Ncert Exemplar Class 11 Maths Solutions

NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions is considered a very important chapter for practical use, and application in various different fields and also for the exams. NCERT Exemplar Class 11 Maths chapter 3 solutions give a brief procedure and explanation about angles along with degree and radian measure. It also establishes a relationship between radian and real numbers, and radian and degree measure. Class 11 Maths NCERT Exemplar solutions chapter 3 also covers a variety of questions relating to a conversion of degrees to radian and vice versa through established relation between them through the use of notational convention.

NCERT Exemplar Class 11 Maths Solutions Chapter 3: Exercise - 1.3

Question:1

Prove that
$\\ \frac{tanA+secA-1}{tanA-secA+1} =\frac{1+sinA}{cosA}$

Answer:

$\\ L.H.S=\frac{tanA+secA-1}{tanA-secA+1} \\\\ =\frac{tanA+secA- \left( \sec ^{2}A-\tan ^{2}A \right) }{tanA-secA+1} \\\\ =\frac{tanA+secA- \left[ \left( secA+tanA \right) \left( secA-tanA \right) \right] }{tanA-secA+1}\\ \\ = \frac{ \left( secA+tanA \right) \left[ 1- \left( secA-tanA \right) \right] }{tanA-secA+1} \\\\$

$\\ = \frac{ \left( secA+tanA \right) \left[ 1-secA+tanA \right] }{tanA-secA+1} \\ \\ =secA+tanA \\ \\ =\frac{1}{cosA}+\frac{sinA}{cosA} \\ \\ =\frac{1+sinA}{cosA}=R.H.S \\ \\$

Question:31

If $f(x) = \cos\textsuperscript{2}x + sec\textsuperscript{2}x$ , then

A. $f(x) < 1\\\\$

B. $f(x) = 1\\\\$

C. $2 < f(x) < 1\\\\$

D. $f(x) \geq 2\\\\$

[Hint: $A.M \geq G.M.$ ]

Answer:

The answer is the option (d)
$\\ f \left( x \right) =\cos ^{2}x+\sec ^{2}x~~ \\\\ We~know~that AM \geq GM \\\\ \frac{ \left( \cos^{2}x+\sec ^{2}x \right) }{2} \geq \sqrt {\cos ^{2}xsec^{2}~x}~~~~ \\\\ ~~\frac{ \left( \cos^{2}x+\sec ^{2}x \right) }{2} \geq 1~ \\\\ ~~\cos ^{2}x+\sec ^{2}x \geq 2~ \\\\ \text{~ f} \left( x \right) \geq 2 \\\\$

Question:34

The value of $\tan 1 ^{\circ} \tan 2 ^{\circ} \tan 3 ^{\circ} \ldots \tan 89 ^{\circ} \: \: \: is\\\\$
A. 0
B. 1
C. $1/2$
D. Not defined

Answer:

The answer is the option (b).
$\\\text{Given that } \tan1^{0}~\tan2^{0} \ldots \ldots \ldots \ldots \ldots .\tan89^{0}~~ \\\\ =\tan1^{0}~\tan2^{0} \ldots \ldots \tan45^{0}\tan \left( 90 - 44^{0} \right) \tan \left( 90 - 43^{0} \right) \ldots .\tan \left( 90 - 1^{0} \right) \\\\ =\tan1^{0}\cot 1^{0}\tan 2^{0}\cot 2^{0} \ldots \ldots \ldots \ldots .\tan89^{0}\cot 89^{0} \\\\ =1.1 \ldots 1 \ldots \ldots 1.1=1 \\\\$

Question:35

Answer:

The answer is the option (c).
$\\Given~that~~\frac{1 - \tan ^{2}15^{0}}{1+\tan ^{2}15^{0}}~~~~ \\\\ Let \: \: \theta =15^{0}~ 2 \theta =30^{0}~~ \\\\ ~ \cos2 \theta =\frac{1 - \tan ^{2} \theta }{1+\tan ^{2} \theta }~~ \\\\ ~\cos 30^{0}=\frac{1 - \tan ^{2}15^{0}}{1+\tan ^{2}15^{0}}=\frac{\sqrt {3}}{2} \\\\$

Question:36

The value of $\cos 1 ^{\circ} \cos 2 ^{\circ} \cos 3 ^{\circ} \ldots \cos 179 ^{\circ} \: \: is\\\\$
A. $1/ \sqrt 2\\\\$
B. 0
C. 1
D. -1

Answer:

The answer is the option (b).
$\\ \cos 1^{0}~\cos2^{0} \ldots \ldots \ldots \ldots \ldots .\cos179^{0} \\\\ = \cos1^{0}~\cos2^{0} \ldots \ldots \ldots \ldots \ldots .\cos 90^{0} \ldots \ldots .\cos179^{0} \] =0 \left( as~~\cos 90^{0}=0 \right) \\\\$

Question:39

Which of the following is correct?

A. $\sin 1 ^{\circ} > \sin 1\\\\$

B. $\sin 1 ^{\circ} < \sin 1\\\\$

C. $\sin 1 ^{\circ} = \sin 1\\\\$

D. $\sin 1^{\circ}=\frac{\pi}{18^{\circ}} \sin 1$

[Hint: $1 radian = 180 ^{\circ} \pi = 57 ^{\circ} 30'$ approx.]

Answer:

The answer is the option (b).
If $~ \theta$ increases then the value of $\sin \theta$ also increases. \\\\

So, $\sin1^{\circ}<\sin 1$
Hence, b is correct.

Question:41

The minimum value of $3 \cos x + 4 \sin x + 8$ is
A. 5
B. 9
C. 7
D. 3

Answer:

The answer is the option (d).

$\\Let\ y=3\cos x+4\sin x+8 \\\\ y - 8= 3\cos x+4\sin x \\\\ \text{Minimum value of } y - 8= - \sqrt { \left( 3 \right) ^{2}+ \left( 4 \right) ^{2}}= - 5 \\\\ y=8 - 5=3 \\\\$

Hence, (d) is the correct option.

Question:43

The value of $\sin (45 ^{\circ} + \theta ) - \cos (45 ^{\circ} - \theta )$ is
A. $2 \cos \theta$
B. $2\sin \theta$
C. 1
D. 0

Answer:

The answer is the option (d).
$\\ \sin \left( 45+ \theta \right) - \cos \left( 45 - \theta \right) \\\\ \sin \left( 45+ \theta \right) =\sin 45\cos \theta +\cos 45\sin \theta =\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta \\\\ \cos \left( 45 - \theta \right) =\cos 45\cos \theta +\sin 45\sin \theta =\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta \\\\ \sin \left( 45+ \theta \right) - \cos \left( 45 - \theta \right) =\frac{1}{\sqrt {2}}\cos \theta +\frac{1}{\sqrt {2}}\sin \theta - \frac{1}{\sqrt {2}}\cos \theta - \frac{1}{\sqrt {2}}\sin \theta =0 \\\\$

Question:44

The value of $\cot \left( \frac{ \pi }{4}+ \theta \right) \cot \left( \frac{ \pi }{4} - \theta \right)$ is
A. –1
B. 0
C. 1
D. Not defined

Answer:

The answer is the option (c).

$\\ \cot \left( \frac{ \pi }{4}+ \theta \right) \cot \left( \frac{ \pi }{4} - \theta \right) =\frac{\cot \frac{ \pi }{4}\cot \theta - 1}{\cot \theta +\cot \frac{ \pi }{4}} \times \frac{\cot \frac{ \pi }{4}\cot \theta +1}{\cot \theta - \cot \frac{ \pi }{4}} \\\\ =\frac{\cot \theta - 1}{\cot \theta +1} \times \frac{\cot \theta +1}{\cot \theta - 1}=1 \\\\$
(c) is correct.

Question:45

$\cos 2 \theta \cos 2 \phi + \sin^2( \theta - \phi ) - \sin^2( \theta + \phi )$ is equal to
A. $\sin 2( \theta + \phi )$
B. $\cos 2( \theta + \phi )$
C. $\sin 2( \theta - \phi )$
D. $\cos 2( \theta - \phi )$
[Hint: Use $\sin2A - \sin2B = \sin (A + B) \sin (A - B)$ ]

Answer:

The answer is the option (b).
$\\ \cos 2 \theta \cos 2 \varnothing +\sin ^{2} \left( \theta - \varnothing \right) +\sin ^{2} \left( \theta + \varnothing \right) \\\\ since,~\sin ^{2}A - \sin ^{2}B=\sin \left( A+B \right) \sin \left( A - B \right) \\\\ =\cos 2 \theta \cos 2 \varnothing +\sin \left( \theta - \varnothing + \theta + \varnothing \right) \sin \left( \theta - \varnothing - \theta - \varnothing \right) \\\\ =\cos 2 \theta \cos 2 \varnothing - \sin 2 \theta \sin 2 \varnothing \\\\ since,~\cos x\cos y - \sin x\sin y=\cos \left( x+y \right) \\\\ =\cos \left( 2 \theta +2 \varnothing \right) \\\\ =\cos 2 \left( \theta + \varnothing \right) \\\\$
Hence, the correct option is (b).

Question:48

The value of $\sin \frac{ \pi }{10}\sin \frac{13 \pi }{10}$ is
A. $\frac{1}{2}$
B. $-\frac{1}{2}$
C. $-\frac{1}{4}$
D. $1$

Answer:

The answer is the option (c).
$\\ \sin \frac{ \pi }{10}\sin \frac{13 \pi }{10}=\sin \frac{ \pi }{10}\sin \left( \pi +\frac{3 \pi }{10} \right) =\sin \frac{ \pi }{10} \left( - \sin \frac{3 \pi }{10} \right) \\\\ = - \sin 18\sin 54= - \left( \frac{\sqrt {5} - 1}{4} \right) \left( \frac{\sqrt {5}+1}{4} \right) \\\\ =\frac{5 - 1}{16}=\frac{4}{16}=\frac{1}{4} \\\\$
Hence, (c) is correct option.

Question:56

Answer:

The answer is the option (a).
$\\\\ \cos ^{2}48 - \sin ^{2}12=\cos \left( 48+12 \right) \cos \left( 48 - 12 \right) =\cos 60\cos 36=\frac{1}{2} \times \frac{\sqrt {5}+1}{4}=\frac{\sqrt {5}+1}{8} \\\\$

Hence, correct option is (a).

Question:57

If $\tan \alpha =\frac{1}{7} , \tan \beta =\frac{1}{3} \\\\$ then $\cos 2 \alpha$ is equal to
A. $\sin 2 \beta$
B. $\sin 4 \beta$
C. $\sin 2 \beta$
D. $\cos 2 \beta$

Answer:

The answer is the option (b).
$\\\\\\ \tan \alpha =\frac{1}{7} \\\\ \tan \beta =\frac{1}{3} \\\\ \cos 2 \alpha =\frac{1 - \tan ^{2} \alpha }{1+\tan ^{2} \alpha }=\frac{1 - \frac{1}{49}}{1+\frac{1}{49}}=\frac{24}{25} \\\\ \tan 2 \beta =\frac{2\tan \beta }{1 - \tan ^{2} \beta }=\frac{2 \times \frac{1}{3}}{1 - \frac{1}{9}}=\frac{3}{4} \\\\$
$\\ \sin 4 \beta =\frac{2\tan 2 \beta }{1+\tan ^{2}2 \beta }=\frac{2 \times \frac{3}{4}}{1+ \left( \frac{3}{4} \right) ^{2}}=\frac{24}{25} \\\\ \cos 2 \alpha =\sin 4 \beta =\frac{24}{25} \\\\$
Hence, correct option is (b).

Question:58

Answer:

The answer is the option (b).
$\\\\ \tan \theta =\frac{a}{b} \\\\ b\cos 2 \theta +a\sin 2 \theta =b \left[ \frac{1 - \tan ^{2} \theta }{1+\tan ^{2} \theta } \right] +a \left[ \frac{2\tan \theta }{1+\tan ^{2} \theta } \right] \\\\ =b \left[ \frac{1 - \frac{a^{2}}{b^{2}}}{1+\frac{a^{2}}{b^{2}}} \right] +a \left[ \frac{2\frac{a}{b}}{1+\frac{a^{2}}{b^{2}}} \right] =b \left[ \frac{b^{2} - a^{2}}{b^{2}+a^{2}} \right] + \left[ \frac{\frac{2a^{2}}{b}}{\frac{b^{2}+a^{2}}{b^2}} \right] \\\\$
$\\ =\frac{b^{3} - a^{2}b}{b^{2}+a^{2}}+\frac{2a^{2}b}{b^{2}+a^{2}} \\\\ =\frac{b \left( b^{2}+a^{2} \right) }{b^{2}+a^{2}}=b \\\\$
Hence, correct option is (b).

Important notes of NCERT Exemplar Class 11 Maths Solutions Chapter 3 Trigonometric Functions

Trigonometry is an ancient concept which in ancient times was used to solve problems relating to triangles and geometry, but it has extended its use to various different fields in the present times including varied areas of studies. It was derived from Greek words meaning, "measuring the sides of a triangle" which has widened its scope to much more than the original meaning. It is basically used to measure length, height and angles of different triangles with its reach in real-life practical situations. NCERT Exemplar solutions for class 11 Maths Chapter 3 extends studying trigonometric ratios to any angle regarding or concerning radian measure and interpreting and representing it as a trigonometric ratio with the help of diagrams for a better understanding.

Students can make use of NCERT Exemplar Class 11 Maths solutions chapter 3 pdf download for further learning.

Main subtopics of NCERT Exemplar Class 11 Maths Solutions Chapter 3

The topics covered in the chapter are as follows:

3.1 Introduction

3.2 Angles

3.2.1 Degree measure

3.2.2 Radian measure

3.2.3 Relation between radian and real numbers

3.2.4 Relation between degree and radian

3.3 Trigonometric Functions

3.3.1Sign of trigonometric functions

3.3.2 Domain and range of trigonometric functions

3.4 Trigonometric Functions of Sum and Difference of Two Angles

3.5 Trigonometric Equations

What will the students learn from NCERT Exemplar Class 11 Maths Solutions Chapter 3?

The students will learn a variety of concepts from NCERT Exemplar solutions for class 11 Maths chapter 3 which has a wide range of application in different fields such as engineering, sound engineers, architects, astronauts, surveyors, physicist, and much more for future references. NCERT Exemplar Class 11 Maths Solutions Chapter 3 is also given importance since it has various applications in real life and could be connected to routine activities that happen around us. It is even used in the gaming industry, IT sector, construction of bridges, buildings, mountains, the inclination of floors, roofs, marine biology, criminal investigations, wide use in physics for derivations and explanations, and much more. The uses of trigonometry in these many fields justify its use for inexhaustible purposes and its importance for students belonging or deciding to enter any field or subject in the future.

NCERT Solutions for Class 11 Mathematics Chapters

Important topics to cover from NCERT Exemplar class 11 Maths solutions Chapter 3 Trigonometric Functions

· NCERT Exemplar Class 11 Maths chapter 3 solutions give explanation, and interpretation of different trigonometric functions for sin x, cos x, sec x, cot x, cosec x and tan x along with values of trigonometric ratios for 0º, 30º, 45º, 60º, 90º, 180º, 270º and 360º along with the sign convention of these trigonometric functions.

· Class 11 Maths NCERT Exemplar solutions chapter 3 also covers the range and domain of trigonometric functions with the help of diagrammatic representation of the same. This chapter also extends to the trigonometric functions of sum and difference of two angles with a variety of questions and illustrations to be done for the same.

· NCERT Exemplar class 11 Maths solutions Chapter 3 concludes with insight on trigonometric equations that involve equations containing trigonometric functions of any variable.

NCERT Exemplar Class 11 Solutions

Ncert Exemplar Class 11 Maths Solutions

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Ncert Exemplar Class 11 Maths Solutions