RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

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RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

Kuldeep MauryaUpdated 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 $(-\infty,-1]\cup [1,\infty)$

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