How can I increase my speed in solving the FBQs in the Inverse Trigonometry chapter of Class 12?

The Class 12 RD Sharma Chapter 3 FBQ will provide you with straightforward ways of solving the questions most simply. Once you practice via this mode, you will complete this portion quickly during your exams.

Where can I find the solutions for Class 12 FBQ online?

You can use the solutions given in the RD Sharma Class 12 Solutions Chapter 3 FBQ available at the Career 360 website. The main advantage is that you need to pay even a single penny to utilize this resource material; everything is available free of cost.

What is the best solution guide for the FBQs in Class 12 chapter 3?

Most of the students use the RD Sharma class 12th FBQ solutions book to find accurate answers for the FBQs.

Is RD Sharma's book suitable for referring to lengthy solutions or FBQs?

The RD Sharma books are the best guide to search for lengthy solutions and one-word answers. In mathematics, every answer is given on how it has arrived.

Which solutions book can the Class 12 students refer to prepare for their public exams?

The RD Sharma solution books are the most recommended materials for students preparing for their class 12 public exams.

RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

Schools
RD Sharma Solutions
RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

Updated on 19 Jan 2022, 10:51 AM IST

RD Sharma's books have been a great companion for the class 12 students preparing for their public examinations. RD Sharma Class 12th FBQ has thirty-five questions in total, having questions like Finding the principal value of trigonometric functions, Measuring angles, Finding the value of x or an expression, and many more. The students who find it hard to solve these questions can use the RD Sharma Class 12 Chapter 3 FBQ Solutions book.

Inverse Trigonometric Functions Excercise:FBQ

Question:38

Fill in the blanks The principle value of $\cos ^{-1}\left ( -\frac{1}{2} \right )$ is ___________.

Answer:

The principal value of $\cos^{-1}\left ( -\frac{1}{2} \right )$ is $\frac{2\pi}{3}$.
Principal value cos^-1 x is [0,$\pi$]
Let, $\cos^{-1}\left ( -1 \right )=\theta$
$\Rightarrow \cos \theta=-\frac{1}{2}$
As, $\cos \frac{2\pi}{3} =-\frac{1}{2}$
So, $\theta= \frac{2\pi}{3}$

Question:39

Fill in the blanks The value of $sin^{-1}\left ( sin \frac{3\pi}{5} \right )$ is_______.

Answer:

The value of $\sin^{-1}\left ( \sin\frac{3\pi}{5} \right )$ is $\frac{2\pi}{5}$
Principal value of $\sin^{-1}$ is $\left [ -\frac{\pi}{2} ,\frac{\pi}{2}\right ]$
now, $\sin^{-1}\left ( \sin\frac{3\pi}{5} \right )$ should be in the given range
$\frac{3\pi}{5}$ is outside the range $\left [ -\frac{\pi}{2} ,\frac{\pi}{2}\right ]$
As, sin (π – x) = sin x
So, $\sin^{-1}\left ( \sin\frac{3\pi}{5} \right )=\sin^{-1}\left ( \sin \left ( \pi-\frac{3\pi}{5} \right ) \right )$
$=\sin^{-1}\left ( \sin\frac{2\pi}{5} \right )$
$=\sin^{-1}\left ( \sin\frac{2\pi}{5} \right )=\frac{2\pi}{5}$

Question:40

Fill in the blanks
If cos (tan^–1x + cot^–1 √3) = 0, then value of x is _________.

Answer:

$\begin{aligned} &\text { If } \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0, \text { then value of } x \text { is } \sqrt{3} \text { . }\\ &\text { Given, } \cos \left(\tan ^{-1} x+\cot ^{-1} \sqrt{3}\right)=0\\ &\Rightarrow \tan ^{-1} x+\cot ^{-1} \sqrt{3}=\frac{\pi}{2}\\ &\text { We know that, } \tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\\ &\text { So, } x=\sqrt{3} \end{aligned}$

Question:41

fill in blanks the set of value of $\sec^{-1}\left (\frac{1}{2} \right )$ is______________.

Answer:

Fill in the blanks the set of value of $\sec^{-1}\left ( \frac{1}{2} \right )$ is $\phi$
Domain of sec^-1 x is R – (-1,1).
As, $-\frac{1}{2}$ is outside domain of sec^-1 x.
Which means there is no set of value of $\sec^{-1}\frac{1}{2}$
So, the solution set of $\sec^{-1}\frac{1}{2}$ is null set or $\phi$

Question:42

Fill in the blanks
The principal value of tan^–1 √3 is _________.

Answer:

The Principal value of $\tan^{-1} \sqrt{3}$ is $\frac{\pi}{3}$
Principal value of tan^-1 x is $\left (-\frac{\pi}{2},\frac{\pi}{2} \right )$
Let, $\tan^{-1}\left ( \sqrt{3} \right )=\theta$
$\Rightarrow \tan \theta=\sqrt{3}$
As $\Rightarrow \tan \frac{\pi}{3}=\sqrt{3}$
so, $\Rightarrow \theta=\sqrt{3}$

Question:43

The value of $\cos^{-1}\left ( \cos \frac{14\pi}{3} \right )$

Answer:

The value of $\cos^{-1}\left ( \cos \frac{14\pi}{3} \right )$ is $\frac{2\pi}{3}$
We needd, $\cos^{-1}\left ( \cos \frac{14\pi}{3} \right )$
Principal value of cos^-1 x is [0,π]
Also, cos (2nπ + θ) = cos θ for all n ? N
$\cos \frac{14\pi}{3}=\cos \left ( 4\pi+\frac{2\pi}{3} \right )$
$\Rightarrow \cos \frac{14\pi}{3}=\cos \frac{2\pi}{3}$
$So, \cos^{-1} \left (\cos \frac{14\pi}{3} \right )=\cos^{-1}\left (\cos \frac{2\pi}{3} \right )$
$\Rightarrow \cos^{-1} \left (\cos \frac{14\pi}{3} \right )=\frac{2\pi}{3}$

Question:44

Fill in the blanks

The value of cos (sin^–1x + cos^–1 x), |x| ≤ 1 is ________.

Answer:

The value of cos (sin^–1 x + cos^–1 x) for |x| ≤ 1 is 0.
cos (sin^–1 x + cos^–1 x), |x| ≤ 1
We know that, (sin^–1 x + cos^–1 x), |x| ≤ 1 is $\frac{\pi}{2}$
So, $\cos\left ( \sin^{-1}x+\cos^{-1}x \right )=\cos\frac{\pi}{2}$
= 0

Question:45

The value of expression $\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x }{2}\right ),$ when $x=\frac{\sqrt{3}}{2}$ is___________.

Answer:

The value of expression $\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x}{2} \right )$ When $X=\frac{\sqrt{3}}{2}$ is 1
$\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x}{2} \right )$ When $X=\frac{\sqrt{3}}{2}$
We know that, (sin^–1 x + cos^–1 x) for all |x| ≤ 1 is $\frac{\pi}{2}$
As, $x=\frac{\sqrt{3}}{2}$ lies in domain
So $\tan\left ( \frac{\sin^{-1}x+\cos^{-1}x}{2} \right )$=$\tan \frac{\pi}{4}$
=1

Question:46

Fill in the blanks if $y= 2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$ for all x, then ______<y<_____.

Answer:

Fill in the blanks if $y= 2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$ for all x, then $-2\pi< y< 2\pi$
$y= 2 \tan^{-1}x+\sin^{-1}\frac{2x}{1+x^{2}}$
We know that,
$2 \tan^{-1}p=\sin^{-1}\frac{2x}{1+x^{2}}$
so
$2 \tan^{1}x+\sin^{-1}\frac{2x}{1+x^{2}}=2 \tan^{1}x+2\tan^{-1}x$
=4 tan^-1 x
So, y = 4 tan^-1 x
As, principal value of tan-1 x is $\left (-\frac{\pi}{2},\frac{\pi}{2} \right )$
So, $4 \tan^{-1}x\epsilon \left ( -2\pi,2\pi \right )$
Hence, -2π < y < 2π

Question:47

The result $\tan^{-1}x-\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$ is true when value of xy is _________.

Answer:

The result $\tan^{-1}x-\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$ is true when value of xy is > -1.
We have,
$\tan^{-1}x-\tan^{-1}=\tan^{-1} \frac{x-y}{1+xy}$
Principal range of tan^-1a is $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
Let tan^-1x = A and tan^-1y = B … (1)
So, A,B $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
We know that, $\tan\left ( A-B \right )=\frac{\tan A - \tan B}{1-\tan A \tan B }$ … (2)
From (1) and (2), we get,
Applying, tan^-1 both sides, we get,
$\tan^{-1}\tan\left ( A-B \right )=\tan^{-1}\frac{x-y}{1-xy}$
As, principal range of tan^-1a is $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
So, for tan^-1tan(A-B) to be equal to A-B,
A-B must lie in $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$– (3)
Now, if both A,B < 0, then A, B $\epsilon \left ( -\frac{\pi}{2},0\right )$
∴ A $\epsilon \left ( -\frac{\pi}{2},0\right )$ and -B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
So, A – B $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{z+xy}$
Now, if both A,B > 0, then A, B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
∴ A $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and -B $\epsilon \left ( -\frac{\pi}{2},0\right )$
So, A – B $\epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan{-1}x-\tan{-1}y=\tan^{-1}\frac{x-y}{z+xy}$
Now, if A > 0 and B < 0,
Then, A $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
∴ A $\epsilon \left ( 0,\frac{\pi}{2}\right )$ and -B $\epsilon \left ( 0,\frac{\pi}{2}\right )$
So, A – B $\epsilon$ (0,π)
But, required condition is A – B $\epsilon$ $\left ( -\frac{\pi}{2},\frac{\pi}{2} \right )$
As, here A – B $\epsilon$ (0,π), so we must have A – B $\epsilon$ $\left ( 0,\frac{\pi}{2} \right )$
$A-B< \frac{\pi}{2}$
$A< \frac{\pi}{2} +B$
Applying tan on both sides,
$\tan A< \tan\left ( \frac{\pi}{2} +B \right )$
As, $\tan\left ( \frac{\pi}{2} +\alpha \right )=-\cot \alpha$
So, tan A < - cot B
Again, $\cot \alpha=\frac{1}{\tan \alpha}$
So, $\tan A< \frac{1}{\tan B}$
⇒ tan A tan B < -1
As, tan B < 0
xy > -1
Now, if A < 0 and B > 0,
Then, A $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$ and B $\epsilon$ $\left ( 0,\frac{\pi}{2} \right )$
∴ A $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$ and -B $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$
So, A – B $\epsilon$ (-π,0)
But, required condition is A – B $\epsilon$ $\left ( -\frac{\pi}{2} ,\frac{\pi}{2}\right )$
As, here A – B $\epsilon$ (0,π), so we must have A – B $\epsilon$ $\left ( -\frac{\pi}{2} ,0\right )$
$\Rightarrow A-B> -\frac{\pi}{2}$
$\Rightarrow A>B -\frac{\pi}{2}$
Applying tan on both sides,
$\tan A>\tan\left (B -\frac{\pi}{2} \right )$
As, $\tan\left (\alpha -\frac{\pi}{2} \right )=-\cot \alpha$
So, tan B > - cot A
Again, $\cot \alpha\frac{1}{\tan \alpha}$
So, $\tan B >-\frac{1}{\tan A}$
⇒ tan A tan B > -1
⇒xy > -1

Question:48

Fill in the blanks

The value of cot^–1(–x) for all x ? R in terms of cot^–1 x is _______.

Answer:

The value of cot^–1(–x) for all x ? R in terms of cot^–1 x is
π – cot^-1 x.
Let cot^–1(–x) = A
⇒ cot A = -x
⇒ -cot A = x
⇒ cot (π – A) = x
⇒ (π – A) = cot^-1 x
⇒ A = π – cot-1 x
So, cot^–1(–x) = π – cot-1 x

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 1

Answer: $15$
Hint:
You must know the value of the trigonometric and inverse trigonometric functions.
Given:
$\sec^{2}\left ( \tan^{-1} 2\right )+cosec^{2}\left ( \cot^{-1}3 \right )$
Solution:
$\sec^{2}\left ( \tan^{-1} 2\right )+cosec^{2}\left ( \cot^{-1}3 \right )$
$\left[\therefore \sec ^{2}x=1+\tan ^{2}x \: \ \&\: \ cosec ^{2}x=1+\cot^{2}x \right ]$
$=1+\tan^{2}\left ( \tan^{-1}2 \right )+1+\cot^{2}\left ( \cot^{-1} 3\right )$
$=1+\left [ \tan\left ( \tan^{-1}2 \right ) \right ]^{2}+1+\left [ \cot\left ( \cot^{-1}3 \right ) \right ]^{2}$
$=1+\left [ 2 \right ]^{2}+1+\left [ 3 \right ]^{2}$
$=1+4+1+9$
$=15$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 2

Answer:
$\frac{\sqrt{3}}{2}$
Hint:
You must know the value of trigonometric and inverse trigonometric function.
Given:
$\sin^{-1}x-\cos^{-1}x= \frac{\pi}{6}$
Solution:
$\sin^{-1}x-\cos^{-1}x= \frac{\pi}{6}$
$\Rightarrow \frac{\pi}{2}-x-x=\frac{\pi}{6}$ $\left [ \because \sin^{-1} x+\cos^{-1}x=\frac{\pi}{2}\right ]$
$\\\Rightarrow-2\cos^{-1}x= \frac{\pi}{6}-\frac{\pi}{2}$
$\Rightarrow$ $-2\cos^{-1}x=-\frac{\pi}{3}$
$\Rightarrow$ $\cos^{-1}x=\frac{\pi}{6}$
$\Rightarrow$ $x=\cos\frac{\pi}{6}$
$\Rightarrow$$\frac{\sqrt{3}}{2}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 3

Answer:
$\left [ \frac{\pi}{4} ,\frac{3\pi}{4}\right ]$
Hint:
You must know the value of trigonometric and inverse trigonometric function.
Given:
Range of $\sin^{-1}x+\cos^{-1}x+\tan^{-1}x$
Solution:
$\sin^{-1}x+\cos^{-1}x+\tan^{-1}x=f\left ( x \right )$
$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$ [Inverse trigonometric function]
So, we get
$\Rightarrow$ $\frac{\pi}{2}+\tan^{-1}x=f\left ( x \right )$
Now, most of us apply the range of $\tan^{-1}x$ as $\left [ \frac{-\pi}{2},\frac{\pi}{2} \right ]$
So, range of $f\left ( x \right )=\left [ 0,\pi \right ]$
But, we know that domain of $f\left ( x \right )$ is $\left [ -1,1 \right ]$
Hence, term $\tan^{-1}x$ does not hold this value$\frac{-\pi}{2}$ and $\frac{\pi}{2}$
So new range becomes,
$\Rightarrow f\left ( -1 \right )=\frac{\pi}{2}-\frac{\pi}{4}=\frac{\pi}{4}$
$\Rightarrow f\left ( 1 \right )=\frac{\pi}{2}+\frac{\pi}{4}=\frac{3\pi}{4}$
So, we get $f\left ( x \right )\epsilon \left [ \frac{\pi}{4},\frac{3\pi}{4} \right ]$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 4

Answer:
$\frac{3\pi}{10}$
Hint:
You must know the rules of trigonometric and inverse trigonometric function.
Given:
$\sin^{-1}x=\frac{\pi}{5}$ for some $x\: \epsilon \left ( -1,1 \right )$, find the value of $\cos^{-1}x$.
Solution:
Inverse trigonometric rule,
$\sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}$
$\Rightarrow$ $\frac{\pi}{5}+cos^{-1}x=\frac{\pi}{2}$
$\Rightarrow$ $cos^{-1}x=\frac{\pi}{2}-\frac{\pi}{5}$
$\Rightarrow$ $cos^{-1}x=\frac{5\pi-2\pi}{10}$
$\Rightarrow$ $cos^{-1}x=\frac{3\pi}{10}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 5

Answer:
$\frac{-\pi}{2}$
Hint:
You must know the rules of trigonometric and inverse trigonometric function.
Given:
$x< 0,\tan^{-1}x+\tan^{-1}\frac{1}{x}$
Solution:
Using the property of inverse trigonometry
$\begin{gathered} \tan ^{-1} x+\tan ^{-1} \frac{1}{x}=\tan ^{-1}\left(\frac{x+\frac{1}{x}}{1-1}\right)\; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \quad\left[\because \tan ^{-1} a+\tan ^{-1} b=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)\right] \\ \end{gathered}$
$=\tan ^{-1}\left(\frac{\frac{x^{2}+1}{x}}{0}\right)$
Since,$\left ( x+\frac{1}{x} \right )$=Negative value $=$ integer $= -a$
$\tan^{-1}\left ( \frac{-a}{0} \right )$
$\Rightarrow \tan^{-1}x+\tan^{-1}\frac{1}{x}=\tan^{-1}\left ( -\infty \right )$
Using value of inverse trigonometry,
$\Rightarrow \tan^{-1}\left ( -\infty \right )=\frac{-\pi}{2}$
Substituting the value, we get
$\Rightarrow \tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{-\pi}{2}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 6

Answer:
$\frac{3\pi}{4}$
Hint:
You must know the value of inverse trigonometric function.
Given:
$\tan^{-1}2+\tan^{-1}3$
Solution:
$\tan^{-1}2+\tan^{-1}3$
$\Rightarrow$ $=\left [ \frac{\pi}{2}-\cot^{-1} 2\right ]+\left [ \frac{\pi}{2}-\cot^{-1} 3\right ]$
$\Rightarrow$ $=\frac{\pi}{2}+\frac{\pi}{2}-\left [ \cot^{-1}2+\cot^{-1}3 \right ]$
$\Rightarrow$ $=\frac{2\pi}{2}-\left [ \tan^{-1}\left ( \frac{1}{2} \right ) +\tan^{-1}\left ( \frac{1}{3} \right )\right ]$
$\Rightarrow$ $=\pi-\left [ \tan^{-1}\left ( \frac{\frac{1}{2}+\frac{1}{3}}{1-\left ( \frac{1}{2} \right )\left ( \frac{1}{3} \right )} \right ) \right ]$ $\left [ \because \tan^{-1}a+\tan^{-1}b =\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]$
$\Rightarrow$ $=\pi-\left [ \tan^{-1}\left ( \frac{\frac{5}{6}}{\frac{5}{6}} \right ) \right ]$
$\Rightarrow$ $=\pi-\left [ \tan^{-1}\left ( 1\right ) \right ]$ $\left [ 1=tan^{-1}(tan\frac{\pi}{4}) \right ]$
$\Rightarrow$ $=\pi-\frac{\pi}{4}$
$\Rightarrow$ $=\frac{4\pi-\pi}{4}$
$\Rightarrow$ $=\frac{3\pi}{4}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 7

Answer:
$\frac{1}{\sqrt3}$
Hint:
You must know the value of inverse trigonometric function.
Given:
$\tan^{-1}\left (\frac{1}{\sqrt3} \right )+\cot^{-1}x=\frac{\pi}{2}$
Solution:
$\tan^{-1}\left (\frac{1}{\sqrt3} \right )+\cot^{-1}x=\frac{\pi}{2}$
$\Rightarrow$ $\tan^{-1}\left (\frac{1}{\sqrt3} \right )=\frac{\pi}{2}-\cot^{-1}x$
$\Rightarrow$ $\tan^{-1}\left (\frac{1}{\sqrt3} \right )=\tan^{-1}x$
$\Rightarrow$ $x=\frac{1}{\sqrt3}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 8

Answer:
$1$
Hint:
You must know the value of inverse trigonometric function.
Given:
$\tan ^{-1}x-\tan^{-1}y=\frac{\pi}{4}$, then find $x-y-xy$
Solution:
$\Rightarrow$ $\tan ^{-1}x-\tan^{-1}y=\frac{\pi}{4}$
$\Rightarrow$ $\tan^{-1}\left ( \frac{x-y}{1+xy} \right )=\frac{\pi}{4}$ $\left [ \because \tan^{-1}a-\tan^{-1}b=\tan^{-1}\left ( \frac{a-b}{1+ab} \right ) \right ]$
$\Rightarrow$ $\left ( \frac{x-y}{1+xy} \right )=\tan\frac{\pi}{4}$
$\Rightarrow$ $\left ( \frac{x-y}{1+xy} \right )=1$
$\Rightarrow$ $\left ( x-y \right )=1+xy$
$\Rightarrow$ $x-y-xy=1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 9

Answer:
$0$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\cot\left ( \tan^{-1} x+\cot^{-1}x\right )$ for all $x\: \epsilon \: R$
Solution:
$= \cot\left ( \tan^{-1} x+\cot^{-1}x\right )$
$\Rightarrow$ $= \cot\left ( \frac{\pi}{2} \right )$ $\left [ \because \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2} \right ]$
$\Rightarrow$ $= \cot\left ( 90^{\circ} \right )$
$\Rightarrow$ $= 0$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 10

Answer:
$\frac{2\pi}{3}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\cos^{-1}x+\cos^{-1}y=\frac{\pi}{3}$, then find $\sin^{-1}x+\sin^{-1}y$
Solution:
We know
$\cos^{-1}x+\sin^{-1}x=\frac{\pi}{2}$ and
$\cos^{-1}y+\sin^{-1}y=\frac{\pi}{2}$
Adding both
$\cos^{-1}x+\sin^{-1}x+\cos^{-1}y+\sin^{-1}y=\frac{\pi}{2}+\frac{\pi}{2}$
$\Rightarrow$ $\cos^{-1}x+\cos^{-1}y+\sin^{-1}x+\sin^{-1}y=\frac{2\pi}{2}$
$\Rightarrow$ $\frac{\pi}{3}+\sin^{-1}x+\sin^{-1}y=\pi$
$\Rightarrow$ $\sin^{-1}x+\sin^{-1}y=\pi-\frac{\pi}{3}$
$\Rightarrow$ $\sin^{-1}x+\sin^{-1}y=\frac{3\pi-\pi}{3}$
$\Rightarrow$ $\sin^{-1}x+\sin^{-1}y=\frac{2\pi}{3}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 11

Answer:
$\pi+\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
Hint:
You must know the values of inverse trigonometric function.
Given:
$x> 0,y> 0,xy> 1$, then find $\tan^{-1}x+\tan^{-1}y$
Solution:
Using inverse trigonometry rule,
$\tan^{-1}x+\tan^{-1}y=\tan^{-1}\left ( \frac{x+y}{1-xy} \right ),xy> 1$
$\tan^{-1}x=a\Rightarrow x=\tan a$
$\tan^{-1}y=b\Rightarrow y=\tan b$
$\Rightarrow$ $0< a+b< \pi,\tan\left ( a+b \right )< 0$
$\frac{\pi}{2}< a+b< \pi$
$\frac{-\pi}{2}< a+b-\pi< 0$
$\Rightarrow$ $\tan\left ( a+b-\pi \right )=-\tan\left [ \pi-\left ( a+b \right ) \right ]$
$\Rightarrow$ $\tan\left ( a+b \right )=\left ( \frac{x+y}{1-xy} \right )$
$\Rightarrow$ $\tan^{-1}\left ( \frac{x+y}{1-xy} \right )=a+b -\pi$
$\Rightarrow$ $a+b= \pi+\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$
$\Rightarrow$ $\tan^{-1}x+\tan^{-1}y= \pi+\tan^{-1}\left ( \frac{x+y}{1-xy} \right )$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 12

Answer:
$\frac{1}{\sqrt2}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$3\sin^{-1}x=\pi-\cos^{-1}x$, then find $x$
Solution:
$3\sin^{-1}x=\pi-\cos^{-1}x$
$\Rightarrow$ $3\sin^{-1}x=\pi-\left [ \frac{-\pi}{2}-\sin^{-1} x\right ]$
$\Rightarrow$ $3\sin^{-1}x=\pi- \frac{\pi}{2}+\sin^{-1} x$
$\Rightarrow$ $3\sin^{-1}x-\sin^{-1} x=\frac{2\pi-\pi}{2}$
$\Rightarrow$ $2\sin^{-1}x=\frac{\pi}{2}$
$\Rightarrow$ $\sin^{-1}x=\frac{\pi}{4}$
$\Rightarrow$ $x=\sin\left (\frac{\pi}{4} \right )$
$\Rightarrow$ $x=\frac{1}{\sqrt{2}}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 13

Answer:
$\frac{\pi}{6}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{-1}x+\tan^{-1}y=\frac{5\pi}{6}$, then find $cot^{-1}x+\cot^{-1}y$
Solution:
Using inverse trigonometric rule,
We know
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$ and
$\tan^{-1}y+\cot^{-1}y=\frac{\pi}{2}$
Adding both equations,
$\begin{aligned} &\Rightarrow \quad \tan ^{-1} x+\cot ^{-1} x+\tan ^{-1} y+\cot ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2} \\\\ &\Rightarrow \quad\left(\tan ^{-1} x+\tan ^{-1} y\right)+\left(\cot ^{-1} x+\cot ^{-1} y\right)=\frac{2 \pi}{2} \\ \\&\Rightarrow \quad \frac{5 \pi}{6}+\left(\cot ^{-1} x+\cot ^{-1} y\right)=\pi \\\\ &\Rightarrow \quad \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{5 \pi}{6} \\ \\&\Rightarrow \quad \cot ^{-1} x+\cot ^{-1} y=\frac{6 \pi-5 \pi}{6} \end{aligned}$
$\Rightarrow \cot^{-1}x+\cot^{-1}y=\frac{\pi}{6}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 14

Answer:
$\tan\frac{5\pi}{12}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{-1}x-\cot^{-1}x=\tan^{-1}\sqrt{3}$, then find $\cot^{-1}x+\cot^{-1}y$
Solution:
$\tan^{-1}x-\cot^{-1}x=\tan^{-1}\sqrt{3}$
$\Rightarrow \tan^{-1}x-\left [ \frac{\pi}{2}-\tan^{-1}x \right ]=\frac{\pi}{3}$
$\Rightarrow \tan^{-1}x-\frac{\pi}{2}+\tan^{-1}x=\frac{\pi}{3}$
$\Rightarrow 2\tan^{-1}x=\frac{\pi}{3}+\frac{\pi}{2}$
$\Rightarrow 2\tan^{-1}x=\frac{5\pi}{6}$
$\Rightarrow \tan^{-1}x=\frac{5\pi}{12}$
$\Rightarrow x=\tan\frac{5\pi}{12}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 15

Answer:
$-1$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{-3\pi}{2}$
Solution:
$\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\frac{-3\pi}{2}$
$\because \frac{-\pi}{2}< \sin^{-1}x\leq \frac{\pi}{2}$
$\because \sin^{-1}x=\frac{-\pi}{2},\sin^{-1}y=\frac{-\pi}{2},\sin^{-1}z=\frac{-\pi}{2}$
$\because x=y=z=-1$
$\because xyz=-1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 16

Answer:
$\frac{\pi}{6}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\cos^{-1}\left \{\sin\left (\cos^{-1}\frac{1}{2} \right ) \right \}$
Solution:
$\cos^{-1}\left \{\sin\left (\cos^{-1}\frac{1}{2} \right ) \right \}$
$\cos^{-1}\left \{\sin\left (\frac{\pi}{3} \right ) \right \}$ $\left [ cos^{-1}\frac{1}{2}=cos^{-1}\left ( cos\frac{\pi}{3} \right )=\frac{\pi}{3} \right ]$
$\cos^{-1}\left \{ \frac{\sqrt{3}}{2} \right \}$ $\left [\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \right ]$
$= \frac{\pi}{6}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 17

Answer:
$1$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan\left [ \cos^{-1}\left \{ \sin\left ( \cot^{-1} 1\right ) \right \} \right ]$
Solution:
$\tan\left [ \cos^{-1}\left \{ \sin\left ( \cot^{-1} 1\right ) \right \} \right ]$ $\left [ \cot^{-1}1=\cot^{-1}\left ( \cot\frac{\pi}{4} \right )=\frac{\pi}{4}\right ]$
$=\tan\left [ \cos^{-1}\left \{ \sin\left ( \frac{\pi}{4}\right ) \right \} \right ]$
$=\tan\left [ \cos^{-1}\left ( \frac{1}{\sqrt2} \right ) \right ]$
$=\tan\left [ \frac{\pi}{4} \right ]$
$=1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 18

Answer:
$23$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{2}\left ( \sec^{-1}3 \right )+\cot^{2}\left ( cosec^{2}4 \right )$
Solution:
$\tan^{2}\left ( \sec^{-1}3 \right )+\cot^{2}\left ( cosec^{2}4 \right )$
$\left [ \tan^{2}x=\sec^{2}x-1 \& \cot^{2}x=cosec^{2}x-1 \right ]$
$= \sec^{2}\left ( \sec^{-1} 3\right )-1+cosec^{2}\left ( cosec^{-1} 4\right )-1$
$=\left ( 3 \right )^{2}+\left ( 4 \right )^{2}-2$
$=9+16-2$
$=23$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 19

Answer:
$\pm \frac{\pi}{6}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{-1}\left ( \cot\theta \right )=2\theta$
Solution:
$\tan^{-1}\left ( \cot\theta \right )=2\theta$
$\Rightarrow \cot\theta=\tan2\theta$
$\Rightarrow \frac{1}{\tan\theta}=\frac{2\tan \theta}{1-\tan^{2}\theta}$
$\Rightarrow 2\tan^{2}\theta=1-\tan^{2}\theta$
$\Rightarrow 3\tan^{2}\theta=1$
$\Rightarrow \tan^{2}\theta=\frac{1}{3}$
$\Rightarrow \tan\theta=\pm \frac{1}{\sqrt{3}}$
$\Rightarrow \theta=\pm \frac{\pi}{6}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 20

Answer:
$\frac{-\pi}{10}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\sin^{-1}\left ( \cos\frac{33\pi}{5} \right )$
Solution:
$\sin^{-1}\left ( \cos\frac{33\pi}{5} \right )$
$=\sin^{-1}\left ( \cos\left ( 6\pi+\frac{3\pi}{5} \right ) \right )$ $\left [ cos(6\pi+\theta) =\cos\theta \right ]$
$=\sin^{-1}\left ( \cos\left ( \frac{3\pi}{5} \right ) \right )$ $\theta =\sin\left ( \frac{\pi}{2}-\theta \right )$
$=\sin^{-1}\left ( \sin\left ( \frac{\pi}{2}-\frac{3\pi}{5} \right ) \right )$
$=\sin^{-1}\left ( \sin\left ( \frac{-\pi}{10} \right ) \right )$
$= \frac{-\pi}{10}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 15

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 21

Answer:
$\frac{\pi}{5}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{-1}x+\tan^{-1}y=\frac{4\pi}{5}$, then find $\cot^{-1}x+\cot^{-1}y$
Solution:
Using inverse rule,
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$ and
$\tan^{-1}y+\cot^{-1}y=\frac{\pi}{2}$
Adding both equations,
$\begin{aligned} &\Rightarrow \quad \tan ^{-1} x+\cot ^{-1} x+\tan ^{-1} y+\cot ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2} \\\\ &\Rightarrow \quad \tan ^{-1} x+\tan ^{-1} y+\cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{2}+\frac{\pi}{2} \\\\ &\Rightarrow \quad \frac{4 \pi}{5}+\cot ^{-1} x+\cot ^{-1} y=\pi \end{aligned}$
$\begin{aligned} \Rightarrow & \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{4 \pi}{5} \\\\ \Rightarrow & \cot ^{-1} x+\cot ^{-1} y=\frac{5 \pi-4 \pi}{5} \\\\ \Rightarrow & \cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{5} \end{aligned}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 22

Answer:
$1$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$3\tan^{-1}x+\cot^{-1}x=\pi$
Solution:
$3\tan^{-1}x+\cot^{-1}x=\pi$ $\left [ \cot^{-1}x+\tan^{-1}x=\frac{\pi}{2} \right ]$
$\begin{aligned} &\Rightarrow \quad 3 \tan ^{-1} x+\left(\frac{\pi}{2}-\tan ^{-1} x\right)= \pi\\\\ &\Rightarrow \quad 3 \tan ^{-1} x-\tan ^{-1} x=\pi-\frac{\pi}{2} \\\\ &\Rightarrow \quad 2 \tan ^{-1} x=\frac{2 \pi-\pi}{2} \\\\ &\Rightarrow \quad 2 \tan ^{-1} x=\frac{\pi}{2} \\\\ &\Rightarrow \quad \tan ^{-1} x=\frac{\pi}{4} \\\\ &\Rightarrow \quad x=\tan \left(\frac{\pi}{4}\right) \end{aligned}$
$\Rightarrow x=1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 23

Answer:

_{$\frac{\pi}{4}$}

Hint:

You must know the sum of three angles of s triangle and inverse trigonometric function.

Given:

_$\tan^{-1}2$ and _$\tan^{-1}3$ are measures of two angles of a triangle, find the third angle.

Solution:

Given two angles are_$\tan^{-1}2$and _$\tan^{-1}3$

Let third angle be_$\theta$

_{$\tan^{-1}2+\tan^{-1}3+\theta=180^{\circ}$}

_{$\Rightarrow \tan^{-1}\left ( \frac{2+3}{1-2\left ( 3 \right )} \right )=180^{\circ}-\theta$} _{$\left [ \because \tan^{-1}a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right ) \right ]$}

_{$\Rightarrow \left (\frac{5}{-5} \right )=\tan\left ( 180-\theta^{\circ} \right )$}

_{$\Rightarrow-1=-\tan\theta$}

_{$\Rightarrow1=\tan\theta$}

_{$\Rightarrow\tan\frac{\pi}{4}=\tan\theta$}

_{$\Rightarrow\theta=\frac{\pi}{4}$}

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 24

Answer:
$\sqrt{ab}$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{-1}\frac{a}{x}+\tan^{-1}\frac{b}{x}=\frac{\pi}{2}$
Solution:
As we know that,
$\tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}$
$\Rightarrow \frac{a}{x}=\frac{\pi}{2}-\frac{a}{x}$ …….. (i)
Substituting the values on equation, we get
$\Rightarrow \frac{\pi}{2}-\frac{a}{x} +\frac{b}{x} =\frac{\pi}{2}$
$\Rightarrow \frac{b}{x} =\frac{a}{x}$
$\Rightarrow \frac{b}{x} =\frac{x}{a}$
$\Rightarrow \frac{b}{x} =\frac{x}{a}$
$\Rightarrow x^{2}=ab$
$\Rightarrow x=\pm \sqrt{ab}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 25

Answer:
$\frac{2}{3}$
Hints:
You must know the rules of inverse trigonometric functions
Given:
$\cos\left ( 2\sin^{-1}x \right )=\frac{1}{9}$
Solution:
$\cos\left ( 2\sin^{-1}x \right )=\frac{1}{9}$
Let $sin^{-1}x=\theta$
$x=sin\theta$
$\therefore \cos2\theta=\left ( \frac{1}{9}\right )$
$1-2\sin^{2}\theta=\frac{1}{9}$
$2\sin^{2}\theta=\frac{8}{9}$
$x^{2}=\frac{4}{9}$
$x=\pm \frac{2}{3}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 26

Answer:
$\pi-2x$
Hints:
You must know the rules of inverse trigonometric functions
Given:
$0<x<\frac{\pi}{2}$ then $\sin^{-1}\left ( \cos x \right )+\cos^{-1}\left ( \sin x\right )$
Solution:
$\sin^{-1}\left ( \cos x \right )+\cos^{-1}\left ( \sin x\right )$
$\sin^{-1}\left ( \sin\left ( \frac{\pi}{2}-x \right ) \right )+\cos^{-1}\left ( \cos\left ( \frac{\pi}{2}-x \right )\right )$ $x=\sin\left ( \frac{\pi}{2}-x \right )]$
$\frac{\pi}{2}-x+\frac{\pi}{2}-x$
$\frac{2\pi}{2}-2x$
$\pi-2x$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 27

Answer:
$\frac{1}{2}$
Hints:
You must know the rules and values of inverse function
Given:
$\tan^{-1}x=\frac{\pi}{4}-\tan^{-1}\frac{1}{3}$
Solution:
$\tan^{-1}x=\frac{\pi}{4}-\tan^{-1}\frac{1}{3}$
$\tan^{-1}x+\tan^{-1}\frac{1}{3}=\frac{\pi}{4}$
$\tan^{-1}\left ( \frac{x+\frac{1}{3}}{1-\frac{x}{3}} \right )=\frac{\pi}{4}$ $\left [ \because \tan^{-1} a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]$
$\frac{3x+1}{3}\times\frac{3}{3-x}=\tan\frac{\pi}{4}$
$\frac{3x+1}{3-x}=1$
$3x+1=3-x$
$3x+x=3-1$
$4x=2$
$x=\frac{1}{2}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 28

Answer:
$\frac{1}{3}$
Hints:
You must know the rules of inverse functions.
Given:
$\tan^{-1}x+\tan^{-1}\frac{1}{2}=\frac{\pi}{4}$

Solution:
$\tan^{-1}x+\tan^{-1}\frac{1}{2}=\frac{\pi}{4}$
$\tan^{-1}\left ( \frac{x+\frac{1}{2}}{1-\frac{x}{2}} \right )=\frac{\pi}{4}$ $\left [ \because \tan^{-1} a+\tan^{-1}b=\tan^{-1}\left ( \frac{a+b}{1-ab} \right )\right ]$
$\frac{2x+1}{2}\times\frac{2}{2-x}=\tan\frac{\pi}{4}$
$2x+1=\left ( 2-x \right )\left ( 1 \right )$
$2x+x=2-1$
$3x=1$
$x=\frac{1}{3}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 29

Answer:
$7$
Hints:
You must know the rules of inverse function.
Given:
$\cot\left ( \frac{\pi}{4} -2\cot^{-1}3\right )$
Solution:
$\cot\left ( \frac{\pi}{4} -2\cot^{-1}3\right )$
Let $\cot^{-1}3=a$
$\Rightarrow \cot a=3$
$\cot \cot\left ( \frac{\pi}{4}-2\left ( \cot a \right ) \right )=\cot \cot\left ( \frac{\pi}{4} -2a\right )$
$=\frac{1}{\tan\left ( \frac{\pi}{4}-2a \right )}$
$=\frac{1+\tan 2a}{1-\tan 2a}$
Now $\cot a=\frac{1}{3}$
$\tan a=\frac{1}{3}$
$\tan 2a=\frac{2\tan a}{1-\tan^{2}a}$
$=\frac{2\times\frac{1}{3}}{1-\frac{1}{9}}$
$=\frac{3}{4}$
So,$\cot\left ( \frac{\pi}{4}-2\cot^{-1} 3\right )$
$=\frac{1+\frac{3}{4}}{1-\frac{3}{4}}$
$=\frac{4+3}{4}\times\frac{4}{4-3}$
$=7$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 30

Answer:
$-\frac{\pi}{3}$
Hints:
You must know the rules and values of inverse trigonometric functions.
Given:
$\tan^{-1}\left ( \tan\frac{2\pi}{3}\right )$
Solution:
$\tan^{-1}\left ( \tan\frac{2\pi}{3}\right )$
We know, $\tan^{-1}\tan x=x,x\: \epsilon \left ( -\frac{\pi}{2},\frac{\pi}{2}\right )$
$\tan^{-1}\left ( \tan\frac{2\pi}{3} \right )=\tan^{-1}\left ( \tan\left ( \pi-\frac{\pi}{3} \right ) \right )$ $\left [ \tan\tan\left ( \pi-\theta \right ) =-\tan\theta \right ]$
$\tan^{-1}\left ( \tan\left ( -\frac{\pi}{3} \right ) \right )$
$\Rightarrow -\frac{\pi}{3}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 31

Answer:
$\left [ -\pi,\pi \right ]$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$y=2\tan^{-1}x+\sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )$
Solution:
$y= 2\tan^{-1}x+\sin^{-1}\left ( \frac{2x}{1+x^{2}} \right )$
Let, $x=\tan \tan\theta$
$\therefore y=2\tan\theta+\left ( \frac{2\tan\tan\theta}{1+\theta} \right )$
$Y=2\theta+\sin^{-1}\left ( \sin2\theta \right )$ $\left [ \therefore \sin2\theta=\left ( \frac{2\tan\theta}{1+\tan^{2}\theta} \right ) \right ]$
$\Rightarrow y=2\theta+2\theta$
$\Rightarrow y=4\theta$ $\left [ \therefore \theta=\tan^{-1} \left ( x \right )\right ]$
$\Rightarrow y=4\tan^{-1}x$
$\because -\frac{\pi}{2}< \tan^{-1}x< \frac{\pi}{2}$
$\because -\frac{4\pi}{2}<4 \tan^{-1}x< \frac{4\pi}{2}$
$\Rightarrow -2\pi< 4\tan^{-1}x< 2\pi$
$\Rightarrow -2\pi< y< 2\pi$ $\left [ y=4\tan^{-1}x \right ]$
Range of $y$ is $\left [ -2\pi,2\pi \right ]$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 32

Answer:
$xy> -1$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$
Solution:
$\tan^{-1}x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$
Principal range of $\tan^{-1}$ is $\left ( \frac{-\pi}{2},\frac{\pi}{2} \right )$
Let $\tan^{-1}x=A$
$\tan^{-1}y=B$
So $\tan^{-1}\left ( \tan\left ( A-B \right ) \right ),A-B$ must be lie in $\left ( \frac{-\pi}{2},\frac{\pi}{2} \right )$
Now, if both $A,B< 0$ , then $A,B\: \epsilon \left ( \frac{-\pi}{2},0 \right )$
$A\: \epsilon \left ( \frac{-\pi}{2},0 \right )$ and $-B\: \epsilon \left ( 0,\frac{-\pi}{2} \right )$
$A-B\: \epsilon \left ( \frac{-\pi}{2},\frac{\pi}{2} \right )$
$\tan^{-1}\left ( \tan\left ( A-B \right ) \right )=A-B$
$\Rightarrow \tan^{-1}x-\tan^{-1}y=\tan^{-1}\left ( \frac{x-y}{1+xy} \right )$
$\Rightarrow A-B< \frac{\pi}{2}$
$\Rightarrow A< \frac{\pi}{2}+B$
Apply $\tan$,
$\Rightarrow \tan A< \tan\left ( \frac{\pi}{2}+B \right )$
$\Rightarrow \tan\left ( \frac{\pi}{2}+\alpha \right )=-\cot\alpha$
$\Rightarrow \cot\alpha=\frac{-1}{\tan\alpha}$
So, $\tan A< \frac{-1}{\tan B}$
$\Rightarrow \tan A\tan B< -1$
$\Rightarrow \tan B< 0$
$\Rightarrow xy> -1$
Similarly,
$\Rightarrow \tan A \tan B> -1$
$\Rightarrow xy> -1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 33

Answer:
$\pi-\cot^{-1}x$
Hints:
You must have know about the rules of inverse functions

Given:
$\cot^{-1}\left ( -x \right )$ for all $x\: \epsilon \: R$ in terms of $\cot^{-1}x$ is
Solutions:
Let $\cot^{-1}\left ( -x \right )=\theta$
$\cot\theta=-x$
$\cot\left ( \pi-\theta \right )=x$
$\pi-\theta=\cot^{-1}x$
$\theta=\pi-\cot^{-1}x$
$\therefore \cot^{-1}\left ( -x \right ) = \pi-\cot^{-1}x$ for $x\: \epsilon \: R$.

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 34

Answer:
$\frac{2\pi}{3}$
Hints:
You must have known about the principal values of inverse functions.
Given:
$\cos^{-1}\left ( -\frac{1}{2}\right )$ is
Solution:
$\cos^{-1}\left ( -\frac{1}{2}\right )=y$
$\cos y=\frac{1}{2}$
$\cos y=\cos\left ( \frac{2\pi}{3} \right )$

Since range of $\cos^{-1}$ is $\left [ 0,\pi \right ]$
Hence, principal value is $\frac{2\pi}{3}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 35

Answer:
$(-\infty,-1]\cup [1,\infty)$
Hints:
You must have known about the principal values and range of inverse trigonometric function
Given:
$y=\sec^{-1}x$
Solution:
$y=\sec^{-1}x$
The range of principal value branch of $y=\sec^{-1}x$
Principal value branch of $y=\sec^{-1}x$ is$\left [ 0,\pi \right ]-\left \{ \frac{\pi}{2} \right \}$
And the range is in $\left ( 1,\infty \right )$ it is increasing and $\left (-\infty,-1 \right )$ it is increasing
The function does not have any inflection points
$\therefore$The range of $y=\sec^{-1}x$ is $(-\infty,-1]\cup [1,\infty)$

Even though the questions are given in objective types, each student must work out the whole sum to find its solution. This consumes time; hence the RD Sharma Class 12 Chapter 3 FBQ solutions will lend a helping hand to the students. Moreover, as the experts deliver the keys, the assurance of the correctness of each answer is strong. And if you belong to the CBSE board institutions, it is a wise option to use the RD Sharma Solutions to increase your ability in solving mathematical solutions.

Moreover, the RD Sharma Class 12th FBQ solutions are available on top educational websites like Career360 for free of cost. Hence, you need not spend hundreds or thousands of rupees in purchasing solution materials. Instead, click on the download button and save the RD Sharma solutions material on your device.

In the RD Sharma Class 12 Solutions Inverse Trigonometry FBQ, there are numerous practice questions that you can try before your tests and exams to increase your efficiency and speed. The students need not search for the answers; it is given in the same order as provided in the textbook. Based on the NCERT pattern, CBSE school students can make the most use of it.

There are many ways to find a solution for a mathematical function. The answers in this book are given in many possible ways. This will make you choose the method that you find easy and prepare accordingly. You can use RD Sharma Class 12th FBQ solutions while doing your homework, making your assignments, and even preparing for the examinations.

Hence, if you start using this book from day 1 in 12th grade, none can stop you from scoring good marks. It is better to make it a practice to work out sums every day from the RD Sharma Class 12 Solutions Chapter 3 FBQ book. Once the students work out the problems from the RD Sharma solution books, they can find themselves with better marks than their previous performance.

There were instances when students had to work out sums from the RD Sharma book during their public exams. Hence, it will be easier to attempt questions from the RD Sharma solutions book if you have practiced. Do not slip the chance of elevating your score by using this free material.

Articles

JAC Board Class 10 Half Yearly Question Paper 2025-26 - Download Solutions PDF

Sep 24, 2025

IGKO Olympiad Sample Papers 2025-26 Download PDF for Class 1 to 10

Sep 24, 2025

HBSE Exam Registration 2026 for 10th & 12th: Steps, Dates & Fees

Sep 24, 2025

NCERT Solutions for Class 10 Maths Chapter 14 Probability

Sep 24, 2025