERT solutions online, which offer homework solutions to students. However, RD Sharma solutions are easily the most trusted book that is followed by thousands of students. Teachers also like to use these books to provide home tasks, so they are very useful when solving homework problem

5. Are RD Sharma solutions expensive?

RD Sharma Solutions are not all expensive. They are entirely free of cost and don't come with any hidden fees. You can download the soft copy of the book from Career360, which has the latest and updated versions of the books. In addition, you can print out the pdf, which can be done for a nominal amount.