2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

   

Students can Download Maths Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise

Find the value of the following:

Question 1.
\(\cos ^{-1}\left(\cos \frac{13 \pi}{6}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 1

Question 2.
\(\tan ^{-1}\left(\tan \frac{7 \pi}{6}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 2

KSEEB Solutions

Prove that

Question 3.
\(2 \sin ^{-1} \frac{3}{5}=\tan ^{-1} \frac{24}{7}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 3

Question 4.
\(\sin ^{-1} \frac{8}{17}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{77}{85}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 4
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 5

Question 5.
\(\cos ^{-1} \frac{4}{5}+\cos ^{-1} \frac{12}{13}=\cos ^{-1} \frac{33}{65}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 6

Question 6.
\(\cos ^{-1} \frac{12}{13}+\sin ^{-1} \frac{3}{5}=\sin ^{-1} \frac{56}{65}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 7
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 8

KSEEB Solutions

Question 7.
\(\tan ^{-1} \frac{63}{16}=\sin ^{-1} \frac{5}{13}+\cos ^{-1} \frac{3}{5}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 9

Question 8.
\(\tan ^{-1} \frac{1}{5}+\tan ^{-1} \frac{1}{7}+\tan ^{-1} \frac{1}{3}+\tan ^{-1} \frac{1}{8}=\frac{\pi}{4}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 10
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 11

Prove that

Question 9.
\(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right), x \in[0,1]\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 12

KSEEB Solutions

Question 10.
\(\cot ^{-1}\left(\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\pi}{4}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 13
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 14

Question 11.
\(\begin{aligned}&\tan ^{-1}\left(\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\right)=\frac{x}{4}-\frac{1}{2} \cos ^{-1} x\\&-\frac{1}{\sqrt{2}} \leq x \leq[\text { Hint: put } x=\cos 2 \theta]\end{aligned}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 15

Question 12.
\(\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 16
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 17

Question 13.
2 tan-1 (cos x) = tan-1 (2cosec x)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 18

KSEEB Solutions

Question 14.
\(\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x,(x>0)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 19
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 20
Choose the correct Answer

Question 15.
\(\sin \left(\tan ^{-1} x\right),|x|<1 \text { is equal to }\)
(A) \(1, \frac{1}{2}\)
[B] \(1, \frac{1}{2}\)
(C) 0
(D)\(\frac{1}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 21

Question 16.
\(\sin ^{-1}(1-x)-2 \sin ^{-1} x=\cfrac{\pi}{2}, \text { then } x\)
(A) \(0, \frac{1}{2}\)
[B] \(1, \frac{1}{2}\)
(C) 0
(D)\(\frac{1}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 22

Question 17.
\(\tan ^{-1}\left(\cfrac{x}{y}\right)-\tan ^{-1} \cfrac{x-y}{x+y} \text { is equal to }\)
(A)\(\frac{\pi}{2}\)
(B)\(\frac{\pi}{3}\)
(C)\(\frac{\pi}{4}\)
(D)\(\frac{-3 \pi}{4}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 23

2nd PUC Maths Inverse Trigonometric Functions Miscellaneous Exercise Additional Questions and Answers

Multiple Choice Questions:

Question 1.
cos-1 cos(x) = x, then x lies in the interval
(a) [0,π]
(b) \(\left[\frac{-\pi}{2}, \frac{\pi}{2}\right]\)
(c) (o,π])
(d) \(\left(\frac{-\pi}{2}, \frac{\pi}{2}\right)\)
Answer:
(a) [0,π]

KSEEB Solutions

Question 2.
\(\cos ^{-1} \cos \left(\cfrac{4 \pi}{3}\right)=\)
(a) \(\frac{4 \pi}{3}\)
(b) \(\frac{2 \pi}{3}\)
(c) \(\frac{-\pi}{3}\)
(d) – π
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 24

Question 3.
\(\sin ^{-1}\left(\cfrac{4}{5}\right)-\sec ^{-1}\left(\cfrac{5}{4}\right)=\sin ^{-1}\left(\cfrac{4}{5}\right)+\cos ^{-1}\left(\cfrac{4}{5}\right)\)
(a) \(\frac{\pi}{2}\)
(b) 0
(c) π
(d) \(\frac{-\pi}{4}\)
Answer:
(a) sin1 x + cos-1 x = \(\frac{\pi}{2}\)

Question 4.
\(\left(\sqrt[12]{(2 x-2)}\left(\cfrac{-1}{3}\right)\right)+(-\sin x\)
Answer:
(a) \(\frac{24}{25}\)
(b) \(-\frac{24}{25}\)
(c) \(-\frac{6}{25}\)
(d) \(\frac{-6}{25}\)
Answer: (b)
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 25

Question 5.
The value of tan-1 (tan 5)
(a) 2π
(b) 5 – 2π
(c) 5
(d) \(\frac{2 \pi}{3}\)
Answer:
(b).
Let 2π – 5 = θ,
⇒ 5 = 2π – θ
tan 5 = tan (2π – θ)
tan-1 (tan 5) = tan-1 tan (2π – θ)
= tan-1 [-tan θ]
= tan-1 [tan (-θ)] = – θ
= – (2π – 5) = 5 – 2π

KSEEB Solutions

Question 6.
If a > b > c,
\(\cot ^{-1}\left(\cfrac{1+a b}{a-b}\right)+\cot ^{-1}\left(\cfrac{1+b c}{b-c}\right)+\cot ^{-1}\left(\cfrac{1+a c}{c-a}\right)\)
(a) 0
(b) π
(c) 2π
(d) \(\frac{\pi}{2}\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 26

2nd PUC Maths Inverse Trigonometric Functions Miscellaneous Exercise Extra Question and Answers

Question 1.
Find
\(\sin ^{-1} \sin \left(\cfrac{3 \pi}{5}\right)\) (CBSE 2010)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 27

Question 2.
Find the principle value of \(^{\cot ^{-1}}\left(\cfrac{-1}{\sqrt{3}}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 28

Question 3.
Write \(\cot ^{-1}\left(\cfrac{1}{\sqrt{x^{2}-1}}\right),|x|>1\) in the simplest form
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 29

KSEEB Solutions

Question 4.
Show that \(\sin ^{-1}(2 x \sqrt{1-x^{2}}\}=2 \sin ^{-1} x\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 30

Question 5.
Find the value of tan (tan-1 x + cot-1 x)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 31
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 32

Question 6.
Solve \(\tan ^{-1}\left\{\cfrac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}\right\}\) (CBSE 09)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 33

Question 7.
Prove that \(\tan ^{-1} x+\tan ^{-1} \cfrac{2 x}{1-x^{2}}=\tan ^{-1}\left(\cfrac{3 x-x^{3}}{1-3 x^{2}}\right)\)
Answer:
2nd PUC Maths Question Bank Chapter 2 Inverse Trigonometric Functions Miscellaneous Exercise 34

error: Content is protected !!