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RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

RD Sharma Solutions Class 12 Mathematics Chapter 3 FBQ

Edited By Kuldeep Maurya | Updated on Jan 19, 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–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\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

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 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}2and \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
\thereforeThe range of y=\sec^{-1}x is (-\infty,-1]\cup [1,\infty)

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