# NCERT Solutions For Class 11 Maths Chapter 3

## NCERT Solutions Class 11 Maths Trigonometric Functions

For students of class 11 who are looking to do their best their class 11 final exam it is important for them to practice their knowledge on questions of NCERT Solutions for Class 11 Maths Chapter 3. Hence we have provided the downloadable form of NCERT Solutions for Class 11 Maths Chapter 3 pdf for ease of access. NCERT Solutions for Class 11 Maths Chapter 3 Trigonometric Functions handles questions and solutions of topics such as measuring angles in radians and operations on them, definition of trigonometric functions with the help of a circle, signs of trigonometric functions, domain and range of trigonometric functions and more.

### NCERT Solutions Class 11 Maths Chapter 3 Exercises

Exercise 3.1

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

(i).  25$25\;^{\circ}$

(ii).  240$240\;^{\circ}$

(iii).  4730'$-47\;^{\circ} \; 30′$

(iv).  520$520\;^{\circ}$

Sol:

(i).  25$25\;^{\circ}$

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

Therefore, 25$25\;^{\circ}$ = π180×25$\frac{\pi }{180} \times 25$ radian = 5π36$\frac{5\pi }{36}$ radian

Hence, 25$25\;^{\circ}$ = 5π36$\frac{5\pi }{36}$ radian

(ii).  240$240\;^{\circ}$

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

Therefore, 240$240\;^{\circ}$ = π180×240$\frac{\pi }{180} \times 240$ radian = 4π3$\frac{4\pi }{3}$ radian

Hence, 240$240\;^{\circ}$ = 4π3$\frac{4\pi }{3}$ radian

(iii).  4730'$-47\;^{\circ} \; 30′$

= 4712$-47\frac{1}{2}$

= 952$\frac{-95}{2}$ degree

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

Therefore, 952$\frac{-95}{2}$ degree = π180×952$\frac{\pi }{180}\times \frac{-95}{2}$ radian = 1936×2π$\frac{-19 }{36\; \times \;2}\pi$ radian = 1972π$\frac{-19 }{72}\pi$

Hence, 4730'$-47\;^{\circ} \; 30′$ = 1972π$\frac{-19 }{72}\pi$

(iv).  520$520\;^{\circ}$

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

Therefore, 520$520\;^{\circ}$ = π180×520$\frac{\pi }{180} \times 520$ radian = 26π9$\frac{26\pi }{9}$ radian

Hence, 520$520\;^{\circ}$ = 26π9$\frac{26\pi }{9}$ radian

Q.2: Calculate the degree measurement of the given degree measurement: [Use π = 227$\frac{22}{7}$]

(i) 1116$\frac{11}{16}$

(ii) -4

(iii) 5π3$\frac{5\pi }{3}$

(iv) 7π6$\frac{7\pi }{6}$

Sol:

(i).  1116$\frac{11}{16}$:

As we know, that Π Radian = 180°

Therefore, 1116$\frac{11}{16}$ radian = 180π×1116$\frac{180}{\pi }\times \frac{11}{16}$ degree

= 45×11π×4$\frac{45\times 11}{\pi \times 4 }$ degree

= 45×11×722×4$\frac{45 \times 11 \times 7}{22 \times 4 }$ degree

= 3158$\frac{315}{8}$ degree

= 3938$39\;\frac{3}{8}$ degree

= 39+3×608$39\;^{\circ} \;+ \; \frac{3\times 60}{8}$ minute  [1$1^{\circ}$ = 60’]

= 39+22'+12$39^{\circ} \;+ \; 22′ \;+ \;\frac{1}{2}$ minute

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

(ii).  -4:

As we know, that Π Radian = 180°

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

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

= 252011$\frac{-2520}{11}$ degree

= 229111$-229\frac{1}{11}$ degree

= 229+1×6011$-229^{\circ} \; + \; \frac{1\times 60}{11}$ minutes     [1$1^{\circ}$ = 60’]

= 229+5'+511$-229^{\circ} \; +\; 5′ +\;\frac{5}{11}$

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

(iii).  5π3$\frac{5\pi }{3}$

As we know, that Π Radian = 180°

Therefore, 5π3$\frac{5\pi}{3}$ radian = 180π×5π3$\frac{180}{\pi }\times \frac{5\pi}{3}$ degree

= 300$300^{\circ}$

(iv).  7π6$\frac{7\pi }{6}$

As we know, that Π Radian = 180°

Therefore, 7π6$\frac{7\pi}{6}$ radian = 180π×7π6$\frac{180}{\pi }\times \frac{7\pi }{6}$

= 210$210^{\circ}$

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

No. of revolutions made in a minute = 360 revolutions

Therefore, no. of revolutions made in a second = 36060$\frac{360}{60}$ = 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 = 227$\frac{22}{7}$]

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=lr$\theta = \frac{l}{r}$

Given:

r = 100 m and L = 22 m

We have,

θ=22100$\theta = \frac{22}{100}$ radian

= 180π×22100$\frac{180}{\pi }\times \frac{22}{100}$ degree

= 180×7×2222×100$\frac{180 \;\times \;7 \;\times \;22}{22 \; \times 100 }$ degree

= 12610$\frac{126}{10 }$ degree

= 1235$12\frac{3}{5}$ degree

= 1236'$12^{\circ} \; 36′$ [1$1^{\circ}$ = 60’]

Therefore, the req angle is 1236'$12^{\circ} \; 36′$

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

Diameter of circle = 40 m

Radius of circle = 4020$\frac{40}{20}$ m = 20 m

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

In ΔOXY$\Delta OXY$, OX = OY  = radius of the circle = 20 m

XY = 20 m

Therefore,

ΔOXY$\Delta OXY$ is an equilateral triangle.

θ=60$\theta \;= \;60^{\circ}$ = π3$\frac{\pi }{3}$ 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 = lr$\frac{l}{r}$

π3$\frac{\pi }{3}$ = arc(AB)20$\frac{arc\;(AB\;)}{20}$

arc(AB)20$\frac{arc\;(AB\;)}{20}$ = 20π3$\frac{20\pi }{3}$ m

The length of the minor arc of the chord is 20π3$\frac{20\pi }{3}$ m.

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

Sol:

Let, the radii of two circles be r1$r_{1}$ and r2$r_{2}$.

Let, an arc of length l subtend an angle of 60$60^{\circ}$ at the center of the circle of radius r1$r_{1}$, while let an arc of length l subtend an angle of 75$75^{\circ}$ at the center of the circle of radius r2$r_{2}$.

60$\\60^{\circ}$ = π3$\frac{\pi }{3}$ radian

75$75^{\circ}$ = 5π12$\frac{5\pi }{12}$ radian

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

θ = lr$\frac{l}{r}$

l = r θ

l = r1π3$\frac{r_{1}\;\pi }{3}$

l = r25π12$\frac{r_{2}\;5\pi }{12}$

r1π3$\frac{r_{1}\;\pi }{3}$ = r25π12$\frac{r_{2}\;5\pi }{12}$

r1$r_{1}$ = r254$\frac{r_{2}5}{4}$

r1r2$\frac{r_{1}}{r_{2}}$ = 54$\frac{5}{4}$

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 θ$\theta$ radian at the center, then:

θ=lr$\theta = \frac{l}{r}\\$

Here, r = 75 cm

(i).  10 cm:

θ$\theta$ = 1075$\frac{10}{75}$ radian = 215$\frac{2}{15}$ radian

(ii).  15 cm:

θ$\theta$ = 1575$\frac{15}{75}$ radian = 15$\frac{1}{5}$ radian

(iii).  21 cm:

θ$\theta$ = 2175$\frac{21}{75}$ radian = 725$\frac{7}{25}$ radian

Exercise 3.2

Q.1: Calculate the values of five trigonometric func. if cosy$\cos y$ = 12$-\frac{1}{2}$ and y lies in 3rd quadrant.

Sol:

(i)  sec y :

Since, cos y = 12$\frac{1}{2}$

Therefore, sec y = 1cosy$\frac{1}{\cos y}$ = 1(12)$\frac{1}{\left (-\frac{1}{2} \right )}$

Hence, sec y = -2

(ii)  sin y :

Since, sin2y+cos2y=1$\sin ^{2}y \;+ \;\cos ^{2}y \;= \;1$

Therefore, sin2y=1cos2y$\sin ^{2}y \;= \;1 \;- \;\cos ^{2}y$

$\Rightarrow$ sin2y=1(12)2$\sin ^{2}y \;= \;1 \;- \;\left (-\frac{1}{2} \right )^{2}$

$\Rightarrow$ sin2y=114$\sin ^{2}y \;= \;1 \;- \;\frac{1}{4}$

$\Rightarrow$ sin2y=34$\sin ^{2}y \;= \;\frac{3}{4}$

$\Rightarrow$ siny=±32$\sin y \;= \;\pm \frac{\sqrt{3}}{2}$

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

Therefore, sin y = 32$\frac{\sqrt{3}}{2}$

(iii)  cosec y = 1siny$\frac{1}{\sin y}$ = 1(32)$\frac{1}{\left (-\frac{\sqrt{3}}{2} \right )}$ = 23$-\frac{2}{\sqrt{3}}$

Therefore, cosec y = 23$-\frac{2}{\sqrt{3}}$

(iv)  tan y = sinycosy$\frac{\sin y}{\cos y}$ = tany=(32)(12)$\tan y=\frac{\left (-\frac{\sqrt{3}}{2} \right )}{\left (-\frac{1}{2} \right )}$ = 3$\sqrt{3}$

Therefore, tan y = 3$\sqrt{3}$

(v)  cot y = 1tany$\frac{1}{\tan y}$ = 13$\frac{1}{\sqrt{3}}$

Therefore, cot y = 13$\frac{1}{\sqrt{3}}$

Q.2: Calculate the other five trigonometric function if we are given the values for sin y = 35$\frac{3}{5}$, where y lies in second quadrant.

Sol:

sin y = 35$\frac{3}{5}$

Therefore, cosec y = 1Siny$\frac{1}{Sin\; y}$ = 135=53$\frac{1}{\frac{3}{5}} = \frac{5}{3}$

Since, sin2y+cos2y=1$sin^{2}\;y \; + \; cos^{2}\;y = 1$

$\Rightarrow$  cos2y=1sin2y$cos^{2} y = 1 – sin^{2} y$

$\Rightarrow$  cos2y=1(35)2$cos^{2} y = 1 – \left ( \frac{3}{5} \right )^{2}$

$\Rightarrow$ cos2y=1925$cos^{2} y = 1 – \frac{9}{25}$

$\Rightarrow$ cos2y=1625$cos^{2} y = \frac{16}{25}$

$\Rightarrow$ cos y = ±45$\pm \frac{4}{5}$

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

Therefore, cos y = – 45$\frac{4}{5}$

$\Rightarrow$ sec y = 1cosy=1(45)=54$\frac{1}{cos \; y} = \frac{1}{\left (- \frac{4}{5} \right )} =\; – \frac{5}{4}$

$\Rightarrow$ tan y = sinycosy=(35)(45)=34$\frac{sin \; y}{cos \; y} = \frac{\left (\frac{3}{5} \right ) }{\left (-\frac{4}{5} \right ) } = \;- \frac{3}{4}$

$\Rightarrow$ cot y = 1tany=43$\frac{1}{tan \; y} =\;- \frac{4}{3}$

Q.3: Find the values of other five trigonometric functions if coty=34$cot \; y = \frac{3}{4}$, where y lies in the third quadrant.

Sol:

cot y = 34$\frac{3}{4}$

Since, tan y = 1coty=134=43$\frac{1}{cot \; y}=\frac{1}{\frac{3}{4}}=\frac{4}{3}$

Since, 1+tan2y=sec2y$1 + tan^{2}y = sec^{2}y$

$\Rightarrow$  1+(43)2=sec2y$1 + \left ( \frac{4}{3} \right )^{2} = sec^{2}y$

$\Rightarrow$  1+169=sec2y$1 + \frac{16}{9} = sec^{2}y$

$\Rightarrow$  259=sec2y$\frac{25}{9} = sec^{2}y$

$\Rightarrow$  sec y = ±53$\pm \frac{5}{3}$

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

Therefore, sec y = 53$\;- \frac{5}{3}$

cos y = 1secy=1(53)=35$\frac{1}{sec \; y} = \frac{1}{\left (- \frac{5}{3} \right )} = – \frac{3}{5}$

Since, tan y = sinycosy$\frac{sin \; y}{cos \; y}$

$\Rightarrow$  43=siny(35)$\frac{4}{3} = \frac{sin \; y}{\left (- \frac{3}{5} \right )}$

$\Rightarrow$ sin y = (43)×(35)=45$\left (\frac{4}{3} \right ) \times \left (- \frac{3}{5} \right ) = -\frac{4}{5}$

$\Rightarrow$  cosec y = 1siny=54$\frac{1}{sin \; y} = -\frac{5}{4}$

Q.4: Find the values of other five trigonometric if secy=135$sec \; y = \frac{13}{5}$, where y lies in the fourth quadrant.

Sol:

sec y = 135$\frac{13}{5}$

cos y = 1secy=1(135)=513$\frac{1}{sec \; y} = \frac{1}{\left (\frac{13}{5}\right )} = \frac{5}{13}$

Since, sin2y+cos2y=1$sin^{2}y + cos^{2}y = 1$

$\Rightarrow$   sin2y=1cos2y$sin^{2}y = 1 – cos^{2}y$

$\Rightarrow$   sin2y=1(513)2$sin^{2}y = 1 – \left ( \frac{5}{13} \right )^{2}$

$\Rightarrow$   sin2y=125169=144169$sin^{2}y = 1 – \frac{25}{169} = \frac{144}{169}$

$\Rightarrow$ sin y = ±1213$\pm \frac{12}{13}$

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

Therefore, sin y = 1213$– \frac{12}{13}$

$\Rightarrow$ cosec y = 1siny=1(1213)=1312$\frac{1}{sin \; y} = \frac{1}{\left (- \frac{12}{13} \right )} = \;- \frac{13}{12}$

$\Rightarrow$ tan y = sinycosy=(1213)(513)=125$\frac{sin \; y}{cos \; y} = \frac{\left (- \frac{12}{13} \right )}{\left (\frac{5}{13} \right )} =\; -\frac{12}{5}$

$\Rightarrow$ cot y = 1tany=1(125)=512$\frac{1}{tan \; y} = \frac{1}{\left ( -\frac{12}{5} \right )} = \;-\frac{5}{12}$

Q.5: Find the values of other five trigonometric function if tan y = 512$– \frac{5}{12}$ and y lies in second quadrant.

Sol:

tan  y = 512$-\frac{5}{12}$   [Given]

And, cot y = 1tany=1(512)=125$\frac{1}{tan \; y} = \frac{1}{ \left (- \frac{5}{12} \right )} = – \frac{12}{5}$

Since, 1+tan2y=sec2y$1 + tan^{2} y = sec^{2} y$

Therefore, 1+(512)2=sec2y$1 + \left ( – \frac{5}{12} \right )^{2} = sec^{2} y$

$\Rightarrow$   sec2y=1+(25144)$sec^{2} y = 1 + \left ( \frac{25}{144} \right )$

$\Rightarrow$  sec2y=169144$sec^{2} y = \frac{169}{144}$

$\Rightarrow$ sec y = ±1312$\pm \frac{13}{12}$

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

Therefore, sec  y = 1312$– \frac{13}{12}$

cos y = 1secy=1(1312)=(1213)$\frac{1}{sec \; y } = \frac{1}{\left (- \frac{13}{12} \right )} = \left (- \frac{12}{13} \right )\\$

Since, tany=sinycosy$tan \; y = \frac{sin \; y}{cos \; y}$

$\Rightarrow$  512=siny(1213)$-\frac{5}{12} = \frac{sin \; y}{\left (- \frac{12}{13} \right )}\\$

$\Rightarrow$ sin y = (512)×(1213)=513$\left (-\frac{5}{12} \right ) \times \left (- \frac{12}{13}\right ) = \frac{5}{13}\\$

$\Rightarrow$ cosec y = 1siny=1(513)=135$\frac{1}{sin \; y} = \frac{1}{\left ( \frac{5}{13} \right )} = \frac{13}{5}$

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° = 12$\frac{1}{\sqrt{2}}$

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 tan19π3$\tan \frac{19\pi }{3}$.

Sol:

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

Therefore, tan19π3$\tan \frac{19\pi }{3}$ = tan613π$\tan 6\frac{1}{3}\pi$ = tan(6π+π3)$\tan \left ( 6\pi + \frac{\pi }{3} \right )$ = tanπ3$\tan \frac{\pi }{3}$ = tan 60° = 3$\sqrt{3}$

Q.9: Calculate the value of the trigonometric function sin(11π3)$\sin \left ( -\frac{11\pi }{3} \right )$.

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

Therefore, sin(11π3)$\sin \left ( -\frac{11\pi }{3} \right )$ = sin(11π3+2×2π)$\sin \left ( -\frac{11\pi }{3} + 2 \times 2\pi \right )$ = sinπ3$\sin \frac{\pi }{3}$ = 32$\frac{\sqrt{3}}{2}$

Q.10: Calculate the value of the trigonometric function cot(15π4)$\cot \left ( -\frac{15\pi }{4} \right )$.

Sol:

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

Therefore, cot(15π4)$\cot \left ( -\frac{15\pi }{4} \right )$ = cot(15π4+4π)$\cot \left ( -\frac{15\pi }{4} + 4\pi \right )$ = cotπ4$\cot \frac{\pi }{4}$ = 1

Exercise 3.3

Q.1: Prove:

sin2π6+cos2π3tan2π4=12$\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4} = -\frac{1}{2}$

Sol:

Now, taking L.H.S.

sin2π6+cos2π3tan2π4$\sin ^{2} \frac{\pi }{6} + \cos ^{2} \frac{\pi }{3} – \tan ^{2} \frac{\pi }{4}$:

= (12)2+(12)2(1)2$\\\left (\frac{1}{2} \right )^{2} + \left (\frac{1}{2} \right )^{2} – \left ( 1 \right )^{2}$

= 14+141$\\\frac{1}{4} + \frac{1}{4} – 1$ = 12$-\frac{1}{2}$

= R.H.S.

Q.2: Prove:

2sin2π6+cosec27π6cos2π3=32$2 \sin ^{2} \frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} \cos ^{2}\frac{\pi }{3} = \frac{3}{2}$

Sol:

Now, taking L.H.S.

2sin2π6+cosec27π6cos2π3$2 \sin ^{2} \frac{\pi }{6} + cosec^{2}\frac{7\pi }{6} \cos ^{2}\frac{\pi }{3}\\$ :

= 2(12)2+cosec2(π+π6)(12)2$\\2 \left (\frac{1}{2} \right )^{2} + cosec ^{2}\left ( \pi + \frac{\pi }{6} \right )\left ( \frac{1}{2} \right )^{2}$

= 2×14+(cosecπ6)2(14)$2 \times \frac{1}{4} + \left (-cosec \frac{\pi }{6} \right )^{2}\left ( \frac{1}{4} \right )$

= 12+(2)2(14)$\frac{1}{2} + \left ( -2 \right )^{2}\left ( \frac{1}{4} \right )$

= 12+44$\frac{1}{2} + \frac{4}{4}$

= 12+1$\frac{1}{2} + 1$

= 32$\frac{3}{2}$

= R.H.S.

Q.3: Prove:

cot2π6+cosec5π6+3tan2π6=6$\cot ^{2}\frac{\pi }{6} + cosec \frac{5\pi }{6} + 3\tan ^{2} \frac{\pi }{6} = 6$

Sol:

Taking L.H.S.

cot2π6+cosec5π6+3tan2π6$\cot ^{2}\frac{\pi }{6} + cosec \frac{5\pi }{6} + 3\tan ^{2} \frac{\pi }{6}\\$ :

= (3)2+cosec(ππ6)+3(13)2$\left (\sqrt{3} \right )^{2} + cosec \left ( \pi – \frac{\pi }{6} \right ) + 3\left ( \frac{1}{\sqrt{3}} \right )^{2}$

= 3+cosecπ6+3×13$3 + cosec \frac{\pi }{6} + 3\times \frac{1}{3}$

= 3 + 2 + 1 = 6

= R.H.S.

Q.4: Prove:

2sin23π4+2cos2π4+2sec2π3=10$2\sin ^{2} \frac{3\pi }{4} + 2\cos ^{2} \frac{\pi }{4} + 2\sec ^{2} \frac{\pi }{3} = 10$

Sol:

Now, taking L.H.S.

2sin23π4+2cos2π4+2sec2π3$2\sin ^{2} \frac{3\pi }{4} + 2\cos ^{2} \frac{\pi }{4} + 2\sec ^{2} \frac{\pi }{3}\\$ :

= 2{sin(ππ4)}2+2(12)2+2(2)2$\\2\left \{ \sin \left ( \pi – \frac{\pi }{4} \right ) \right \}^{2} + 2\left (\frac{1}{\sqrt{2}} \right )^{2} + 2\left ( 2 \right )^{2}$

= 2{sinπ4}2+2×12+8$2\left \{ \sin \frac{\pi }{4} \right \}^{2} + 2\times \frac{1}{2} + 8$

= 2(12)2$2\left ( \frac{1}{\sqrt{2}} \right )^{2}$ + 1 + 8

= 1 + 1 + 8 = 10

= R.H.S.

Q.5: Calculate the value of:

(i).  sin75$\sin 75^{\circ}$

(ii).  tan15$\tan 15^{\circ}$

Sol:

(i).  sin75$\sin 75^{\circ}$:

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

= sin45cos30+cos45sin30$\sin 45^{\circ} \cos 30^{\circ} + \cos 45^{\circ} \sin 30^{\circ}$

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

= (12)(32)+(12)(12)$\left (\frac{1}{\sqrt{2}} \right )\left (\frac{\sqrt{3}}{2} \right ) + \left (\frac{1}{\sqrt{2}} \right )\left ( \frac{1}{2} \right )$

= 322+122$\frac{\sqrt{3}}{2\sqrt{2}} + \frac{1}{2\sqrt{2}}$

= 3+122$\frac{\sqrt{3} + 1}{2\sqrt{2}}$

(ii).  tan15$\tan 15^{\circ}$:

= tan(4530)$\tan \left ( 45^{\circ} – 30^{\circ}\right )$

= tan45tan301+tan45tan30$\frac{\tan 45^{\circ} – \tan 30^{\circ}}{1 + \tan 45^{\circ}\tan 30^{\circ}}$

Since, [tan (x – y) = tanxtany1+tanxtany$\frac{\tan x – \tan y}{1 + \tan x \tan y}$]

= 1131+1(13)$\\\frac{1 – \frac{1}{\sqrt{3}}}{1 + 1\left ( \frac{1}{\sqrt{3}} \right )}$

= 3133+13$\\\frac{\frac{\sqrt{3} – 1}{\sqrt{3}}}{\frac{\sqrt{3 + 1}}{\sqrt{3}}}$

= 313+1$\\\frac{\sqrt{3} – 1}{\sqrt{3} + 1}$

= (31)2(3+1)(31)$\\\frac{\left ( \sqrt{3} – 1\right )^{2}}{\left ( \sqrt{3} + 1\right )\left ( \sqrt{3} – 1 \right )}$

= 3+123(3)2(1)2$\\\frac{3 + 1 – 2\sqrt{3}}{\left ( \sqrt{3} \right )^{2} – \left ( 1 \right )^{2}}$

= 42331$\frac{4 – 2\sqrt{3}}{3 – 1}$

= 23$2 – \sqrt{3}$

Q.6:Prove:

cos(π4x)cos(π4y)sin(π4x)sin(π4y)=sin(x+y)$\cos \left ( \frac{\pi }{4} – x\right ) \cos \left ( \frac{\pi }{4} – y\right ) – \sin \left ( \frac{\pi }{4} – x\right ) \sin \left ( \frac{\pi }{4} – y\right ) = \sin \left ( x+y \right )$

Sol:

Now, taking L.H.S.

cos(π4x)cos(π4y)sin(π4x)sin(π4y):$\cos \left ( \frac{\pi }{4} – x\right ) \cos \left ( \frac{\pi }{4} – y\right ) – \sin \left ( \frac{\pi }{4} – x\right ) \sin \left ( \frac{\pi }{4} – y\right ):\\$

= 12[2cos(π4x)cos(π4y)]+12[2sin(π4x)sin(π4y)]$\\\frac{1}{2}\left [ 2\cos \left ( \frac{\pi }{4} – x \right ) \cos \left ( \frac{\pi }{4} – y \right ) \right ] + \frac{1}{2}\left [ -2\sin \left ( \frac{\pi }{4} – x \right ) \sin \left ( \frac{\pi }{4} – y \right ) \right ]\\$

=12[cos{(π4x)+(π4y)}+cos{(π4x)(π4y)}]+12[cos{(π4x)+(π4y)}cos{(π4x)(π4y)}]$=\frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right )\right \} + \cos \left \{ \left ( \frac{\pi }{4} – x \right ) – \left ( \frac{\pi }{4} – y \right )\right \}\right ] + \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right )\right \} – \cos \left \{ \left ( \frac{\pi }{4} – x \right ) – \left ( \frac{\pi }{4} – y \right )\right \}\right ]\\$

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

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

= 2×12[cos{(π4x)+(π4y)}]$\\2\times \frac{1}{2}\left [ \cos \left \{ \left ( \frac{\pi }{4} – x \right ) + \left ( \frac{\pi }{4} – y \right ) \right \} \right ]\\$

= cos[π2(x+y)]$\\\cos \left [ \frac{\pi }{2} – \left ( x + y \right )\right ]$

= sin (x + y)

= R.H.S.

Q.7: Prove:

tan(π4+x)tan(π4x)=(1+tanx