NCERT Solutions For Class 11 Maths Chapter 3

NCERT Solutions Class 11 Maths Trigonometric Functions

Ncert Solutions For Class 11 Maths Chapter 3 PDF Download

NCERT Solutions For Class 11 Maths Chapter 3 Trigonometric Functions are given below in a very detailed way. These solutions for NCERT class 11 maths trigonometric identities (chapter 3) given here are available for free and are extremely easy to understand. All the exercises from trigonometric identities are covered here in easy steps which the students can easily comprehend. These solutions can help the students to not only clear their doubts but also to help them to know the best answers to the respective questions.

The class 11 NCERT solutions for maths chapter 3 given here can also help the students to have a better understanding of the concepts. With these solutions, students no longer need to pile up their doubts and can clear instantly. At BYJU’S, students are provided with the NCERT solutions of all the classes for maths and science even in PDF format so that students can download and study offline. Check the comprehensive NCERT Solutions Class 11 Maths Trigonometric Functions (chapter 3) below.

NCERT Solutions Class 11 Maths Chapter 4 Exercises

Exercise 3.1

Q.1: Calculate the radian measurement of the given degree measurement:

(i).  \(25\;^{\circ}\)

(ii).  \(240\;^{\circ}\)

(iii).  \(-47\;^{\circ} \; 30′\)

(iv).  \(520\;^{\circ}\)

 

Sol:

(i).  \(25\;^{\circ}\)

As we know that, \(180\;^{\circ}\) = π radian

Therefore, \(25\;^{\circ}\) = \(\frac{\pi }{180} \times 25\) radian = \(\frac{5\pi }{36}\) radian

Hence, \(25\;^{\circ}\) = \(\frac{5\pi }{36}\) radian

 

(ii).  \(240\;^{\circ}\)

As we know that \(180\;^{\circ}\) = π radian

Therefore, \(240\;^{\circ}\) = \(\frac{\pi }{180} \times 240\) radian = \(\frac{4\pi }{3}\) radian

Hence, \(240\;^{\circ}\) = \(\frac{4\pi }{3}\) radian

 

(iii).  \(-47\;^{\circ} \; 30′\)

= \(-47\frac{1}{2}\)

= \(\frac{-95}{2}\) degree

As we know that, \(180\;^{\circ}\) = π radian

Therefore, \(\frac{-95}{2}\) degree = \(\frac{\pi }{180}\times \frac{-95}{2}\) radian = \(\frac{-19 }{36\; \times \;2}\pi\) radian = \(\frac{-19 }{72}\pi\)

Hence, \(-47\;^{\circ} \; 30′\) = \(\frac{-19 }{72}\pi\)

 

(iv).  \(520\;^{\circ}\)

As we know that, \(180\;^{\circ}\) = n radian

Therefore, \(520\;^{\circ}\) = \(\frac{\pi }{180} \times 520\) radian = \(\frac{26\pi }{9}\) radian

Hence, \(520\;^{\circ}\) = \(\frac{26\pi }{9}\) radian

 

 

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = \(\frac{22}{7}\)]

(i) \(\frac{11}{16}\)

 

(ii) -4

 

(iii) \(\frac{5\pi }{3}\)

 

(iv) \(\frac{7\pi }{6}\)

 

Sol:

(i).  \(\frac{11}{16}\):

As we know, that Π Radian = 180°

Therefore, \(\frac{11}{16}\) radian = \(\frac{180}{\pi }\times \frac{11}{16}\) degree

= \(\frac{45\times 11}{\pi \times 4 }\) degree

= \(\frac{45 \times 11 \times 7}{22 \times 4 }\) degree

= \(\frac{315}{8}\) degree

= \(39\;\frac{3}{8}\) degree

= \(39\;^{\circ} \;+ \; \frac{3\times 60}{8}\) minute  [\(1^{\circ}\) = 60’]

= \(39^{\circ} \;+ \; 22′ \;+ \;\frac{1}{2}\) minute

= \(39^{\circ} \;+ \; 22′ \;+ \;30”\)  [1’ = 60’’]

 

(ii).  -4:

As we know, that Π Radian = 180°

Therefore, -4 radian = \(\frac{180}{\pi } \times (-4)\) degree

= \(\frac{180 \times 7(-4)}{22}\) degree

= \(\frac{-2520}{11}\) degree

= \(-229\frac{1}{11}\) degree

= \(-229^{\circ} \; + \; \frac{1\times 60}{11}\) minutes     [\(1^{\circ}\) = 60’]

= \(-229^{\circ} \; +\; 5′ +\;\frac{5}{11}\)

= \(-229^{\circ} \; +\; 5′ +\;27”\) [1’ = 60’’]

 

(iii).  \(\frac{5\pi }{3}\)

As we know, that Π Radian = 180°

Therefore, \(\frac{5\pi}{3}\) radian = \(\frac{180}{\pi }\times \frac{5\pi}{3}\) degree

= \(300^{\circ}\)

 

(iv).  \(\frac{7\pi }{6}\)

As we know, that Π Radian = 180°

Therefore, \(\frac{7\pi}{6}\) radian = \(\frac{180}{\pi }\times \frac{7\pi }{6}\)

= \(210^{\circ}\)

 

 

Q.3: In a minute, wheel makes 360 revolutions. Through how many radians does it turn in 1 second?

Answer:

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = \(\frac{360}{60}\) = 6

In one revolution, the wheel rotates an angle of 2π radian.

Therefore, in 6 revolutions, it will turn an angle of 12π radian.

 

 

Q.4: Calculate the degree measurement of the angle subtended at the centre of a circle of radius 100 m by an arc of length 22 m.

[Use π = \(\frac{22}{7}\)]

Answer:

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, \(\theta = \frac{l}{r}\)

Given:

r = 100 m and L = 22 m

We have,

\(\theta = \frac{22}{100}\) radian

= \(\frac{180}{\pi }\times \frac{22}{100}\) degree

= \(\frac{180 \;\times \;7 \;\times \;22}{22 \; \times 100 }\) degree

= \(\frac{126}{10 }\) degree

= \(12\frac{3}{5}\) degree

= \(12^{\circ} \; 36′\) [\(1^{\circ}\) = 60’]

Therefore, the req angle is \(12^{\circ} \; 36′\)

 

 

Q.5: In a circle of diameter 40 m, the length of the chord 20 m. Find the length of minor arc of chord.

Answer:

13

Diameter of circle = 40 m

Radius of circle = \(\frac{40}{20}\) m = 20 m

Let XY be the chord (length = 20 m) of the circle.

In \(\Delta OXY\), OX = OY  = radius of the circle = 20 m

XY = 20 m

Therefore,

\(\Delta OXY\) is an equilateral triangle.

\(\theta \;= \;60^{\circ}\) = \(\frac{\pi }{3}\) radian

As we know,

In a circle of radius r unit, if an arc of length l unit subtends an angle θ radian at the centre, then, θ = \(\frac{l}{r}\)

\(\frac{\pi }{3}\) = \(\frac{arc\;(AB\;)}{20}\)

\(\frac{arc\;(AB\;)}{20}\) = \(\frac{20\pi }{3}\) m

The length of the minor arc of the chord is \(\frac{20\pi }{3}\) m.

 

 

Q.6: In two circles, arcs which has same length subtended at an angle of \(60^{\circ}\) and \(75^{\circ}\) at the center. Calculate the ratio of their radii.

 Sol:

Let, the radii of two circles be \(r_{1}\) and \(r_{2}\).

Let, an arc of length l subtend an angle of \(60^{\circ}\) at the center of the circle of radius \(r_{1}\), while let an arc of length l subtend an angle of \(75^{\circ}\) at the center of the circle of radius \(r_{2}\).

\(\\60^{\circ}\) = \(\frac{\pi }{3}\) radian

\(75^{\circ}\) = \(\frac{5\pi }{12}\) radian

In a circle of radius r unit, if an arc of length l unit subtends an angle θ then:

θ = \(\frac{l}{r}\)

l = r θ

l = \(\frac{r_{1}\;\pi }{3}\)

l = \(\frac{r_{2}\;5\pi }{12}\)

\(\frac{r_{1}\;\pi }{3}\) = \(\frac{r_{2}\;5\pi }{12}\)

\(r_{1}\) = \(\frac{r_{2}5}{4}\)

\(\frac{r_{1}}{r_{2}}\) = \(\frac{5}{4}\)

Therefore, the ratio of radii is 5: 4

 

Q.7: Calculate the angle in radian through which a pendulum swings if the length is 75 cm and the tip describes an arc of length

(i) 10 cm

 

(ii) 15 cm

 

(iii) 21 cm

 

Sol:

As we know that, in a circle of radius ‘r’ unit, if an arc of length ‘l’ unit subtends an angle \(\theta\) radian at the center, then:

\(\theta = \frac{l}{r}\\\)

Here, r = 75 cm

(i).  10 cm:

\(\theta\) = \(\frac{10}{75}\) radian = \(\frac{2}{15}\) radian

 

(ii).  15 cm:

\(\theta\) = \(\frac{15}{75}\) radian = \(\frac{1}{5}\) radian

 

(iii).  21 cm:

\(\theta\) = \(\frac{21}{75}\) radian = \(\frac{7}{25}\) radian

 

 

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if \(\cos y\) = \(-\frac{1}{2}\) and y lies in 3rd quadrant.

Sol:

(i)  sec y :

Since, cos y = \(\frac{1}{2}\)

Therefore, sec y = \(\frac{1}{\cos y}\) = \(\frac{1}{\left (-\frac{1}{2} \right )}\)

Hence, sec y = -2

 

(ii)  sin y :

Since, \(\sin ^{2}y \;+ \;\cos ^{2}y \;= \;1\)

Therefore, \(\sin ^{2}y \;= \;1 \;- \;\cos ^{2}y\)

\(\Rightarrow\) \(\sin ^{2}y \;= \;1 \;- \;\left (-\frac{1}{2} \right )^{2}\)

\(\Rightarrow\) \(\sin ^{2}y \;= \;1 \;- \;\frac{1}{4}\)

\(\Rightarrow\) \(\sin ^{2}y \;= \;\frac{3}{4}\)

\(\Rightarrow\) \(\sin y \;= \;\pm \frac{\sqrt{3}}{2}\)

Since, y lies in the third quadrant, the value of sin y will be negative.

Therefore, sin y = \(\frac{\sqrt{3}}{2}\)

 

(iii)  cosec y = \(\frac{1}{\sin y}\) = \(\frac{1}{\left (-\frac{\sqrt{3}}{2} \right )}\) = \(-\frac{2}{\sqrt{3}}\)

Therefore, cosec y = \(-\frac{2}{\sqrt{3}}\)

 

(iv)  tan y = \(\frac{\sin y}{\cos y}\) = \(\tan y=\frac{\left (-\frac{\sqrt{3}}{2} \right )}{\left (-\frac{1}{2} \right )}\) = \(\sqrt{3}\)

Therefore, tan y = \(\sqrt{3}\)

 

(v)  cot y = \(\frac{1}{\tan y}\) = \(\frac{1}{\sqrt{3}}\)

Therefore, cot y = \(\frac{1}{\sqrt{3}}\)

 

 

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = \(\frac{3}{5}\), where y lies in second quadrant.

Sol:

sin y = \(\frac{3}{5}\)

Therefore, cosec y = \(\frac{1}{Sin\; y}\) = \(\frac{1}{\frac{3}{5}} = \frac{5}{3}\)

Since, \(sin^{2}\;y \; + \; cos^{2}\;y = 1\)

\(\Rightarrow\)  \( cos^{2} y = 1 – sin^{2} y\)

\(\Rightarrow\)  \(cos^{2} y = 1 – \left ( \frac{3}{5} \right )^{2}\)

\(\Rightarrow\) \(cos^{2} y = 1 – \frac{9}{25}\)

\(\Rightarrow\) \(cos^{2} y = \frac{16}{25}\)

\(\Rightarrow\) cos y = \(\pm \frac{4}{5}\)

Since, y lies in the 2nd quadrant, the value of cos y will be negative,

Therefore, cos y = – \(\frac{4}{5}\)

\(\Rightarrow\) sec y = \(\frac{1}{cos \; y} = \frac{1}{\left (- \frac{4}{5} \right )} =\; – \frac{5}{4}\)

\(\Rightarrow\) tan y = \(\frac{sin \; y}{cos \; y} = \frac{\left (\frac{3}{5} \right ) }{\left (-\frac{4}{5} \right ) } = \;- \frac{3}{4}\)

\(\Rightarrow\) cot y = \(\frac{1}{tan \; y} =\;- \frac{4}{3}\)

 

 

 Q.3: Find the values of other five trigonometric functions if \(cot \; y = \frac{3}{4}\), where y lies in the third quadrant.

Sol:

cot y = \(\frac{3}{4}\)

Since, tan y = \(\frac{1}{cot \; y}=\frac{1}{\frac{3}{4}}=\frac{4}{3}\)

Since, \(1 + tan^{2}y = sec^{2}y\)

\(\Rightarrow\)  \(1 + \left ( \frac{4}{3} \right )^{2} = sec^{2}y\)

\(\Rightarrow\)  \(1 + \frac{16}{9} = sec^{2}y\)

\(\Rightarrow\)  \(\frac{25}{9} = sec^{2}y\)

\(\Rightarrow\)  sec y = \(\pm \frac{5}{3}\)

Since, y lies in the 3rd quadrant, the value of sec y will be negative.

Therefore, sec y = \(\;- \frac{5}{3}\)

cos y = \(\frac{1}{sec \; y} = \frac{1}{\left (- \frac{5}{3} \right )} = – \frac{3}{5}\)

Since, tan y = \(\frac{sin \; y}{cos \; y}\)

\(\Rightarrow\)  \( \frac{4}{3} = \frac{sin \; y}{\left (- \frac{3}{5} \right )}\)

\(\Rightarrow \) sin y = \(\left (\frac{4}{3} \right ) \times \left (- \frac{3}{5} \right ) = -\frac{4}{5}\)

\(\Rightarrow\)  cosec y = \(\frac{1}{sin \; y} = -\frac{5}{4}\)

 

 

Q.4: Find the values of other five trigonometric if \(sec \; y = \frac{13}{5}\), where y lies in the fourth quadrant.

Sol:

sec y = \(\frac{13}{5}\)

cos y = \(\frac{1}{sec \; y} = \frac{1}{\left (\frac{13}{5}\right )} = \frac{5}{13}\)

Since, \(sin^{2}y + cos^{2}y = 1\)

\(\Rightarrow\)   \(sin^{2}y = 1 – cos^{2}y\)

\(\Rightarrow\)   \(sin^{2}y = 1 – \left ( \frac{5}{13} \right )^{2}\)

\(\Rightarrow\)   \(sin^{2}y = 1 – \frac{25}{169} = \frac{144}{169}\)

\(\Rightarrow\) sin y = \(\pm \frac{12}{13}\)

Since, y lies in the 4th quadrant, the value of sin y will be negative.

Therefore, sin y = \(- \frac{12}{13}\)

\(\Rightarrow\) cosec y = \(\frac{1}{sin \; y} = \frac{1}{\left (- \frac{12}{13} \right )} = \;- \frac{13}{12}\)

\(\Rightarrow\) tan y = \(\frac{sin \; y}{cos \; y} = \frac{\left (- \frac{12}{13} \right )}{\left (\frac{5}{13} \right )} =\; -\frac{12}{5}\)

\(\Rightarrow\) cot y = \(\frac{1}{tan \; y} = \frac{1}{\left ( -\frac{12}{5} \right )} = \;-\frac{5}{12}\)

 

 

Q.5: Find the values of other five trigonometric function if tan y = \(- \frac{5}{12}\) and y lies in second quadrant.

Sol:

tan  y = \(-\frac{5}{12}\)   [Given]

And, cot y = \(\frac{1}{tan \; y} = \frac{1}{ \left (- \frac{5}{12} \right )} = – \frac{12}{5}\)

Since, \(1 + tan^{2} y = sec^{2} y\)

Therefore, \(1 + \left ( – \frac{5}{12} \right )^{2} = sec^{2} y\)

\(\Rightarrow\)   \(sec^{2} y = 1 + \left ( \frac{25}{144} \right )\)

\(\Rightarrow \)  \(sec^{2} y = \frac{169}{144}\)

\(\Rightarrow\) sec y = \(\pm \frac{13}{12}\)

Since, y lies in the 2nd quadrant, the value of sec y will be negative.

Therefore, sec  y = \(- \frac{13}{12}\)

cos y = \(\frac{1}{sec \; y } = \frac{1}{\left (- \frac{13}{12} \right )} = \left (- \frac{12}{13} \right )\\\)

Since, \(tan \; y = \frac{sin \; y}{cos \; y}\)

\(\Rightarrow\)  \( -\frac{5}{12} = \frac{sin \; y}{\left (- \frac{12}{13} \right )}\\\)

\(\Rightarrow\) sin y = \(\left (-\frac{5}{12} \right ) \times \left (- \frac{12}{13}\right ) = \frac{5}{13}\\\)

\(\Rightarrow\) cosec y = \(\frac{1}{sin \; y} = \frac{1}{\left ( \frac{5}{13} \right )} = \frac{13}{5}\)

 

 

Q.6: Calculate the value of trigonometric function sin 765°.

Sol:

The values of sin y repeat after an interval of 360° or 2n.

Therefore, sin 765° = sin ( 2 × 360° + 45° ) = sin 45° = \(\frac{1}{\sqrt{2}}\)

 

 

Q.7: Calculate the value of trigonometric function cosec [-1410°]

 Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, cosec [-1410°] = cosec [-1410° + 4 × 360°] = cosec [ -1410° + 1440°] = cosec 30° = 2

 

 

Q.8: Calculate the value of the trigonometric function \(\tan \frac{19\pi }{3}\).

Sol:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, \(\tan \frac{19\pi }{3}\) = \(\tan 6\frac{1}{3}\pi\) = \(\tan \left ( 6\pi + \frac{\pi }{3} \right )\) = \(\tan \frac{\pi }{3}\) = tan 60° = \(\sqrt{3}\)

 

 

Q.9: Calculate the value of the trigonometric function \(\sin \left ( -\frac{11\pi }{3} \right )\).

 Answer:

It is known that the values of tan y repeat after an interval of 360° or 2n.

Therefore, \(\sin \left ( -\frac{11\pi }{3} \right )\) = \(\sin \left ( -\frac{11\pi }{3} + 2 \times 2\pi \right )\) = \(\sin \frac{\pi }{3}\) = \(\frac{\sqrt{3}}{2}\)

 

 

Q.10: Calculate the value of the trigonometric function \(\cot \left ( -\frac{15\pi }{4} \right )\).

 Sol:

It is known that the values of \(\tan y\) repeat after an interval of \(180^{\circ}\) or n.

Therefore, \(\cot \left ( -\frac{15\pi }{4} \right )\) = \(\cot \left ( -\frac{15\pi }{4} + 4\pi \right )\) = \(\cot \frac{\pi }{4}\) = 1

 

 

Exercise 3.3

Q.1: Prove:

\(\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4} = -\frac{1}{2}\)

 

Sol:

Now, taking L.H.S.

\(\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4}\):

= \(\\\left (\frac{1}{2} \right )^{2} + \left (\frac{1}{2} \right )^{2} – \left ( 1 \right )^{2}\)

= \(\\\frac{1}{4} + \frac{1}{4} – 1\) = \(-\frac{1}{2}\)

= R.H.S.

 

 

Q.2: Prove:

\(2 \sin ^{2} \frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} \cos ^{2}\frac{\pi }{3} = \frac{3}{2}\)

 

Sol:

Now, taking L.H.S.

\(2 \sin ^{2} \frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} \cos ^{2}\frac{\pi }{3}\\\) :

= \(\\2 \left (\frac{1}{2} \right )^{2} + cosec ^{2}\left ( \pi + \frac{\pi }{6} \right )\left ( \frac{1}{2} \right )^{2}\)

= \(2 \times \frac{1}{4} + \left (-cosec \frac{\pi }{6} \right )^{2}\left ( \frac{1}{4} \right )\)

= \(\frac{1}{2} + \left ( -2 \right )^{2}\left ( \frac{1}{4} \right )\)

= \(\frac{1}{2} + \frac{4}{4}\)

= \(\frac{1}{2} + 1\)

= \(\frac{3}{2}\)

= R.H.S.

 

 

Q.3: Prove:

\(\cot ^{2}\frac{\pi }{6} + cosec \frac{5\pi }{6} + 3\tan ^{2} \frac{\pi }{6} = 6\)

 

Sol:

Taking L.H.S.

\(\cot ^{2}\frac{\pi }{6} + cosec \frac{5\pi }{6} + 3\tan ^{2} \frac{\pi }{6}\\\) :

= \(\left (\sqrt{3} \right )^{2} + cosec \left ( \pi – \frac{\pi }{6} \right ) + 3\left ( \frac{1}{\sqrt{3}} \right )^{2}\)

= \(3 + cosec \frac{\pi }{6} + 3\times \frac{1}{3}\)

= 3 + 2 + 1 = 6

= R.H.S.

 

 

Q.4: Prove:

\(2\sin ^{2} \frac{3\pi }{4} + 2\cos ^{2} \frac{\pi }{4} + 2\sec ^{2} \frac{\pi }{3} = 10\)

 

Sol:

Now, taking L.H.S.

\(2\sin ^{2} \frac{3\pi }{4} + 2\cos ^{2} \frac{\pi }{4} + 2\sec ^{2} \frac{\pi }{3}\\\) :

= \(\\2\left \{ \sin \left ( \pi – \frac{\pi }{4} \right ) \right \}^{2} + 2\left (\frac{1}{\sqrt{2}} \right )^{2} + 2\left ( 2 \right )^{2}\)

= \(2\left \{ \sin \frac{\pi }{4} \right \}^{2} + 2\times \frac{1}{2} + 8\)

= \(2\left ( \frac{1}{\sqrt{2}} \right )^{2}\) + 1 + 8

= 1 + 1 + 8 = 10

= R.H.S.

 

 

Q.5: Calculate the value of:

(i).  \(\sin 75^{\circ}\)

 

(ii).  \(\tan 15^{\circ}\)

 

Sol:

(i).  \(\sin 75^{\circ}\):

= \(\sin \left ( 45^{\circ} + 30^{\circ} \right )\)

= \(\sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}\)

Since, [sin (x + y) = sin x cos y + cos x sin y]

= \(\left (\frac{1}{\sqrt{2}} \right )\left (\frac{\sqrt{3}}{2} \right ) + \left (\frac{1}{\sqrt{2}} \right )\left ( \frac{1}{2} \right )\)

= \(\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}\)

= \(\frac{\sqrt{3} + 1}{2\sqrt{2}}\)

 

(ii).  \(\tan 15^{\circ}\):

= \(\tan \left ( 45^{\circ} – 30^{\circ}\right )\)

= \(\frac{\tan 45^{\circ} – \tan 30^{\circ}}{1 + \tan 45^{\circ}\tan 30^{\circ}}\)

Since, [tan (x – y) = \(\frac{\tan x – \tan y}{1 + \tan x \tan y}\)]

= \(\\\frac{1 – \frac{1}{\sqrt{3}}}{1 + 1\left ( \frac{1}{\sqrt{3}} \right )}\)

= \(\\\frac{\frac{\sqrt{3} – 1}{\sqrt{3}}}{\frac{\sqrt{3 + 1}}{\sqrt{3}}}\)

= \(\\\frac{\sqrt{3} – 1}{\sqrt{3} + 1}\)

= \(\\\frac{\left ( \sqrt{3} – 1\right )^{2}}{\left ( \sqrt{3} + 1\right )\left ( \sqrt{3} – 1 \right )}\)

= \(\\\frac{3 + 1 – 2\sqrt{3}}{\left ( \sqrt{3} \right )^{2} – \left ( 1 \right )^{2}}\)

= \(\frac{4 – 2\sqrt{3}}{3 – 1}\)

= \(2 – \sqrt{3}\)

 

 

Q.6:Prove:

\(\cos \left ( \frac{\pi }{4} – x\right ) \cos \left ( \frac{\pi }{4} – y\right ) – \sin \left ( \frac{\pi }{4} – x\right ) \sin \left ( \frac{\pi }{4} – y\right ) = \sin \left ( x+y \right )\)

 

Sol:

Now, taking L.H.S.

\(\cos \left ( \frac{\pi }{4} – x\right ) \cos \left ( \frac{\pi }{4} – y\right ) – \sin \left ( \frac{\pi }{4} – x\right ) \sin \left ( \frac{\pi }{4} – y\right ):\\\)

= \(\\\frac{1}{2}\left [ 2\cos \left ( \frac{\pi }{4} – x \right ) \cos \left ( \frac{\pi }{4} – y \right ) \right ] + \frac{1}{2}\left [ -2\sin \left ( \frac{\pi }{4} – x \right ) \sin \left ( \frac{\pi }{4} – y \right ) \right ]\\\)

\(=\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right )\right \} + \cos \left \{ \left ( \frac{\pi }{4} – x \right ) – \left ( \frac{\pi }{4} – y \right )\right \}\right ] + \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right )\right \} – \cos \left \{ \left ( \frac{\pi }{4} – x \right ) – \left ( \frac{\pi }{4} – y \right )\right \}\right ]\\\)

Since, [2cos A cos B = cos (A + B) + cos (A – B)]

And, [2sin A sin B = cos (A + B) – cos (A – B)]

= \(\\2\times \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right ) \right \} \right ]\\\)

= \(\\\cos \left [ \frac{\pi }{2} – \left ( x + y \right )\right ]\)

= sin (x + y)

= R.H.S.

 

 

Q.7: Prove:

\(\frac{\tan \left ( \frac{\pi }{4} + x\right )}{\tan \left ( \frac{\pi }{4} – x\right )} = \left (\frac{1 + \tan x}{1 – \tan x} \right )^{2}\)

 

Sol:

Since, tan (A + B)= \(\frac{\tan A + \tan B}{1 – \tan A\tan B}\\\)

And, tan (A – B) = \(\frac{\tan A – \tan B}{1 + \tan A\tan B}\)

Now, taking L.H.S.

\(\frac{\tan \left ( \frac{\pi }{4} + x\right )}{\tan \left ( \frac{\pi }{4} – x\right )}\\\):

= \(\\\frac{\left (\frac{\tan \frac{\pi }{4} + \tan x}{1 – \tan \frac{\pi }{4} \tan x} \right )}{\left (\frac{\tan \frac{\pi }{4} – \tan x}{1 + \tan \frac{\pi }{4} \tan x} \right )}\\\)

= \(\\\frac{\left (\frac{1 + \tan x}{1 – \tan x} \right )}{\left (\frac{1 – \tan x}{1 + \tan x} \right )}\\\)

= \(\\\left (\frac{1 + \tan x}{1 – \tan x} \right )^{2}\)

= R.H.S.

 

 

Q.8: Prove:

\(\frac{\cos \left ( \pi + x \right ) \cos \left ( -x \right )}{\sin \left ( \pi – x \right ) \cos \left ( \frac{\pi }{2} + x \right )} = \cot ^{2} x\)

 

Sol:

Now, taking L.H.S.

\(\frac{\cos \left ( \pi + x \right ) \cos \left ( -x \right )}{\sin \left ( \pi – x \right ) \cos \left ( \frac{\pi }{2} + x \right )}\\\):

= \(\\\frac{\left [ -\cos x \right ]\left [ \cos x \right ]}{\left ( \sin x \right )\left ( -\sin x \right )}\\\)

= \(\\\frac{-\cos ^{2} x}{-\sin ^{2} x}\) = \(\cot ^{2} x\)

= R.H.S.

 

 

Q.9: Prove:

\(\cos \left ( \frac{3\pi }{2} + x\right ) \cos \left ( 2\pi + x \right )\left [ \cot \left ( \frac{3\pi }{2} – x\right ) + \cot \left ( 2\pi + x \right ) \right ] = 1\)

 

Sol:

Now, taking L.H.S.

\(\cos \left ( \frac{3\pi }{2} + x\right ) \cos \left ( 2\pi + x \right )\left [ \cot \left ( \frac{3\pi }{2} – x\right ) + \cot \left ( 2\pi + x \right ) \right ]\\\):

= \(\\\sin x\cos x \left [ \tan x + \cot x \right ]\\\)

= \(\\\sin x\cos x \left ( \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} \right )\\\)

= \(\\\left (\sin x\cos x \right )\left [ \frac{\sin ^{2} x + \cos ^{2} x}{\sin x \cos x} \right ]\\\)

= \(\\\sin ^{2} x + \cos ^{2} x\) = 1

= R.H.S.

 

 

Q.10: Prove:

\(\sin \!\left ( n + 1 \right )\!x \; \sin\! \left ( n + 2 \right )\!x + \cos \! \left ( n + 1 \right )\! x \cos \! \left ( n + 2 \right )\! x = \cos x\)

 

Sol:

Now, taking L.H.S.

\(\sin \!\left ( n + 1 \right )\!x \; \sin\! \left ( n + 2 \right )\!x + \cos \! \left ( n + 1 \right )\! x \cos \! \left ( n + 2 \right )\! x:\\\)

=\(\\\frac{1}{2}\left [ 2\sin \! \left ( n + 1 \right ) \! x \sin \! \left ( n + 2 \right ) \! x + 2\cos \! \left ( n + 1 \right ) \! x \cos \! \left ( n + 2 \right ) \! x\right ]\\\)

=\(\\\frac{1}{2}\left [ \cos \left \{ \left ( n + 1 \right )x – \left ( n + 2 \right )x \right \} – \cos \left \{ \left ( n + 1 \right )x + \left ( n + 2 \right )x \right \} + \cos \left \{ \left ( n + 1 \right )x + \left ( n + 2 \right )x \right \} + \cos \left \{ \left ( n + 1 \right )x – \left ( n + 2 \right )x \right \}\right ]\\\)

Since, [2sin A sin B = cos (A + B) – cos (A – B)]

And, [2cos A cos B = cos (A + B) + cos (A – B)]

= \(\\\frac{1}{2}\times 2 \cos \left \{ \left ( n + 1 \right )x – \left ( n + 2 \right )x \right \}\\\)

= \(\\\cos \left ( -x \right )\) = cos x

= R.H.S.

 

 

Q.11 Prove:

\(\cos \left ( \frac{3\pi }{4} + x\right ) – \cos \left ( \frac{3\pi }{4} – x\right ) = -\sqrt{2}\sin x\)

 

Sol:

Since, cos A – cos B = \(-2\sin \left ( \frac{A + B}{2} \right )\sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\cos \left ( \frac{3\pi }{4} + x\right ) – \cos \left ( \frac{3\pi }{4} – x\right ):\\\)

= \(\\-2\sin \left \{ \frac{\left ( \frac{3\pi }{4} + x \right ) + \left ( \frac{3\pi }{4} – x \right )}{2} \right \} sin \left \{ \frac{\left ( \frac{3\pi }{4} + x \right ) – \left ( \frac{3\pi }{4} – x \right )}{2} \right \}\\\)

= \(\\-2\sin \left ( \frac{3\pi }{4} \right )\sin x\)

= \(\\-2\sin \left ( \pi – \frac{\pi }{4} \right )\sin x\)

= \(\\-2\sin \frac{\pi }{4} \sin x\)

= \(\\-2 \times \frac{1}{\sqrt{2}} \times \sin x\)

= \(\\-\sqrt{2} \sin x\)

= R.H.S.

 

 

Q.12: Prove:

\(\sin ^{2} 6x – \sin ^{2} 4x = \sin\! 2x \; \sin \! 10x\)

 

Sol:

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\sin ^{2} 6x – \sin ^{2} 4x\\\):

= \(\\\left (\sin 6x + \sin 4x \right )\left ( \sin 6x – \sin 4x \right )\)

= \(\\\left [ 2\sin \left ( \frac{6x + 4x}{2} \right ) \cos \left ( \frac{6x – 4x}{2} \right ) \right ] \left [ 2\cos \left ( \frac{6x + 4x}{2} \right ) \sin \left ( \frac{6x – 4x}{2} \right ) \right ]\)

= (2sin 5x cos x) (2cos 5x sin x)

= (2 sin 5x cos 5x) (2 cos x sin x)

= sin 10x sin 2x

= R.H.S.

 

 

Q.13: Prove:

\(\cos ^{2} 2x – \cos ^{2} 6x = \sin 4x \sin 8x\)

 

Sol:

Since, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, cos A – cos B = \(2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\cos ^{2} 2x – \cos ^{2} 6x\):

= (cos 2x + cos 6x)  (cos 2x – cos 6x)

= \(\\\left [ 2\cos \left ( \frac{2x + 6x}{2} \right ) \cos \left ( \frac{2x – 6x}{2} \right ) \right ] \left [ -2 \sin \left ( \frac{2x + 6x}{2} \right ) \sin \left ( \frac{2x – 6x}{2} \right ) \right ]\\\)

= [2cos 4x cos (-2x)]   [ -2sin 4x sin ( -2x)]

= [2cos 4x cos 2x]  [-2sin 4x ( -sin2x)]

= [2sin 4x cos 4x]  [2sin 2x cos 2x]

= sin 8x sin 4x

= R.H.S.

 

 

Q.14:Prove:

\(\sin 2x + 2\sin 4x + \sin 6x = 4 \cos ^{2} x \sin 4x\)

 

Sol:

Now, taking L.H.S.

sin 2x + 2sin 4x + sin 6x:

= [sin 2x + sin 6x] + 2 sin 4x

= \(\\\left [ 2 \sin \left (\frac{2x + 6x}{2} \right ) \left (\frac{2x – 6x}{2} \right ) \right ] + 2 \sin 4x\\\)

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

= 2sin 4x cos(-2x) + 2sin 4x

= 2sin 4x cos 2x + 2sin 4x

= 2sin 4x (cos 2x + 1)

= \(2 \sin 4x \; \left ( 2 \cos ^{2} x – 1 + 1\right )\)

= \(2 \sin 4x \; \left ( 2 \cos ^{2} x \right )\)

= \(4 \cos ^{2} x \sin 4x\)

= R.H.S.

 

 

Q.15: Prove:

\(\cot 4x \left ( \sin 5x + \sin 3x \right ) = \cot x \left ( \sin 5x – \sin 3x \right )\)

 

Sol:

Now, taking L.H.S.

cot 4x (sin 5x + sin 3x):

= \(\frac{\cos 4x}{\sin 4x} \left [ 2 \sin \left ( \frac{5x + 3x}{2} \right ) \cos \left ( \frac{5x – 3x}{2} \right ) \right ]\)

Since, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

= \(\left ( \frac{\cos 4x}{\sin 4x} \right )\left [ 2 \sin 4x \; \cos x \right ]\)

=2cos 4x cos x . . . . . . . . . . . . . . . (1)

Now, taking R.H.S.

cot x (sin 5x – sin 3x):

= \(\frac{\cos x}{\sin x} \left [ 2 \cos \left ( \frac{5x + 3x}{2} \right ) \sin \left ( \frac{5x – 3x}{2} \right ) \right ]\)

\(\sin A – \sin B = 2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

= \(\frac{\cos x}{\sin x} \left [ 2 \cos 4x\; \sin x \right ]\)

= 2 cos 4x cos x . . . . . . . . . . . . . . . . . . . . (2)

From equation (1) and (2):

L.H.S. = R.H.S.

 

 

Q.16: Prove:

\(\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x} = -\frac{\sin 2x}{\cos 10x}\)

 

Sol:

Since, cos A – cos B = \(2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

And, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\frac{\cos 9x – \cos 5x}{\sin 17x – \sin 3x}:\\\)

= \(\\\frac{-2 \sin \left (\frac{9x + 5x}{2} \right )\; \sin \left (\frac{9x – 5x}{2} \right )}{2\cos \left (\frac{17x + 3x}{2} \right ) \; \sin \left (\frac{17x – 3x}{2} \right )}\\\)

= \(\\\frac{-2 \sin 7x \; \sin 2x}{2 \cos 10x \; \sin 7x}\\\)

= \(\\-\frac{\sin 2x}{\cos 10x}\)

= R.H.S.

 

 

Q.17: Prove:

\(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x} = \tan 4x\)

 

Sol:

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\frac{\sin 5x + \sin 3x}{\cos 5x + \cos 3x}\\\):

= \(\\\frac{2 \sin \left ( \frac{5x + 3x}{2} \right ) \; \cos \left ( \frac{5x – 3x}{2} \right ) }{2 \cos \left ( \frac{5x + 3x}{2} \right ) \; \cos \left ( \frac{5x – 3x}{2} \right )}\\\)

= \(\\\frac{2 \sin 4x \cos x}{2 \cos 4x \cos x}\) = tan 4x

=R.H.S.

 

 

Q.18: Prove:

\(\frac{\sin x – \sin y}{\cos x + \cos y} = \tan \frac{x – y}{2}\)

 

Sol:

Since, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

Now taking L.H.S.

\(\frac{\sin x – \sin y}{\cos x + \cos y}:\\\)

= \(\\\frac{2 \cos \left ( \frac{x + y}{2} \right ) \;\sin \left ( \frac{x – y}{2} \right )}{2 \cos \left ( \frac{x + y}{2} \right ) \;\cos \left ( \frac{x – y}{2} \right )}\\\)

= \(\\\frac{\sin \left ( \frac{x – y}{2} \right )}{\cos \left ( \frac{x – y}{2} \right )}\)

= \(\\\tan \left ( \frac{x – y}{2} \right )\)

= R.H.S.

 

 

Q.19: Prove:

\(\frac{\sin x + \sin 3x}{\cos x + \cos 3x} = \tan 2x\)

 

Sol::

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

And, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

Now, taking L.H.S.

\(\frac{\sin x + \sin 3x}{\cos x + \cos 3x}:\\\)

= \(\\\frac{2 \sin \left ( \frac{x + 3x}{2} \right ) \;\cos \left ( \frac{x – 3x}{2} \right )}{2 \cos \left ( \frac{x + 3x}{2} \right ) \;\cos \left ( \frac{x – 3x}{2} \right )}\)

= \(\\\frac{\sin 2x}{\cos 2x}\) = tan 2x

= R.H.S.

 

 

Q.20: Prove:

\(\frac{\sin x – \sin 3x}{\sin ^{2} x – \cos ^{2} x} = 2\sin x\)

 

Answer:

Since, sin A – sin B = \(2\cos \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

\(\cos ^{2} A – \sin ^{2} A = \cos 2A\\\)

Now, taking L.H.S.

\(\frac{\sin x – \sin 3x}{\sin ^{2} x – \cos ^{2} x}\\\):

= \(\\\frac{2 \cos \left (\frac{x + 3x}{2} \right ) \sin \left ( \frac{x – 3x}{2} \right )}{ – \cos 2x}\\\)

= \(\\\frac{2 \cos 2x \;\sin \left ( -x \right )}{-\cos 2x}\\\)

= \(\\-2x \left ( -\sin x \right )\) = 2 sin x

= R.H.S.

 

 

Q.21: Prove:

\(\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x} = \cot 3x\)

 

Sol:

Taking L.H.S.

\(\frac{\cos 4x + \cos 3x + \cos 2x}{\sin 4x + \sin 3x + \sin 2x}\):

= \(\\\frac{\left (\cos 4x + \cos 2x \right )+ \cos 3x}{\left (\sin 4x + \sin 2x \right ) + \sin 3x}\\\)

= \(\\\frac{2 \cos \left ( \frac{4x + 2x}{2} \right ) \cos \left ( \frac{4x – 2x}{2} \right ) + \cos 3x}{2 \sin \left ( \frac{4x + 2x}{2} \right ) \cos \left ( \frac{4x – 2x}{2} \right ) + \sin 3x}\\\)

\(\\\begin{bmatrix} \sin A + \sin B = 2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\\* \cos A + \cos B = 2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\\ \end{bmatrix}\)

= \(\frac{2 \cos 3x \; \cos x + \cos 3x}{2 \sin 3x \; \cos x + \sin 3x}\\\)

= \(\\\frac{\cos 3x \left ( 2 \cos x + 1 \right )}{\sin 3x \left ( 2 \cos x + 1 \right )}\) = cot 3x

= R.H.S.

 

 

Q.22: Prove:

\(\cot x \;\cot 2x – \cot 2x \;\cot 3x – \cot 3x \;\cot x = 1\)

 

Sol:

Now, taking L.H.S.

cot x cot 2x – cot 2x cot 3x – cot 3x cot x :

= \(\\\cot x \;\cot 2x – \cot 3x\left ( \cot 2x + \cot x \right )\\\)

= \(\\\cot x \;\cot 2x – \cot\left ( 2x + x \right )\left ( \cot 2x + \cot x \right )\\\)

= \(\\\cot x \;\cot 2x – \left [ \frac{\cot 2x \;\cot x – 1}{\cot x + \cot 2x} \right ]\left ( \cot 2x + \cot x \right )\\\)

= \(\\\left [\cot \left ( A + B \right ) = \frac{\cot A \; \cot B – 1}{\cot A + \cot B} \right ]\\\)

= \(\\\cot x \;\cot 2x – \left ( \cot 2x \cot x – 1 \right )\) = 1

= R.H.S

 

 

Q.23: Prove:

\(\tan 4x = \frac{4 \tan x \left ( 1 – \tan ^{2}x \right )}{1 – 6\tan ^{2} x + \tan ^{4} x}\)

 

Sol:

Since, tan 2A = \(\frac{2\tan A}{1 – \tan ^{2}A}\)

Now, taking L.H.S.

tan 4x :

= \(\tan 2\left (2x \right )\)

= \(\\\frac{2 \tan 2x}{1 – \tan ^{2}\left (2x \right )}\\\)

= \(\\\frac{2 \left ( \frac{2 \tan x}{1 – \tan ^{2}x} \right )}{1 – \left ( \frac{2 \tan x}{1 – \tan ^{2}x} \right )^{2}}\\\)

= \(\\\frac{\left ( \frac{4 \tan x}{1 – \tan ^{2}x} \right )}{1 – \left ( \frac{4 \tan ^{2}x}{\left (1 – \tan ^{2}x \right )^{2}} \right )}\\\)

=  \(\\\frac{\left (\frac{4\tan x}{1 – \tan ^{2}x} \right )}{\left [\frac{\left ( 1 – \tan ^{2}x \right )^{2} – 4 \tan ^{2}x}{\left ( 1 – \tan ^{2}x \right )^{2}} \right ]}\\\)

= \(\\\frac{4 \tan x \left ( 1 – \tan ^{2} x\right )}{\left ( 1 – \tan ^{2}x \right )^{2} – 4 \tan ^{2} x}\\\)

= \(\\\frac{4 \tan x \left ( 1 – \tan ^{2}x \right )}{1 + \tan ^{4}x – 2 \tan ^{2}x – 4 \tan ^{2}x}\\\)

= \(\\\frac{4 \tan x \left ( 1 – \tan ^{2}x \right )}{1 – 6 \tan ^{2}x + \tan ^{4}x }\)

= R.H.S.

 

 

Q.24: Prove:

\(\cos 4x = 1 – 8 \sin ^{2}x \;\cos ^{2}x\)

 

Sol:

Now, taking L.H.S.

cos 4x:

= \(\\\cos 2\left (2x \right )\\\)

= \(\\1 – 2 \sin ^{2} 2x \;\;\;\;\left [ \cos 2A = 1 – 2 \sin ^{2}A \right ]\\\)

= \(\\1 – 2 \left ( 2 \sin x \cos x \right )^{2} \;\;\;\;\;\left [ \sin 2A = 2 \sin A \cos A \right ]\\\)

= \(\\1 – 8 \sin ^{2}x \cos ^{2}x\\\)

= R.H.S.

 

 

Q.25: Prove:

\(\cos 6x = 32 \cos ^{6}x – 48 \cos ^{4}x + 18 \cos ^{2}x -1\)

 

Sol:

Now, taking L.H.S.

cos 3 (2x ):

= \(\\4 \cos ^{3}2x – 3 \cos 2x \;\;\;\;\; \left [ \cos 3A = 4\cos ^{3}A – 3 \cos A \right ]\\\)

= \(\\4 \left ( 2 \cos ^{2}x – 1 \right )^{3} – 3 \left ( 2 \cos ^{2}x -1 \right ) \;\;\;\;\; \left [ \cos 2x = 2 \cos ^{2}x -1 \right ]\\\)

= \(\\4 \left [ \left ( 2 \cos ^{2}x \right )^{3} – \left (1 \right )^{3} – 3 \left ( 2 \cos ^{2}x \right )^{2} + 3 \left ( 2 \cos ^{2}x \right )\right ] – 6 \cos ^{2}x + 3\\\)

= \(\\4 \left [ 8 \cos ^{6}x – 1 – 12\cos ^{4}x + 6 \cos ^{2}x \right ] – 6 \cos ^{2}x + 3\\\)

= \(\\32 \cos ^{6}x – 4 – 48 \cos ^{4}x + 24 \cos ^{2}x – 6 \cos ^{2}x + 3\\\)

= \(\\32 \cos ^{6}x – 48 \cos ^{4}x + 18 \cos ^{2}x -1\)

= R.H.S.

 

 

Exercise 3.4

 Q.1: Find general solutions and the principle solutions of the given equation: tan x = \(\sqrt{3}\)

 

Sol:

tan x = \(\sqrt{3}\)    [Given]

As we know that, \(\tan \frac{\pi }{3} = \sqrt{3}\)

And, \(\tan \frac{4\pi }{3}\) = \(\tan \left ( \pi + \frac{\pi }{3} \right )\) = \(\tan \left ( \frac{\pi }{3} \right )\) = \(\sqrt{3}\)

Therefore, the principle solutions are \(x = \frac{\pi }{3}\) and \(\frac{4\pi }{3}\)

Now, \(\tan x = \tan \frac{\pi }{3}\\\)

\(x = n\pi + \frac{\pi }{3}\), where n \(\in\) Z

Therefore, the general solution is \(x = n\pi + \frac{\pi }{3}\), where n \(\in\) Z.

 

 

Q.2: Find general solutions and the principle solutions of the given equation: sec x = 2

 

Sol:

sec x = 2   [Given]

As we know that, \(\sec \frac{\pi }{3} = 2\)

And, \(\sec \frac{5\pi }{3}\) = \(\sec \left (2\pi – \frac{\pi }{3} \right )\) = \(\sec \frac{\pi }{3}\) = 2

Therefore, the principle solutions are \(x = \frac{\pi }{3}\) and \(\frac{5\pi }{3}\).

Now, \(\sec x = \sec \frac{\pi }{3}\)

And, \(\cos x = \cos \frac{\pi }{3} \;\;\;\;\; \left [ \sec x = \frac{1}{\cos x} \right ]\\\)

\(x = 2n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z

Therefore, the general solution is \(x = 2n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z.

 

 

Q.3: Find general solutions and the principle solutions of the given equation:

cot = \( -\sqrt{3}\)

 

Sol:

cot = \(-\sqrt{3}\)          [Given]

As we know that, \(\cot \frac{\pi }{6} = \sqrt{3}\;\;\Rightarrow \;\;\cot \left (\pi – \frac{\pi }{6} \right ) = -\cot \frac{\pi }{6} = -\sqrt{3}\)

And, \(\cot \left ( 2\pi – \frac{\pi }{6} \right )\) = \(-\cot \frac{\pi }{6}\) = \(-\sqrt{3}\)

That is \(\cot \frac{5\pi }{6} = -\sqrt{3}\) and \(\cot \frac{11\pi }{6} = -\sqrt{3}\)

Therefore, the principle solutions are \(x = \frac{5\pi }{6}\) and \(\frac{11\pi }{6}\).

Now,\(\cot x = \cot \frac{5\pi }{6}\)

And, \(\tan x = \tan \frac{5\pi }{6} \;\;\;\;\;\left [ \cot x = \frac{1}{\tan x} \right ]\)

\(\\x = n\pi + \frac{5\pi }{6}\), where n \(\in\) Z

Therefore, the general solution is \(x = n\pi + \frac{5\pi }{6}\), where n \(\in\) Z.

 

 

Q.4: Find general solutions and the principle solutions of the given equation: cosec x = -2

 

Sol:

cosec x = -2     [Given]

As we know that, \(cosec \frac{\pi }{6} = 2\)

Hence, \(cosec \left (\pi + \frac{\pi }{6} \right )\) = \(-cosec \frac{\pi }{6}\) = -2

And, \(cosec \left (2\pi – \frac{\pi }{6} \right )\) = \(-cosec \frac{\pi }{6}\) = -2

That is \(cosec \frac{7\pi }{6} = -2\) and \(cosec \frac{11\pi }{6} = -2\).

Therefore, the principle solutions are \(x = \frac{7\pi }{6}\) and \(\frac{11\pi }{6}\).

Now,\(cosec \: x = cosec \frac{7\pi }{6}\)

And, \(\sin x = \sin \frac{7\pi }{6} \;\;\;\;\; \left [ cosec x = \frac{1}{\sin x} \right ]\)

\(\\x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z

Therefore, the general solution is \(x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z.

 

 

Q.5: Find the general solution of the given equation: cos 4x = cos 2x

Sol:

cos 4x = cos 2x           [Given]

i.e.  cos 4x – cos 2x = 0

\(-2\sin \left ( \frac{4x + 2x}{2} \right ) \sin \left ( \frac{ 4x – 2x }{2}\right ) = 0\)

Since, cos A – cos B = \(2\sin \left ( \frac{A + B}{2} \right ) \sin \left ( \frac{A – B}{2} \right )\)

(sin 3x)  (sin x) = 0

sin 3x = 0  or  sin x = 0

sin 3x = 0

\(3x = n\pi\\\)

\(x = \frac{n\pi}{3}\), where n \(\in\) Z

sin x = 0

\(\Rightarrow\)  \(x = n\pi\), where n \(\in\) Z

Q.6: Find the general solution of the given equation: cos 3x + cos x – cos 2x = 0

Sol:

cos 3x + cos x – cos 2x = 0        [Given]

\(2 \cos \left ( \frac{3x + x}{2} \right ) \cos \left ( \frac{3x – x}{2} \right ) – \cos 2x = 0\)

Since, cos A + cos B = \(2\cos \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

2cos 2x cos x – cos 2x = 0

cos 2x (2cos x – 1) = 0

cos 2x = 0  or  2cos x -1 = 0

cos 2x = 0

\(2x = \left ( 2n + 1 \right )\frac{\pi }{2}\), where n \(\in\) Z

\(\\x = \left ( 2n + 1 \right )\frac{\pi }{4}\), where n \(\in\) Z

2cos x -1 = 0

\(\cos x = \frac{1}{2}\\\)

\(\cos x = \cos \frac{\pi }{3}\), where n \(\in\) Z

\(x = 2n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z

 

 

Q.7: Find the general solution of the given equation:  sin 2x + cos x = 0

 

Sol:

sin 2x + cos x = 0          [Given]

2sin x cos x + cos x = 0

cos x (2sin x + 1) = 0

cos x = 0   or   2sin x + 1 = 0

cos x = 0

\(\cos x = \left ( 2n + 1 \right )\frac{\pi }{2}\), where n \(\in\) Z

2sin x + 1 = 0

= \(-\sin \frac{\pi }{6}\) = \(\sin \left (\pi + \frac{\pi }{6} \right )\) = \(\sin \frac{7\pi }{6}\)

\(\Rightarrow\)  \(x = n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z

Therefore, the general solution is \(\\\left ( 2n + 1 \right )\frac{\pi }{2}\) or \(n\pi + \left ( -1 \right )^{n}\frac{7\pi }{6}\), where n \(\in\) Z.

 

 

Q.8: Find the general solution of the given equation:

\(\sec ^{2}2x = 1 – \tan 2x\)

 

Sol:

\(\sec ^{2}2x = 1 – \tan 2x\)           [Given]

\(1 + \tan ^{2}2x = 1 – \tan 2x\)

\(\tan ^{2}2x + \tan 2x = 0\)

tan 2x ( tan 2x + 1) = 0

tan 2x = 0  or  tan 2x + 1 = 0

tan 2x = 0

tan 2x = tan 0

2x = n∏ + 0, where n \(\in\) Z

\(x = \frac{n\pi}{2}\), where n \(\in\) Z

tan 2x + 1 = 0

tan 2x = -1= \(-\tan \frac{\pi }{4}\)= \(\tan \left (\pi – \frac{\pi }{4} \right )\)= \(\tan \frac{3\pi }{4}\)

\(\Rightarrow\)  \(2x = n\pi + \frac{3\pi }{4}\), where n \(\in\) Z

\(\Rightarrow\)  \(x = \frac{n\pi}{2} + \frac{3\pi }{8}\), where n \(\in\) Z

Therefore, the general solution is \(\frac{n\pi}{2}\) or \(\frac{n\pi}{2} + \frac{3\pi }{8}\), where n \(\in\) Z.

 

 

Q.9: Find the general solution of the given equation:  sin x + sin 3x + sin 5x = 0

 

Sol:

sin x + sin 3x + sin 5x = 0              [Given]

(sin x + sin 5x) + sin 3x = 0

\(\left [ 2\sin \left ( \frac{x + 5x}{2} \right ) \cos \left ( \frac{x – 5x}{2} \right )\right ] + \sin 3x = 0\)

Since, sin A + sin B = \(2\sin \left ( \frac{A + B}{2} \right ) \cos \left ( \frac{A – B}{2} \right )\)

2sin 3x cos (-2x) + \sin 3x = 0

2sin 3x cos 2x + sin 3x = 0

sin 3x (2cos 2x + 1) = 0

sin 3x = 0 or  2cos 2x + 1 = 0

Now,

sin 3x = 0

\(3x = n\pi\), where n \(\in\) Z

\(x = \frac{n\pi }{3}\), where n \(\in\) Z

2cos 2x + 1 = 0

\(\cos 2x = -\frac{1}{2}\) = \(-\cos \frac{\pi }{3}\) = \(\cos \left (\pi – \frac{\pi }{3} \right )\) = \(\cos \frac{2\pi }{3}\)

\(2x = 2n\pi \pm \frac{2\pi }{3}\), where n \(\in\) Z

\(x = n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z

Therefore, the general solution is \(\frac{n\pi }{3}\) or \(n\pi \pm \frac{\pi }{3}\), where n \(\in\) Z.

 

 

Miscellaneous Exercise

 Q.1: Prove that:

\(2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + \cos \frac{3\pi }{13}\; + \cos \frac{5\pi }{13} = 0\)

 

Sol:

Taking L.H.S.

\(2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + \cos \frac{3\pi }{13}\; + \cos \frac{5\pi }{13}\\\) :

= \(\\2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + 2\cos \left (\frac{\frac{3\pi }{13} + \frac{5\pi }{13}}{2} \right )\; \cos \left (\frac{\frac{3\pi }{13} – \frac{5\pi }{13}}{2} \right )\\\)

Since, \(\\\boldsymbol{\left [\cos x + \cos y = 2\cos \left ( \frac{x + y}{2} \right )\cos \left ( \frac{x – y}{2} \right ) \right ]}\\\)

= \(\\2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + 2\cos \frac{4\pi }{13} \; \cos \left (\frac{-\pi }{13} \right )\\\)

= \(\\2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + 2\cos \frac{4\pi }{13} \; \cos \frac{\pi }{13}\\\)

= \(\\2\cos \frac{\pi }{13}\; \left [\cos \frac{9\pi }{13}\; + \cos \frac{4\pi }{13} \right ]\\\)

= \(\\2\cos \frac{\pi }{13}\; \left [2\cos \left ( \frac{\frac{9\pi }{13} + \frac{4\pi }{13}}{2} \right )\; \cos \left ( \frac{\frac{9\pi }{13} – \frac{4\pi }{13}}{2} \right ) \right ]\\\)

= \(\\2\cos \frac{\pi }{13}\; \left [ 2\cos \frac{\pi }{2} \; \cos \frac{5\pi }{26}\right ]\\\)

= \(\\2\cos \frac{\pi }{13} \times 2 \times 0 \times \cos \frac{5\pi }{26}\)

= 0

= R.H.S.

Therefore, \(\boldsymbol{2\cos \frac{\pi }{13}\; \cos \frac{9\pi }{13}\; + \cos \frac{3\pi }{13}\; + \cos \frac{5\pi }{13} = 0}\)

 

 

Q.2: Prove that:

\(\left ( \sin 3x + \sin x \right )\sin x + \left ( \cos 3x – \cos x \right )\cos x = 0\)

 

Sol:

Taking L.H.S.

\(\\\left ( \sin 3x + \sin x \right )\sin x + \left ( \cos 3x – \cos x \right )\cos x\\\) :

= \(\\\sin 3x\; \sin x + \sin ^{2}x + \cos 3x \;\cos x – \cos ^{2}x\\\)

= \(\\\cos 3x \;\cos x + \sin 3x\; \sin x – \left (\cos ^{2}x – \sin ^{2}x \right )\\\)

= \(\\\cos \left ( 3x – x \right ) – \cos 2x\\\)

Since, \(\\\boldsymbol{\left [ \cos \left ( A -B \right ) = \cos A\; \cos B + \sin A \; \sin B \right ]}\\\)

= \(\\\cos 2x – \cos 2x\)

= 0

= R.H.S.

Therefore, \(\boldsymbol{\left ( \sin 3x + \sin x \right )\sin x + \left ( \cos 3x – \cos x \right )\cos x = 0}\)

 

 

Q-3: Prove that:

\(\left (\cos x + \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\cos ^{2}\frac{x + y}{2}\)

 

Sol:

Taking L.H.S.

\(\left (\cos x + \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2}\\\) :

= \(\\\cos ^{2}x + \cos ^{2}y + 2\cos x\; \cos y + \sin ^{2}x + \sin ^{2}y – 2\sin x\; \sin y\\\)

= \(\\\left (\cos ^{2}x + \sin ^{2}x \right ) + \left (\cos ^{2}y + \sin ^{2}y \right ) + 2 \left (\cos x\; \cos y – \sin x\; \sin y \right )\\\)

= \(\\1 + 1 + 2\cos \left ( x +y \right )\\\)

Since, \(\\\boldsymbol{\left [ \cos \left ( A + B \right ) = \cos A\; \cos B – \sin A \; \sin B \right]}\\\)

= \(\\2 + 2\cos \left ( x +y \right )\\\)

= \(\\2\left [1 + \cos \left ( x +y \right ) \right ]\\\)

= \(\\2\left [1 + 2\cos ^{2}\left ( \frac{x + y }{2}\right ) – 1 \right ]\\\)

Since, \(\\\boldsymbol{\left [\cos 2A = 2\cos ^{2}A – 1 \right ]}\\\)

= \(\\4\cos ^{2}\left ( \frac{x + y}{2} \right )\\\)

= R.H.S.

Therefore, \(\boldsymbol{\left (\cos x + \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\cos ^{2}\frac{x + y}{2}}\)

 

 

Q-4: Prove that:

\(\left ( \cos x – \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\sin ^{2}\frac{x – y}{2}\)

 

Sol:

Taking L.H.S.

\(\left ( \cos x – \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2}\\\) :

= \(\\\cos ^{2}x + \cos ^{2}y – 2\cos x \; \cos y + \sin ^{2}x + \sin ^{2}y – 2\sin x\; \sin y\\\)

= \(\\\left (\cos ^{2}x + \sin ^{2}x \right ) + \left (\cos ^{2}y + \sin ^{2}y \right ) – 2 \left (\cos x\; \cos y – \sin x\; \sin y \right )\\\)

= \(\\1 + 1 – 2\left [ \cos \left ( x – y \right ) \right ]\\\)

Since, \(\\\boldsymbol{\left [ \cos \left ( A -B \right ) = \cos A\; \cos B + \sin A \; \sin B \right ]}\\\)

= \(\\2\left [ 1 – \cos \left ( x – y \right ) \right ]\\\)

= \(\\2\left [ 1 – \left \{ 1 – 2 \sin ^{2}\left ( \frac{x – y}{2} \right ) \right \} \right ]\\\)

Since, \(\\\boldsymbol{\left [\cos 2A = 1 – 2\sin ^{2}A \right ]}\\\)

= \(\\4\sin ^{2}\frac{x – y}{2}\)

= R.H.S.

Therefore, \(\boldsymbol{\left ( \cos x – \cos y \right )^{2} + \left ( \sin x – \sin y \right )^{2} = 4\sin ^{2}\frac{x – y}{2}}\)

 

 

Q-5: Prove that:

\(\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \; \cos 2x \; \cos 4x\)

 

Sol:

Taking L.H.S.

\(\sin x + \sin 3x + \sin 5x + \sin 7x\\\) :

= \(\\\left (\sin x + \sin 5x \right ) + \left (\sin 3x + \sin 7x \right )\\\)

= \(\\2 \sin \left ( \frac{x + 5x}{2} \right )\; \cos \left ( \frac{x – 5x}{2} \right ) + 2 \sin \left ( \frac{3x + 7x}{2} \right )\; \cos \left ( \frac{3x + 7x}{2} \right )\\\)

Since, \(\\\boldsymbol{\left [\sin A + \sin B = 2\sin \left ( \frac{A +B}{2} \right ).\cos \left ( \frac{A – B}{2} \right ) \right ]}\\\)

= \(\\2\sin 3x \cos \left ( -2x \right ) + 2\sin 5x \cos \left ( -2x \right )\\\)

= \(\\2\sin 3x \cos 2x + 2\sin 5x \cos 2x\\\)

= \(\\2 \cos 2x \left [ \sin 3x + \sin 5x\right ]\\\)

= \(\\2 \cos 2x \left [2 \sin \left ( \frac{3x + 5x}{2} \right ). \cos \left ( \frac{3x – 5x}{2} \right )\right ]\\\)

= \(\\2 \cos 2x \left [2 \sin 4x. \cos \left ( -x \right )\right ]\\\)

= \(\\4 \cos 2x\; \sin 4x \; \cos x\)

= R.H.S.

Therefore, \(\boldsymbol{\sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x \; \cos 2x \; \cos 4x}\)

 

 

Q-6: Prove that:

\(\frac{\left ( \sin 7x + \sin 5x \right ) + \left ( \sin 9x + \sin 3x \right )}{\left ( \cos 7x + \cos 5x \right ) + \left ( \cos 9x + \cos 3x \right )} = \tan 6x\)

 

Sol:

Since, \(\boldsymbol{\sin A + \sin B = 2\sin \left ( \frac{A + B}{2} \right ). \cos \left ( \frac{A -B}{2} \right )}\\\)

And, \(\\\boldsymbol{\cos A + \cos B = 2\cos \left ( \frac{A + B}{2} \right ). \cos \left ( \frac{A -B}{2} \right )}\)

Taking L.H.S.

\(\\\frac{\left ( \sin 7x + \sin 5x \right ) + \left ( \sin 9x + \sin 3x \right )}{\left ( \cos 7x + \cos 5x \right ) + \left ( \cos 9x + \cos 3x \right )}\\\) :

= \(\\\frac{\left [2\sin \left ( \frac{7x + 5x}{2} \right ). \cos \left ( \frac{7x – 5x}{2} \right ) \right ] + \left [ 2 \sin \left ( \frac{9x + 3x}{2}\right ). \cos \left ( \frac{9x – 3x}{2}\right ) \right ]}{\left [2\cos \left ( \frac{7x + 5x}{2} \right ). \cos \left ( \frac{7x – 5x}{2} \right ) \right ] + \left [ 2 \cos \left ( \frac{9x + 3x}{2}\right ). \cos \left ( \frac{9x – 3x}{2}\right ) \right ]}\\\)

= \(\\\frac{\left [2\sin 6x. \cos x \right ] + \left [ 2 \sin 6x.\cos 3x \right ]}{\left [2\cos 6x.\cos x \right ] + \left [ 2\cos 6x. \cos 3x \right ]}\\\)

= \(\\\frac{2\sin 6x \left [\cos x + \cos 3x \right ]}{2\cos 6x \left [\cos x + \cos 3x \right ] }\\\)

= \(\tan 6x\)

= R.H.S.

Therefore, \(\boldsymbol{\frac{\left ( \sin 7x + \sin 5x \right ) + \left ( \sin 9x + \sin 3x \right )}{\left ( \cos 7x + \cos 5x \right ) + \left ( \cos 9x + \cos 3x \right )} = \tan 6x}\)

 

 

Q-7: Show that: \(\sin 3y + \sin 2y – \sin y = 4\sin y \cos\frac{y}{2} \cos\frac{3y}{2}\)

 

Sol:

Here, L.H.S = \(\sin 3y + \sin 2y\; – \sin y\\\)

= \(\;\\\sin 3y + (\sin 2y \;- \sin y) = \sin 3y + \left [ 2\cos\left ( \frac{2y + y}{2} \right ) \sin\left ( \frac{2y – y}{2} \right ) \right ]\\\)

Since,  \(\boldsymbol{(\sin x – \sin y) = \left [ 2\cos\left ( \frac{x + y}{2} \right ) \sin\left ( \frac{x – y}{2} \right ) \right ]}\)

= \(\\\sin 3y + \left [ 2\cos\left ( \frac{3y}{2} \right ) \sin\left ( \frac{y}{2} \right ) \right ]\\\)

=\( \left [ 2\sin\left ( \frac{3y}{2} \right ) \cos\left ( \frac{3y}{2} \right ) \right ] + \left [ 2\cos\left ( \frac{3y}{2} \right ) \sin\left ( \frac{y}{2} \right ) \right ]\\\)

Since, sin 2x = 2 sin x cos x

=\(\\2\cos\frac{3y}{2}\cdot [ \sin\ \frac{3y}{2} + \sin \frac{y}{2}]\\\)

=\(2\cos\frac{3y}{2}\cdot [ \sin\ \frac{(\frac{3y}{2} + \frac{y}{2})}{2}][ \cos\ \frac{(\frac{3y}{2} – \frac{y}{2})}{2}]\)

Since, sin x + sin y = \(\boldsymbol{\\2\sin\left ( \frac{x + y}{2} \right )\cos\left ( \frac{x – y}{2} \right )}\)

\(\\= 2\cos\left ( \frac{3y}{2} \right )2\sin y \cos\left ( \frac{y}{2} \right )\)

\(\\= 4\sin y\cos\left ( \frac{y}{2} \right ) \cos\left ( \frac{3y}{2} \right )\)

= R.H.S

 

 

Q-8: The value of \(\tan y = -\frac{4}{3}\) where y in in 2nd quadrant then find out the values of \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\).

 

Sol:

Here, y is in 2nd quadrant.

So, \(\frac{\pi}{2} < y < \pi \\ \\ \Rightarrow \frac{\pi}{4} < \frac{y}{2} < \frac{\pi}{2}\)

Thus, \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\) lies in 1st quadrant.

Now,

\(\tan y = -\frac{4}{3}\)

\(\sec^{2} y = 1 + \tan^{2} y = 1 + (\frac{4}{3})^{2} = 1 + \frac{16}{9} = \frac{25}{9}\\\)

So, \(\cos^{2} y = \frac{9}{25}\)

\(\Rightarrow \cos y = \pm \frac{3}{5}\)

As y is in 2nd quadrant, cos y is negative.

\(\cos y = \frac{-3}{5}\)

So, cos y = \(2\cos^{2} \frac{y}{2} – 1\)

\(\Rightarrow \frac{-3}{5} = 2\cos^{2} \frac{y}{2} – 1 \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = 1 – \frac{-3}{5} \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = \frac{2}{5} \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = \frac{1}{5} \\ \\ \Rightarrow 2\cos^{2} \frac{y}{2} = \frac{3}{\sqrt{5}}\)  [Since, \(\boldsymbol{\cos\frac{y}{2}}\) is positive]

\(\Rightarrow \cos\frac{y}{2} = \frac{\sqrt{5}}{5} \\ \\ \sin^{2} \frac{y}{2} + \cos^{2} \frac{y}{2} = 1 \\ \\ \Rightarrow \sin^{2} \frac{y}{2} + \cos^{2} \frac{1}{\sqrt{5}} = 1 \\ \\ \Rightarrow \sin^{2} \frac{y}{2} = 1 – \frac{1}{5} = \frac{4}{5} \\ \\ \Rightarrow \sin^{2} \frac{y}{2} = \frac{2}{\sqrt{5}}\)  [Since,  \(\boldsymbol{\sin\frac{y}{2}}\) is positive]

\(\Rightarrow \sin^{2} \frac{y}{2} = \frac{2\sqrt{5}}{5} \\ \\ \tan \frac{y}{2} = \frac{\sin\frac{y}{2}}{\cos\frac{y}{2}} = \frac{\frac{2}{\sqrt{5}}}{\frac{1}{\sqrt{5}}} = 2\)

 

 

Q-9: The value of \(\cos y = -\frac{1}{3}\) where y in in 3rd quadrant then find out the values of \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\).

 

Sol:

Here, y is in 3rd  quadrant.

So, \(\pi < y < \frac{3\pi}{2} \\ \\ \Rightarrow \frac{\pi}{2} < \frac{y}{2} < \frac{3\pi}{4}\)

Thus, \(\cos \frac{y}{2} \;and\; \tan\frac{y}{2}\) are negative and \(\sin \frac{y}{2}\) is positive.

Now,  \(\cos y = -\frac{1}{3}\)  [Given]

\(\cos y = 1 – 2\sin^{2}\frac{y}{2} \Rightarrow \sin^{2}\frac{y}{2} = \frac{1 – \cos y}{2} \Rightarrow \sin^{2}\frac{y}{2} = \frac{1 – \frac{-1}{3}}{2} = \frac{1 + \frac{1}{3}}{2} = \frac{2}{3}\\\)

\(\\\Rightarrow \boldsymbol{\sin \frac{y}{2}} = \frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{\sqrt{6}}{3}\\\)   [Since, \(\sin\frac{y}{2}\) is positive]

Now, \(\cos y = 2\cos^{2}\frac{y}{2} – 1\)

\(2\cos^{2}\frac{y}{2} = \frac{1 + \cos y}{2} = \frac{1 – \frac{1}{3}}{2} = \frac{1}{3}\\\)

\(\\\Rightarrow\)   \(\boldsymbol{\cos \frac{y}{2}}= \frac{-1}{\sqrt{3}} = \frac{-1}{\sqrt{3}}\times \frac{-\sqrt{3}}{3} = \frac{-1}{\sqrt{3}}\)

Therefore, \(\boldsymbol{\tan\frac{y}{2}} = \frac{\sin\frac{y}{2}}{\cos\frac{y}{2}} = \frac{\frac{\sqrt{2}}{\sqrt{3}}}{\frac{-1}{\sqrt{3}}} = -\sqrt{2}\)

 

 

Q-10: The value of \(\sin y = \frac{1}{4}\) where y in in 2nd quadrant then find out the values of \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\).

 

Sol:

Here, y is in 2nd quadrant.

So, \(\frac{\pi}{2} < y < \pi \\ \\ \Rightarrow \frac{\pi}{4} < \frac{y}{2} < \frac{\pi}{2}\)

Thus, \(\sin \frac{y}{2}, \cos \frac{y}{2} \;and\; \tan\frac{y}{2}\) lies in 1st quadrant.

Now, \(\sin y = \frac{1}{4}\)

\(\\\cos^{2} y = 1 – 2\sin^{2} y = 1 – (\frac{1}{4})^{2} = 1 – \frac{1}{16} = \frac{15}{16} \\ \\ \Rightarrow \cos y = -\frac{\sqrt{15}}{4}\)  [Since, \(\cos y\) is negative]

\(\sin^{2} \frac{y}{2} = \frac{1 – (\frac{-\sqrt{15}}{4})}{2} = \frac{4 + \sqrt{15}}{8}\\\)

\(\\\Rightarrow \)   \(\boldsymbol{\sin \frac{y}{2}} = \sqrt{\frac{4 + \sqrt{15}}{8}}\)  [ Since, \(\sin\frac{y}{2}\\\) is positive ]

=\(\\\sqrt{\frac{4 + \sqrt{15}}{8}\times \frac{2}{2}}= \sqrt{\frac{8 + 2\sqrt{15}}{16}} =\boldsymbol{\frac{\sqrt{8 + 2\sqrt{15}}}{4}}\\\)

\(\\\cos^{2} y = \frac{1 + \cos y}{2} = \frac{1 + (-\frac{\sqrt{15}}{4})}{2} = \frac{4 – \sqrt{15}}{8}\\\)

Therefore, \(\boldsymbol{\cos \frac{y}{2}}= \sqrt{\frac{4 – \sqrt{15}}{8}}\) = \(\sqrt{\frac{4 – \sqrt{15}}{8}\times \frac{2}{2}}\\\)

= \(\sqrt{\frac{8 – 2\sqrt{15}}{16}}=\boldsymbol{\frac{\sqrt{8 – 2\sqrt{15}}}{4}}\)

Now, \(\boldsymbol{tan \frac{y}{2}} = \frac{\sin \frac{y}{2}}{\cos \frac{y}{2}} = \frac{\left ( \frac{\sqrt {8\; + \;2\sqrt{15}}}{4} \right )}{\left ( \frac{\sqrt {8\; -\; 2\sqrt{15}}}{4} \right )}\)

= \(\\\frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{8 – 2\sqrt{15}}} = \frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{8 – 2\sqrt{15}}}\times \frac{\sqrt{8 + 2\sqrt{15}}}{\sqrt{8 + 2\sqrt{15}}}\\\\\)

= \(\\\frac{\sqrt{(8 + 2\sqrt{15})^{2}}}{\sqrt{64\; -\; 60}} = \frac{8 + 2\sqrt{15}}{2} =\boldsymbol{4 + \sqrt{15}}\)

Trigonometric identities are considered to be one of the most important topics in the entire CBSE class 11 syllabus. The basics of trigonometry are included in the class 10 syllabus and some of the advanced concepts are discussed in detail here. This chapter is not only crucial for the class 11 exam but these concepts are also included in class 12 and several questions related to trigonometric identities are included in the competitive exams like JEE.

In this chapter, students are first introduced to the different terminologies like angles, degrees, radians, the relations between degrees and radians, etc. Then, the trigonometric functions like sin, cos, tan, cosec, cot, and sec angles are discussed with their values. Also, the domain and range of the trigonometric functions are explained. After that, Trigonometric Functions of Sum and Difference of Two Angles are discussed which is an integral part of this chapter.

Students are suggested to go through the different examples provided in this chapter and are also required to solve the exercise problems to have an in-depth understanding of the concepts. Students can also refer to these NCERT Solutions For Class 11 Maths Chapter 3 (trigonometric functions) to clear any doubts.

Keep visiting BYJU’S to get the complete NCERT Solutions For Class 11 Maths for all the chapter. Also, get additional assistance for the exams at BYJU’S as several notes, sample papers and question papers are provided here. Students are also suggested to download BYJU’S- The Learning App to learn in a more engaging way.