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

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

Edited By Kuldeep Maurya | 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.

## 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)

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.

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) .

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}
$\therefore \quad \tan ^{-1}\left(\tan \frac{5 \pi}{6}\right)+\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)=0$

The RD Sharma class 12 solutions chapter 3 ex 3.3 seeks to provide answers and solutions to all the questions present in the NCERT books. The class 10 RD Sharma chapter 3 exercise 3.3 solution has only expert-created solutions for students. The experts who have written and compiled the answer are highly skilled and have a lot of knowledge in mathematics. Moreover, they have provided some new and improved ways of solving mathematical problems, which would make it easier for you to answer the questions. These unique methods you will find only in class 12 RD Sharma chapter 3 exercise 3.3 solution.

The answers provided in RD Sharma class 12 solutions Inverse Trigonometry Function ex 3.3 are simple to follow and easily understood by students. The detailed answers and steps will be of immense help to students to improve their knowledge of Inverse Trigonometry. Using these solutions will be able to clear their doubts and focus on their weak points to make improvements. With the help of the RD Sharma class 12th exercise 3.3, you will learn everything in detail and score high in your exams.

Since RD Sharma is a trusted source for solutions, teachers like to use these books to give homework to students. Students having difficulty completing homework on their own can use these books to get some help. They will find all the answers to the questions that are in the NCERT book. The pdf of RD Sharma class 12 solutions chapter 3 ex 3.3 is updated every year to include the latest syllabus so that you are not left wanting in any way.

RD Sharma class 12th exercise 3.3 Solutions is the top choice for students for board exam preparations and self-practice at home. You can easily trust them to help you improve and enhance your knowledge so that you won't have to purchase too many study materials. You will find the pdf of class 12 RD Sharma chapter 3 exercise 3.3 solution online on Career360, which is the one-stop destination for RD Sharma solutions. These pdfs are available to download at any time and are completely free of cost. With all these benefits of using RD Sharma solutions, you can be sure that the book will help you.

Chapter-wise RD Sharma Class 12 Solutions

1. How can I avail myself the class 12 RD Sharma chapter 3 exercise 3.3 solution?

Finding the downloading the class 12 RD Sharma chapter 3 exercise 3.3 solution is very easy. All you have to do is navigate to the Career360 website and download the free pdf of the book from there. The pdf of the book is always available and does not require any subscription fees at all.

2. What are the benefits of using RD Sharma class 12th exercise 3.3 solutions?

Students and teachers have trusted the RD Sharma class 12th exercise 3.3 solutions for a long time. They have already asserted that students who have practiced the NCERT solutions have found common questions in the board exams. The solutions will also help you to test your knowledge at home and improve upon your weak areas.

3. Are our RD Sharma solutions updated according to the latest syllabus?

RD Sharma seeks to include all the answers to the questions in the NCERT book. For this reason, they are updated every year to include the latest syllabus of CBSE class 12. Therefore, you can be entirely sure that the book will have all the answers you require. All you have to do is download the correct file according to your exam year.

4. Which is the best NCERT solution for solving homework?

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.

## Upcoming School Exams

#### National Means Cum-Merit Scholarship

Application Date:01 August,2024 - 16 September,2024

#### National Rural Talent Scholarship Examination

Application Date:05 September,2024 - 20 September,2024