RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

Question:38

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

Answer:

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

Question:39

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

Answer:

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

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 $\varphi$
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 $\varphi$

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 $(-\frac{\pi }{2},\frac{\pi }{2})$
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 ${\cos }^{-1}(\cos \frac{14\pi }{3})$

Answer:

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

Question:44

Fill in the blanks

The value of cos (sin^–1 x + 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, $\cos ({\sin }^{-1}x+{\cos }^{-1}x)=\cos \frac{\pi }{2}$
= 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 $(-\frac{\pi }{2},\frac{\pi }{2})$
So, $4{\tan }^{-1}xϵ(-2\pi ,2\pi )$
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 $(-\frac{\pi }{2},\frac{\pi }{2})$
Let tan^-1x = A and tan^-1y = B … (1)
So, A,B $ϵ(-\frac{\pi }{2},\frac{\pi }{2})$
We know that, $\tan (A-B)=\frac{\tan A-\tan B}{1-\tan A\tan B}$ … (2)
From (1) and (2), we get,
Applying, tan^-1 both sides, we get,
${\tan }^{-1}\tan (A-B)={\tan }^{-1}\frac{x-y}{1-xy}$
As, principal range of tan^-1a is $(-\frac{\pi }{2},\frac{\pi }{2})$
So, for tan^-1tan(A-B) to be equal to A-B,
A-B must lie in $(-\frac{\pi }{2},\frac{\pi }{2})$– (3)
Now, if both A,B < 0, then A, B $ϵ(-\frac{\pi }{2},0)$
∴ A $ϵ(-\frac{\pi }{2},0)$ and -B $ϵ(0,\frac{\pi }{2})$
So, A – B $ϵ(-\frac{\pi }{2},\frac{\pi }{2})$
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 $ϵ(0,\frac{\pi }{2})$
∴ A $ϵ(0,\frac{\pi }{2})$ and -B $ϵ(-\frac{\pi }{2},0)$
So, A – B $ϵ(-\frac{\pi }{2},\frac{\pi }{2})$
So, from (3),
tan^-1tan(A-B) = A-B
$\Rightarrow \tan -1x-\tan -1y={\tan }^{-1}\frac{x-y}{z+xy}$
Now, if A > 0 and B < 0,
Then, A $ϵ(0,\frac{\pi }{2})$ and B $ϵ(0,\frac{\pi }{2})$
∴ A $ϵ(0,\frac{\pi }{2})$ and -B $ϵ(0,\frac{\pi }{2})$
So, A – B $ϵ$ (0,π)
But, required condition is A – B $ϵ$ $(-\frac{\pi }{2},\frac{\pi }{2})$
As, here A – B $ϵ$ (0,π), so we must have A – B $ϵ$ $(0,\frac{\pi }{2})$
$A-B<\frac{\pi }{2}$
$A<\frac{\pi }{2}+B$
Applying tan on both sides,
$\tan A<\tan (\frac{\pi }{2}+B)$
As, $\tan (\frac{\pi }{2}+\alpha )=-\cot \alpha$
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 $ϵ$ $(-\frac{\pi }{2},0)$ and B $ϵ$ $(0,\frac{\pi }{2})$
∴ A $ϵ$ $(-\frac{\pi }{2},0)$ and -B $ϵ$ $(-\frac{\pi }{2},0)$
So, A – B $ϵ$ (-π,0)
But, required condition is A – B $ϵ$ $(-\frac{\pi }{2},\frac{\pi }{2})$
As, here A – B $ϵ$ (0,π), so we must have A – B $ϵ$ $(-\frac{\pi }{2},0)$
$\Rightarrow A-B>-\frac{\pi }{2}$
$\Rightarrow A>B-\frac{\pi }{2}$
Applying tan on both sides,
$\tan A>\tan (B-\frac{\pi }{2})$
As, $\tan (\alpha -\frac{\pi }{2})=-\cot \alpha$
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:
${\sec }^2({\tan }^{-1}2)+cose{c}^2({\cot }^{-1}3)$
Solution:
${\sec }^2({\tan }^{-1}2)+cose{c}^2({\cot }^{-1}3)$
$[\therefore {\sec }^{2}x=1+{\tan }^{2}x\mathrm{\&}cose{c}^{2}x=1+{\cot }^{2}x]$
$=1+{\tan }^2({\tan }^{-1}2)+1+{\cot }^2({\cot }^{-1}3)$
$=1+{[\tan ({\tan }^{-1}2)]}^2+1+{[\cot ({\cot }^{-1}3)]}^2$
$=1+{\left[2\right]}^2+1+{\left[3\right]}^2$
$=1+4+1+9$
$=15$

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:
${\sin }^{-1}x-{\cos }^{-1}x=\frac{\pi }{6}$
$\Rightarrow \frac{\pi }{2}-x-x=\frac{\pi }{6}$ $[\because {\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}]$
$\Rightarrow -2{\cos }^{-1}x=\frac{\pi }{6}-\frac{\pi }{2}$
$\Rightarrow$ $-2{\cos }^{-1}x=-\frac{\pi }{3}$
$\Rightarrow$ ${\cos }^{-1}x=\frac{\pi }{6}$
$\Rightarrow$ $x=\cos \frac{\pi }{6}$
$\Rightarrow$$\frac{\sqrt{3}}{2}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 3

Answer:
$[\frac{\pi }{4},\frac{3\pi }{4}]$
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 $[\frac{-\pi }{2},\frac{\pi }{2}]$
So, range of $f\left(x\right)=[0,\pi ]$
But, we know that domain of $f\left(x\right)$ is $[-1,1]$
Hence, term ${\tan }^{-1}x$ does not hold this value$\frac{-\pi }{2}$ and $\frac{\pi }{2}$
So new range becomes,
$\Rightarrow f(-1)=\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$
$\Rightarrow f\left(1\right)=\frac{\pi }{2}+\frac{\pi }{4}=\frac{3\pi }{4}$
So, we get $f\left(x\right)ϵ[\frac{\pi }{4},\frac{3\pi }{4}]$

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 $xϵ(-1,1)$, find the value of ${\cos }^{-1}x$.
Solution:
Inverse trigonometric rule,
${\sin }^{-1}x+{\cos }^{-1}x=\frac{\pi }{2}$
$\Rightarrow$ $\frac{\pi }{5}+co{s}^{-1}x=\frac{\pi }{2}$
$\Rightarrow$ $co{s}^{-1}x=\frac{\pi }{2}-\frac{\pi }{5}$
$\Rightarrow$ $co{s}^{-1}x=\frac{5\pi -2\pi }{10}$
$\Rightarrow$ $co{s}^{-1}x=\frac{3\pi }{10}$

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
$\begin{array}{c}{\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}={\tan }^{-1}\left(\frac{x+\frac{1}{x}}{1-1}\right)[\because {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)]\end{array}$
$={\tan }^{-1}\left(\frac{\frac{x^2+1}{x}}{0}\right)$
Since,$(x+\frac{1}{x})$=Negative value $=$ integer $=-a$
${\tan }^{-1}\left(\frac{-a}{0}\right)$
$\Rightarrow {\tan }^{-1}x+{\tan }^{-1}\frac{1}{x}={\tan }^{-1}(-\mathrm{\infty })$
Using value of inverse trigonometry,
$\Rightarrow {\tan }^{-1}(-\mathrm{\infty })=\frac{-\pi }{2}$
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:
${\tan }^{-1}2+{\tan }^{-1}3$
$\Rightarrow$ $=[\frac{\pi }{2}-{\cot }^{-1}2]+[\frac{\pi }{2}-{\cot }^{-1}3]$
$\Rightarrow$ $=\frac{\pi }{2}+\frac{\pi }{2}-[{\cot }^{-1}2+{\cot }^{-1}3]$
$\Rightarrow$ $=\frac{2\pi }{2}-[{\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)]$
$\Rightarrow$ $=\pi -[{\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)}\right)]$ $[\because {\tan }^{-1}a+{\tan }^{-1}b={\tan }^{-1}\left(\frac{a+b}{1-ab}\right)]$
$\Rightarrow$ $=\pi -[{\tan }^{-1}\left(\frac{\frac{5}{6}}{\frac{5}{6}}\right)]$
$\Rightarrow$ $=\pi -[{\tan }^{-1}\left(1\right)]$ $[1=ta{n}^{-1}(tan\frac{\pi }{4})]$
$\Rightarrow$ $=\pi -\frac{\pi }{4}$
$\Rightarrow$ $=\frac{4\pi -\pi }{4}$
$\Rightarrow$ $=\frac{3\pi }{4}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 7

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

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:
$\Rightarrow$ ${\tan }^{-1}x-{\tan }^{-1}y=\frac{\pi }{4}$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{x-y}{1+xy}\right)=\frac{\pi }{4}$ $[\because {\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+ab}\right)]$
$\Rightarrow$ $\left(\frac{x-y}{1+xy}\right)=\tan \frac{\pi }{4}$
$\Rightarrow$ $\left(\frac{x-y}{1+xy}\right)=1$
$\Rightarrow$ $(x-y)=1+xy$
$\Rightarrow$ $x-y-xy=1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 9

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

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,
${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),xy>1$
${\tan }^{-1}x=a\Rightarrow x=\tan a$
${\tan }^{-1}y=b\Rightarrow y=\tan b$
$\Rightarrow$ $0<a+b<\pi ,\tan (a+b)<0$
$\frac{\pi }{2}<a+b<\pi$
$\frac{-\pi }{2}<a+b-\pi <0$
$\Rightarrow$ $\tan (a+b-\pi )=-\tan [\pi -(a+b)]$
$\Rightarrow$ $\tan (a+b)=\left(\frac{x+y}{1-xy}\right)$
$\Rightarrow$ ${\tan }^{-1}\left(\frac{x+y}{1-xy}\right)=a+b-\pi$
$\Rightarrow$ $a+b=\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$
$\Rightarrow$ ${\tan }^{-1}x+{\tan }^{-1}y=\pi +{\tan }^{-1}\left(\frac{x+y}{1-xy}\right)$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 12

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

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 $co{t}^{-1}x+{\cot }^{-1}y$
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,
$\begin{array}{rl}& \Rightarrow {\tan }^{-1}x+{\cot }^{-1}x+{\tan }^{-1}y+{\cot }^{-1}y=\frac{\pi }{2}+\frac{\pi }{2}\\ \\ & \Rightarrow ({\tan }^{-1}x+{\tan }^{-1}y)+({\cot }^{-1}x+{\cot }^{-1}y)=\frac{2\pi }{2}\\ \\ & \Rightarrow \frac{5\pi }{6}+({\cot }^{-1}x+{\cot }^{-1}y)=\pi \\ \\ & \Rightarrow {\cot }^{-1}x+{\cot }^{-1}y=\pi -\frac{5\pi }{6}\\ \\ & \Rightarrow {\cot }^{-1}x+{\cot }^{-1}y=\frac{6\pi -5\pi }{6}\end{array}$
$\Rightarrow {\cot }^{-1}x+{\cot }^{-1}y=\frac{\pi }{6}$

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:
${\tan }^{-1}x-{\cot }^{-1}x={\tan }^{-1}\sqrt{3}$
$\Rightarrow {\tan }^{-1}x-[\frac{\pi }{2}-{\tan }^{-1}x]=\frac{\pi }{3}$
$\Rightarrow {\tan }^{-1}x-\frac{\pi }{2}+{\tan }^{-1}x=\frac{\pi }{3}$
$\Rightarrow 2{\tan }^{-1}x=\frac{\pi }{3}+\frac{\pi }{2}$
$\Rightarrow 2{\tan }^{-1}x=\frac{5\pi }{6}$
$\Rightarrow {\tan }^{-1}x=\frac{5\pi }{12}$
$\Rightarrow x=\tan \frac{5\pi }{12}$

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:
${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{-3\pi }{2}$
$\because \frac{-\pi }{2}<{\sin }^{-1}x\le \frac{\pi }{2}$
$\because {\sin }^{-1}x=\frac{-\pi }{2},{\sin }^{-1}y=\frac{-\pi }{2},{\sin }^{-1}z=\frac{-\pi }{2}$
$\because x=y=z=-1$
$\because xyz=-1$

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:
${\cos }^{-1}\{\sin ({\cos }^{-1}\frac{1}{2})\}$
Solution:
${\cos }^{-1}\{\sin ({\cos }^{-1}\frac{1}{2})\}$
${\cos }^{-1}\{\sin \left(\frac{\pi }{3}\right)\}$ $[co{s}^{-1}\frac{1}{2}=co{s}^{-1}\left(cos\frac{\pi }{3}\right)=\frac{\pi }{3}]$
${\cos }^{-1}\left\{\frac{\sqrt{3}}{2}\right\}$ $[\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}]$
$=\frac{\pi }{6}$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 17

Answer:
$1$
Hint:
You must know the rules of inverse trigonometric function.
Given:
$\tan [{\cos }^{-1}\{\sin ({\cot }^{-1}1)\}]$
Solution:
$\tan [{\cos }^{-1}\{\sin ({\cot }^{-1}1)\}]$ $[{\cot }^{-1}1={\cot }^{-1}(\cot \frac{\pi }{4})=\frac{\pi }{4}]$
$=\tan [{\cos }^{-1}\{\sin \left(\frac{\pi }{4}\right)\}]$
$=\tan [{\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)]$
$=\tan \left[\frac{\pi }{4}\right]$
$=1$

Inverse Trigonometric Functions Excercise Fill in the Blanks Question 18

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