NCERT Solutions Class 11 Maths Chapter 3

Q.1 Find the radian measures corresponding to the following degree measures:
(i) 25°
(ii) – 47°30′
(iii) 240°
(iv) 520°

Ans.

$\begin{array}{l}\text{Since,\hspace{0.33em}1°}=\frac{\text{π}}{180\text{°}}\\ \left(\text{i}\right)25\text{°}=25\text{°}\times \frac{\text{π}}{180\text{°}}\\ =\frac{5\text{π}}{36\text{°}}\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}radians}\\ \left(\text{ii}\right)-47\text{°}30‘=-47.5\text{°\hspace{0.33em}}\left[\because 30‘=0.5\text{°}\right]\\ =-47.5\text{°}\times \frac{\text{π}}{180\text{°}}\text{\hspace{0.33em}\hspace{0.33em}radians}\\ =-\frac{19\text{π}}{72}\text{\hspace{0.33em}\hspace{0.33em}radians}\\ \left(\text{iii}\right)240\text{°}=240\text{°}\times \frac{\text{π}}{180\text{°}}\text{\hspace{0.33em}\hspace{0.33em}radians}\\ =\frac{4\text{π}}{3}\text{\hspace{0.33em}\hspace{0.33em}radians}\\ \left(\text{iv}\right)520\text{°}=520\text{°}\times \frac{\text{π}}{180\text{°}}\text{\hspace{0.33em}\hspace{0.33em}radians}\\ =\frac{26\text{π}}{9}\text{\hspace{0.33em}\hspace{0.33em}radians}\end{array}$

Q.2

$\begin{array}{l}\text{Find\hspace{0.33em}the\hspace{0.33em}degree\hspace{0.33em}measures\hspace{0.33em}corresponding\hspace{0.33em}to\hspace{0.33em}the\hspace{0.33em}following}\\ \text{radian\hspace{0.33em}measures\hspace{0.33em}}\left(\text{Use\hspace{0.33em}π}=\frac{22}{7}\right).\\ \left(\text{i}\right)\frac{11}{16}\\ \left(\text{ii}\right)-4\\ \left(\text{iii}\right)\frac{5\text{π}}{3}\\ \left(\text{iv}\right)\frac{7\text{π}}{6}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{1 radian}=\frac{180\mathrm{°}}{\mathrm{\pi }}\\ \left(\mathrm{i}\right)\frac{11}{16}\text{\hspace{0.17em}\hspace{0.17em}radians}=\frac{11}{16}\times \frac{180\mathrm{°}}{\mathrm{\pi }}\\ =\frac{11}{4}\times \frac{45\mathrm{°}}{\mathrm{\pi }}\\ =\frac{11}{4}\times \frac{45\mathrm{°}\times 7}{22}\\ =\frac{1}{4}\times \frac{45\mathrm{°}\times 7}{2}\\ =\left(39\frac{3}{8}\right)°\\ =39\mathrm{°}+\frac{3}{8}\times 60‘\\ =39\mathrm{°}+22‘+\frac{1}{2}‘\\ =39\mathrm{°}+22‘+\frac{1}{2}\times 60‘‘\\ =39\mathrm{°}22‘30‘‘\end{array}$ $\begin{array}{l}\left(\mathrm{ii}\right)-4\text{\hspace{0.17em}\hspace{0.17em}radians}=-4\times \frac{180\mathrm{°}}{\mathrm{\pi }}\\ =-4\times \frac{180\mathrm{°}\times 7}{22}\\ =-\left(229\frac{1}{11}\right)°\\ =-(229\mathrm{°}+\frac{1}{11}\times 60‘)\\ =-(229\mathrm{°}+5‘+\frac{5}{11}\times 60‘‘)\\ =-(229\mathrm{°}+5‘+27‘‘)\\ =-229\mathrm{°}5‘27‘‘(\mathrm{Approx}.)\\ \left(\mathrm{iii}\right)\frac{5\mathrm{\pi }}{3}\text{\hspace{0.17em}\hspace{0.17em}radians}=\frac{5\mathrm{\pi }}{3}\times \frac{180\mathrm{°}}{\mathrm{\pi }}\\ =\frac{5}{3}\times 180\mathrm{°}\\ =300\mathrm{°}\\ \left(\mathrm{iv}\right)\frac{7\mathrm{\pi }}{6}\text{\hspace{0.17em}\hspace{0.17em}radians}=\frac{7\mathrm{\pi }}{6}\times \frac{180\mathrm{°}}{\mathrm{\pi }}\\ =\frac{7}{6}\times 180\mathrm{°}\\ =210\mathrm{°}\end{array}$

Q.3 A wheel makes 360 revolutions in one minute. Through how many radians does it turn in one second?

Ans.

$\begin{array}{l}\mathrm{Number}\text{of revolutions made by a wheel in 1 minute}=360\\ \mathrm{Number}\text{of revolutions made by a wheel in 1 second}=\frac{360}{60}\\ =6\\ \mathrm{Angle}\text{covered in 1 revolution by wheel}=2\text{π\hspace{0.33em}}\mathrm{Radians}\\ \mathrm{Angle}\text{covered in 6 revolutions by wheel}=6\times 2\text{π}\\ =12\text{π\hspace{0.33em}}\mathrm{Radians}\\ \mathrm{Thus},\text{wheel turns 12π radians in 1 second.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{degree}\mathrm{measure}\mathrm{of}\mathrm{the}\mathrm{angle}\mathrm{subtended}\mathrm{at}\mathrm{the}\\ \mathrm{centre}\mathrm{of}\mathrm{a}\mathrm{circle}\mathrm{of}\mathrm{radius}100\mathrm{cm}\mathrm{by}\mathrm{an}\mathrm{arc}\mathrm{of}\mathrm{length}22\mathrm{cm}\\ (\mathrm{Use}\mathrm{\pi }=\frac{22}{7}).\end{array}$

Ans.

$\begin{array}{l}\mathrm{Radius}\text{of circle}=100\text{cm}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length of arc}=22\mathrm{cm}\\ \mathrm{Angle}\left(\mathrm{\theta }\right)\text{subtended at the centre by an arc}\\ =\frac{\mathrm{Arc}}{\mathrm{Radius}}\mathrm{radians}\\ \mathrm{\theta }=\frac{22}{100}\mathrm{radians}\\ \mathrm{\theta }=\frac{22}{100}\times \frac{180\mathrm{°}}{\mathrm{\pi }}[\because 1\mathrm{radian}=\frac{180\mathrm{°}}{\mathrm{\pi }}]\\ \mathrm{\theta }=\frac{22}{100}\times \frac{180\mathrm{°}\times 7}{22}\\ =1.8\mathrm{°}\times 7\\ =12.6\mathrm{°}\\ =12\mathrm{°}+0.6\times 60‘\end{array}$ $\begin{array}{l}=12\mathrm{°}+36‘\\ =12\mathrm{°}36‘\\ \text{Thus, angle subtended by an arc of 22 cm at the centre is 12°36′ .}\end{array}$

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

Ans.

$\begin{array}{l}\mathrm{Diameter}\text{of a circle}=40\mathrm{cm}\\ \mathrm{Radius}\text{ of a circle}=\frac{40}{2}\mathrm{cm}\\ =20\mathrm{cm}\\ \mathrm{Length}\text{of chord}=20\mathrm{cm}\end{array}$

$\begin{array}{l}\mathrm{Since},\text{}\mathrm{\Delta }\text{OAB is equilateral triangle.}\\ \mathrm{So},\text{ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\angle \text{AOB}=60\mathrm{°}\\ =\frac{\mathrm{\pi }}{3}\mathrm{radians}\\ \mathrm{Since},\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}angle}=\frac{\mathrm{Length}\text{of arc}}{\mathrm{Radius}}\\ \frac{\mathrm{\pi }}{3}\mathrm{Radian}=\frac{\mathrm{Length}\text{of arc AB}}{20\mathrm{cm}}\end{array}$ \begin{array}{l}\end{array} $\begin{array}{l}\mathrm{Length}\text{of arc AB}=20\mathrm{cm}\times \frac{\mathrm{\pi }}{3}\\ =\frac{20\mathrm{\pi }}{3}\mathrm{cm}\end{array}$

Q.6 If in two circles, arcs of the same length subtend angles 60° and 75° at the centre, find the ratio of their radii.

Ans.

$\begin{array}{l}\mathrm{L}{\text{et radii of circles be r}}_{\text{1}}{\text{and r}}_{\text{2}}\text{respectively.}\\ \text{Angle subscribed by arc in first circle}=60\mathrm{°}\\ =60\mathrm{°}\times \frac{\mathrm{\pi }}{180\mathrm{°}}\mathrm{radians}\\ =\frac{\mathrm{\pi }}{3}\mathrm{radians}\\ \text{Angle subscribed by arc in second circle}\\ =75\mathrm{°}\\ =75\mathrm{°}\times \frac{\mathrm{\pi }}{180\mathrm{°}}\mathrm{radians}\\ =\frac{5\mathrm{\pi }}{12}\mathrm{radians}\\ \mathrm{Since},\text{length of both arcs is same.}\\ \text{So,\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}Length of arc of first circle}=\text{Length of arc of}\\ \text{second circle}\\ \frac{\mathrm{\pi }}{3}\mathrm{radian}\times {\mathrm{r}}_1=\frac{5\mathrm{\pi }}{12}\mathrm{radian}\times {\mathrm{r}}_2\end{array}$ $\begin{array}{l}\Rightarrow \frac{{\mathrm{r}}_1}{{\mathrm{r}}_2}=\frac{\left(\frac{5\mathrm{\pi }}{12}\right)}{\left(\frac{\mathrm{\pi }}{3}\right)}\\ =\frac{5}{4}\\ \Rightarrow {\mathrm{r}}_1:{\mathrm{r}}_2=5:4\\ \mathrm{Thus},\text{the ratio of radii of circles is 5:4.}\end{array}$

Q.7 Find the angle in radian through which a pendulum swings if its length is 75 cm and the tip describes an arc of length
(i) 10 cm
(ii) 15 cm
(iii) 21 cm

Ans.

$\begin{array}{l}\mathrm{Length}\text{of pendulum}=\text{75 cm}\\ \left(\text{i}\right)\mathrm{Length}\text{of arc described by pendulum}=10\mathrm{cm}\\ \mathrm{Angle}\text{swept by pendulum}=\frac{10}{75}\mathrm{Rad}.\left(\because \text{\hspace{0.33em}θ}=\frac{\mathrm{arc}}{\text{r}}\right)\\ =\frac{2}{15}\mathrm{radians}\\ \left(\mathrm{ii}\right)\mathrm{Length}\text{of arc described by pendulum}=15\mathrm{cm}\\ \mathrm{Angle}\text{swept by pendulum}=\frac{15}{75}\mathrm{Rad}.\left(\because \text{\hspace{0.33em}θ}=\frac{\mathrm{arc}}{\text{r}}\right)\\ =\frac{1}{5}\mathrm{radians}\\ \left(\mathrm{iii}\right)\mathrm{Length}\text{of arc described by pendulum}\\ =21\mathrm{cm}\\ \mathrm{Angle}\text{swept by pendulum}=\frac{21}{75}\mathrm{Rad}.\left(\because \text{\hspace{0.33em}θ}=\frac{\mathrm{arc}}{\text{r}}\right)\\ =\frac{7}{25}\mathrm{radians}\end{array}$

Q.8 Find the values of other five trigonometric functions in Exercises 1 to 5.

$1.\cos \text{x}=-\frac{1}{2},\text{x}\mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}.$ $2.\sin \text{x}=\frac{3}{5},\text{x}\mathrm{lies}\mathrm{in}\mathrm{second}\mathrm{quadrant}.$ $3.\cot \text{x}=\frac{3}{4},\text{x}\mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}.$ $4.\sec \text{x}=\frac{13}{5},\text{x}\mathrm{lies}\mathrm{in}\mathrm{fourth}\mathrm{quadrant}.$ $5.\tan \text{x}=-\frac{5}{12},\text{x}\mathrm{lies}\mathrm{in}\mathrm{second}\mathrm{quadrant}.$

Ans.

$\begin{array}{l}1.\\ \mathrm{secx}=\frac{1}{\mathrm{cosx}}\\ =\frac{1}{-\frac{1}{2}}\\ =-2\left(\mathrm{x}\mathrm{lies}\mathrm{in}\mathrm{third}\mathrm{quadrant}.\right)\\ {\sin }^2\text{x}+{\cos }^2\text{x}=1\\ \mathrm{sinx}=\sqrt{1-{\cos }^2\text{x}}\\ =\sqrt{1-{\left(-\frac{1}{2}\right)}^2}\\ \mathrm{sinx}=\sqrt{\frac{3}{4}}\\ \mathrm{sinx}=-\frac{\sqrt{3}}{2}\left(\because \text{\hspace{0.33em}\hspace{0.33em}x\hspace{0.33em}}\mathrm{lies}\text{in third quadrant.}\right)\\ \mathrm{cosecx}=\frac{1}{\mathrm{sinx}}\\ =\frac{1}{\left(-\frac{\sqrt{3}}{2}\right)}\\ =-\frac{2}{\sqrt{3}}\\ \mathrm{tanx}=\frac{\mathrm{sinx}}{\mathrm{cosx}}\\ =\frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}}\\ =\sqrt{3}\left(\because \text{x\hspace{0.33em}}\mathrm{lies}\text{in third quadrant.}\right)\\ \mathrm{cotx}=\frac{1}{\mathrm{tanx}}\\ =\frac{1}{\sqrt{3}}\left(\because \text{x\hspace{0.33em}}\mathrm{lies}\text{in third quadrant.}\right)\\ \mathrm{Thus},\text{sinx}=-\frac{\sqrt{3}}{2},\mathrm{cosecx}=-\frac{2}{\sqrt{3}},\mathrm{secx}=-2\text{,tanx}=\sqrt{3}\mathrm{and}\mathrm{cotx}=\frac{1}{\sqrt{3}}.\end{array}$ $\begin{array}{l}\text{2.}\mathrm{sinx}=3/5\\ \mathrm{cosecx}=1/\mathrm{sinx}\\ =\frac{1}{\left(\frac{3}{5}\right)}\\ \mathrm{cosec}\mathrm{x}=\frac{5}{3}\\ {\sin }^2\text{x}+{\cos }^2\text{x}=1\\ \cos \mathrm{x}=\pm \sqrt{1-{\sin }^2\text{x}}\\ =\pm \sqrt{1-{\left(\frac{3}{5}\right)}^2}\\ \cos \mathrm{x}=\pm \sqrt{\frac{16}{25}}\\ =-\frac{4}{5}\left[\because \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x lies in II quadrant.}\right]\\ \sec \mathrm{x}=\frac{1}{\cos \mathrm{x}}\\ =\frac{1}{-\left(\frac{4}{5}\right)}\\ =-\frac{5}{4}\\ \tan \mathrm{x}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}}\\ \tan \mathrm{x}=\frac{\left(\frac{3}{5}\right)}{-\left(\frac{4}{5}\right)}\\ =-\frac{3}{4}\left[\because \text{\hspace{0.33em}\hspace{0.33em}x lies in II quadrant.}\right]\\ \cot \mathrm{x}=\frac{1}{\tan \mathrm{x}}\\ =\frac{1}{\left(-\frac{3}{4}\right)}\\ =-\frac{4}{3}\left[\because \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}x lies in II quadrant.}\right]\\ \mathrm{Thus},\text{cos\hspace{0.33em}x}=-\frac{4}{5},\sec \mathrm{x}=-\frac{5}{4}\text{\hspace{0.33em},}\mathrm{cosec}\mathrm{x}=\frac{5}{3},\text{tan\hspace{0.33em}x}=-\frac{3}{4}\\ \mathrm{and}\cot \mathrm{x}=-\frac{4}{3}.\end{array}$ $\begin{array}{l}3.\\ \tan \mathrm{x}=\frac{1}{\cot \mathrm{x}}\\ =\frac{1}{\left(\frac{3}{4}\right)}\\ =\frac{4}{3}\\ {\mathrm{cosec}}^2\text{x}=1+{\cot }^2\text{x}\\ =1+{\left(\frac{3}{4}\right)}^2\\ =1+\frac{9}{16}\\ \mathrm{cosec}\mathrm{x}=\pm \sqrt{\frac{25}{16}}\\ =-\frac{5}{4}\left[\text{x lies in third quadrant.}\right]\\ \sin \mathrm{x}=\frac{1}{\mathrm{cosec}\mathrm{x}}\\ =\frac{1}{-\frac{5}{4}}\\ =-\frac{4}{5}\\ \sec \mathrm{x}=1+{\tan }^2\text{x}\\ =1+{\left(\frac{4}{3}\right)}^2\\ =1+\frac{16}{9}\\ \sec \mathrm{x}=\pm \sqrt{\frac{25}{9}}\\ =-\frac{5}{3}\left[\text{x lies in third quadrant.}\right]\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}cos x}=-\frac{3}{5}\left[\text{x lies in third quadrant.}\right]\\ \mathrm{Thus},\sin \mathrm{x}=-\frac{4}{5},\mathrm{cosec}\mathrm{x}=-\frac{5}{4},\cos \mathrm{x}=-\frac{3}{5},\sec \mathrm{x}=-\frac{5}{3}\mathrm{and}\tan \mathrm{x}=\frac{4}{3}.\end{array}$ $\begin{array}{l}4.\\ \cos \text{x}=\frac{1}{\sec \text{x}}\\ =\frac{1}{\left(\frac{13}{5}\right)}\\ =\frac{5}{13}\left[\because \text{x lies in fourth quadrant.}\right]\\ \sin \text{x}=\sqrt{1-{\cos }^2\text{x}}\\ =\pm \sqrt{1-{\left(\frac{5}{13}\right)}^2}\\ =-\frac{12}{13}\left[\because \text{x lies in fourth quadrant.}\right]\\ \mathrm{cosec}\text{x}=\frac{1}{\sin \text{x}}\\ =\frac{1}{\left(-\frac{12}{13}\right)}\\ =-\frac{13}{12}\\ \tan \text{x}=\frac{\sin \text{x}}{\cos \text{x}}\end{array}$ $\begin{array}{l}=\frac{-\left(\frac{12}{13}\right)}{\left(\frac{5}{13}\right)}\\ =-\frac{12}{5}[\because \mathrm{x}\text{lies in fourth quadrant.}]\\ \cot \mathrm{x}=\frac{1}{\mathrm{tanx}}\\ =\frac{1}{-\left(\frac{12}{5}\right)}\\ =-\frac{5}{12}\\ \mathrm{Thus},\sin \mathrm{x}=-\frac{12}{13},\cos \mathrm{x}=\frac{5}{13},\tan \mathrm{x}=-\frac{12}{5},\cot \mathrm{x}=-\frac{5}{12}\\ \mathrm{and}\mathrm{cosec}\mathrm{x}=-\frac{13}{12}.\end{array}$ $\begin{array}{l}5.\\ \cot \text{\hspace{0.33em}x}=\frac{1}{\left(-\frac{5}{12}\right)}\left[\because \cot \text{x}=\frac{1}{\tan \text{\hspace{0.33em}x}}\right]\\ =-\frac{12}{5}\\ {\sec }^2\text{x}=1+{\tan }^2\text{x}\\ =1+{\left(-\frac{5}{12}\right)}^2\\ =1+\frac{25}{144}\\ \sec \mathrm{x}=\pm \sqrt{\frac{169}{144}}\\ =-\frac{13}{12}\left[\because \text{\hspace{0.33em}x lies in second quadrant.}\right]\\ \cos \mathrm{x}=\frac{1}{\mathrm{secx}}\\ =\frac{1}{-\left(\frac{13}{12}\right)}\\ =-\frac{12}{13}\left[\because \text{\hspace{0.33em}x lies in second quadrant.}\right]\\ {\mathrm{cosec}}^2\text{x}=1+{\cot }^2\text{x}\\ =1+{\left(-\frac{12}{5}\right)}^2\\ =1+\frac{144}{25}\\ \mathrm{cosec}\mathrm{x}=\pm \sqrt{\frac{169}{25}}\\ =\frac{13}{5}\left[\because \text{\hspace{0.33em}x lies in second quadrant.}\right]\\ \sin \mathrm{x}=\frac{1}{\mathrm{cosec}\mathrm{x}}\\ =\frac{1}{\left(\frac{13}{5}\right)}\\ \sin \mathrm{x}=\frac{5}{13}\left[\because \text{\hspace{0.33em}x lies in second quadrant.}\right]\\ \mathrm{Thus},\sin \mathrm{x}=\frac{5}{13},\cos \mathrm{x}=-\frac{12}{13},\cot \mathrm{x}=-\frac{12}{5},\sec \mathrm{x}=-\frac{13}{12}\\ \mathrm{and}\mathrm{cosec}\mathrm{x}=\frac{13}{5}.\end{array}$

Q.9 Find the values of the trigonometric functions in Exercises 6 to 10.

6. sin 765°
7. cosec (– 1410°)

$8.\tan \frac{19\mathrm{\pi }}{3}$

$9.\sin (-\frac{11\mathrm{\pi }}{3})$

$10.\cot (-\frac{15\mathrm{\pi }}{4})$

Ans.

$\begin{array}{l}6.\\ \text{sin}\left(\text{765°}\right)=\text{sin}\left(\text{2 x 36}0\text{°}+\text{45°}\right)\\ =\text{sin 45°}\left[\because \sin \left(\text{n}\times 360\text{°}+\text{θ}\right)=\text{sinθ}\right]\\ =\frac{1}{\sqrt{2}}\end{array}$ $\begin{array}{l}7.\\ \text{}\begin{array}{l}\text{cosec}\left(–\text{141}0\text{°}\right)=-\mathrm{cosec}\text{141}0\text{°\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\left[\mathrm{cosec}\left(-\text{θ}\right)=-\mathrm{cosec\theta }\right]\\ =-\mathrm{cosec}\left(4\times 360\text{°}-30\text{°}\right)\\ =-\left(-\mathrm{cosec}30\text{°}\right)\left[\mathrm{cosec}\left(\text{n}\times 360\text{°}-\text{θ}\right)=-\mathrm{cosec\theta }\right]\\ =2\end{array}\end{array}$ $\begin{array}{l}8.\\ \tan \frac{19\text{π}}{3}=\tan \left(6\text{π}+\frac{\text{π}}{3}\right)\end{array}$ $\begin{array}{l}\tan \frac{19\mathrm{\pi }}{3}=\tan \frac{\mathrm{\pi }}{3}[\because \tan (2\mathrm{n\pi }+\mathrm{\theta })=\mathrm{tan\theta }]\\ =\sqrt{3}\end{array}$ $\begin{array}{l}9.\\ \text{}\begin{array}{l}\sin \left(-\frac{11\text{π}}{3}\right)=-\sin \left(\frac{11\text{π}}{3}\right)\left[\because \sin \left(-\text{x}\right)=-\sin \text{x}\right]\\ =-\sin \left(4\text{π}-\frac{\text{π}}{3}\right)\\ =-\left(-\sin \frac{\text{π}}{3}\right)\left[\because \sin \left(2\mathrm{n\pi }-\text{x}\right)=-\sin \text{x}\right]\\ \sin \left(-\frac{11\text{π}}{3}\right)=\frac{\sqrt{3}}{2}\end{array}\end{array}$ $\begin{array}{l}10.\\ \text{}\begin{array}{l}\cot \left(-\frac{15\text{π}}{4}\right)=-\cot \left(\frac{15\text{π}}{4}\right)\left[\because \cot \left(-\text{x}\right)=-\cot \text{x}\right]\\ =-\cot \left(4\text{π}-\frac{\text{π}}{4}\right)\\ =-\left(-\cot \frac{\text{π}}{4}\right)\left[\because \cot \left(2\mathrm{n\pi }-\text{x}\right)=-\cot \text{x}\right]\\ =1\end{array}\end{array}$

Q.10

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\sin }^2\frac{\mathrm{\pi }}{6}+{\cos }^2\frac{\mathrm{\pi }}{3}-{\tan }^2\frac{\mathrm{\pi }}{4}=-\frac{1}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ {\sin }^2\frac{\mathrm{\pi }}{6}+{\cos }^2\frac{\mathrm{\pi }}{3}-{\tan }^2\frac{\mathrm{\pi }}{4}={\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2-{\left(1\right)}^2\\ =\frac{1}{4}+\frac{1}{4}-1\\ =\frac{1+1-4}{4}\\ =-\frac{2}{4}\\ =-\frac{1}{2}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.11

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ 2{\sin }^2\left(\frac{\mathrm{\pi }}{6}\right)+{\mathrm{cosec}}^2\left(\frac{7\mathrm{\pi }}{6}\right){\cos }^2\left(\frac{\mathrm{\pi }}{3}\right)=\frac{3}{2}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ 2{\sin }^2\frac{\mathrm{\pi }}{6}+{\mathrm{cosec}}^2\frac{7\mathrm{\pi }}{6}{\cos }^2\frac{\mathrm{\pi }}{3}=2{\left(\frac{1}{2}\right)}^2+{\mathrm{cosec}}^2(\mathrm{\pi }+\frac{\mathrm{\pi }}{6})\times {\left(\frac{1}{2}\right)}^2\\ =2\times \frac{1}{4}+\frac{1}{4}\times {\mathrm{cosec}}^2\left(\frac{\mathrm{\pi }}{6}\right)\end{array}$ $\begin{array}{l}=\frac{1}{2}+\frac{1}{4}\times {\left(2\right)}^2\\ =\frac{1}{2}+1\\ =\frac{3}{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.12

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ {\cot }^2\frac{\mathrm{\pi }}{6}+\mathrm{cosec}\frac{5\mathrm{\pi }}{3}+3{\tan }^2\frac{\mathrm{\pi }}{6}=6\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\\ {\cot }^2\frac{\mathrm{\pi }}{6}+\mathrm{cosec}\frac{5\mathrm{\pi }}{6}+3{\tan }^2\frac{\mathrm{\pi }}{6}={\left(\sqrt{3}\right)}^2+\mathrm{cosec}(\mathrm{\pi }-\frac{\mathrm{\pi }}{6})+3{\left(\frac{1}{\sqrt{3}}\right)}^2\\ =3+\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)+3\times \frac{1}{3}\\ [\because \mathrm{cosec}(\mathrm{\pi }-\mathrm{x})=\mathrm{cosec}\mathrm{x}]\\ =3+2+1\\ =6=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.13

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ 2{\sin }^2\left(\frac{3\mathrm{\pi }}{4}\right)+2{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right)+2{\sec }^2\left(\frac{\mathrm{\pi }}{3}\right)=10\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ 2{\sin }^2\left(\frac{3\mathrm{\pi }}{4}\right)+2{\cos }^2\left(\frac{\mathrm{\pi }}{4}\right)+2{\sec }^2\left(\frac{\mathrm{\pi }}{3}\right)=2{\sin }^2(\mathrm{\pi }-\frac{\mathrm{\pi }}{4})+2{\left(\frac{1}{\sqrt{2}}\right)}^2+2{\left(2\right)}^2\end{array}$ $\begin{array}{l}=2{\sin }^2\left(\frac{\mathrm{\pi }}{4}\right)+2\left(\frac{1}{2}\right)+8\\ =2{\left(\frac{1}{\sqrt{2}}\right)}^2+1+8\\ =2\left(\frac{1}{2}\right)+1+8\\ =1+1+8\\ =10=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.14 Find the value of: (i) sin75° (ii) tan15°

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\sin 75\mathrm{°}=\sin (45\mathrm{°}+30\mathrm{°})\\ =\sin 45\mathrm{°}\cos 30\mathrm{°}+\sin 30\mathrm{°}\cos 45\mathrm{°}\\ [\because \sin (\mathrm{x}+\mathrm{y})=\sin \mathrm{x}\cos \mathrm{y}+\sin \mathrm{y}\cos \mathrm{x}]\\ =\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{2}\times \frac{1}{\sqrt{2}}\\ =\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}}\\ =\frac{\sqrt{3}+1}{2\sqrt{2}}\\ \left(\mathrm{ii}\right)\tan 15\mathrm{°}=\tan (45\mathrm{°}-30\mathrm{°})\\ =\frac{\tan 45\mathrm{°}-\tan 30\mathrm{°}}{1+\tan 45\mathrm{°}\tan 30\mathrm{°}}\\ =\frac{1-\frac{1}{\sqrt{3}}}{1+1\times \frac{1}{\sqrt{3}}}\end{array}$ $\begin{array}{l}=\frac{\sqrt{3}-1}{\sqrt{3}+1}\times \frac{\sqrt{3}-1}{\sqrt{3}-1}\\ =\frac{{(\sqrt{3}-1)}^2}{3-1}\\ =\frac{3-2\sqrt{3}+1}{2}\\ \tan 15\mathrm{°}=\frac{4-2\sqrt{3}}{2}\\ =2-\sqrt{3}\end{array}$

Q.15

$\cos (\frac{\mathrm{\pi }}{4}-\mathrm{x})\cos (\frac{\mathrm{\pi }}{4}-\mathrm{y})-\sin (\frac{\mathrm{\pi }}{4}-\mathrm{x})\sin (\frac{\mathrm{\pi }}{4}-\mathrm{y})=\sin (\mathrm{x}+\mathrm{y})$

Ans.

$\begin{array}{l}\mathrm{Since},\hspace{0.17em}\mathrm{cosA}\mathrm{cosB}-\mathrm{sinA}\mathrm{sinB}=\cos (\mathrm{A}+\mathrm{B})\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.=\\ \cos (\frac{\mathrm{\pi }}{4}-\mathrm{x})\cos (\frac{\mathrm{\pi }}{4}-\mathrm{y})-\sin (\frac{\mathrm{\pi }}{4}-\mathrm{x})\sin (\frac{\mathrm{\pi }}{4}-\mathrm{y})\\ =\cos \{(\frac{\mathrm{\pi }}{4}-\mathrm{x})+(\frac{\mathrm{\pi }}{4}-\mathrm{y})\}\\ =\cos (\frac{\mathrm{\pi }}{2}-\mathrm{x}-\mathrm{y})\\ =\cos \{\frac{\mathrm{\pi }}{2}-(\mathrm{x}+\mathrm{y})\}\\ =\sin (\mathrm{x}+\mathrm{y})[\because \cos (\frac{\mathrm{\pi }}{2}-\mathrm{\theta })=\sin \mathrm{\theta }]\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.16

$\begin{array}{l}\mathrm{Prove}\mathrm{the}\mathrm{following}:\\ \frac{\tan (\frac{\mathrm{\pi }}{4}+\mathrm{x})}{\tan (\frac{\mathrm{\pi }}{4}-\mathrm{x})}={\left(\frac{1+\tan \mathrm{x}}{1-\tan \mathrm{x}}\right)}^2\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\\ \frac{\tan (\frac{\mathrm{\pi }}{4}+\mathrm{x})}{\tan (\frac{\mathrm{\pi }}{4}-\mathrm{x})}=\frac{\left(\frac{\tan \frac{\mathrm{\pi }}{4}+\tan \mathrm{x}}{1-\tan \frac{\mathrm{\pi }}{4}\tan \mathrm{x}}\right)}{\left(\frac{\tan \frac{\mathrm{\pi }}{4}-\tan \mathrm{x}}{1+\tan \frac{\mathrm{\pi }}{4}\tan \mathrm{x}}\right)}\left[\begin{array}{l}\because \tan (\mathrm{A}+\mathrm{B})=\frac{\tan \mathrm{A}+\tan \mathrm{B}}{1-\tan \mathrm{A}\tan \mathrm{B}}\\ \tan (\mathrm{A}-\mathrm{B})=\frac{\tan \mathrm{A}-\tan \mathrm{B}}{1+\tan \mathrm{A}\tan \mathrm{B}}\end{array}\right]\\ =\frac{\left(\frac{1+\tan \mathrm{x}}{1-1.\tan \mathrm{x}}\right)}{\left(\frac{1-\tan \mathrm{x}}{1+1.\tan \mathrm{x}}\right)}\\ =\left(\frac{1+\tan \mathrm{x}}{1-\tan \mathrm{x}}\right)\times \left(\frac{1+\tan \mathrm{x}}{1-\tan \mathrm{x}}\right)\\ ={\left(\frac{1+\tan \mathrm{x}}{1-\tan \mathrm{x}}\right)}^2\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.17

$\begin{array}{l}\mathrm{Prove}\mathrm{the}\mathrm{following}:\\ \frac{\cos (\mathrm{\pi }+\mathrm{x})\cos (-\mathrm{x})}{\sin (\mathrm{\pi }-\mathrm{x})\cos (\frac{\mathrm{\pi }}{2}+\mathrm{x})}={\cot }^2\mathrm{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\cos (\mathrm{\pi }+\mathrm{x})\cos (-\mathrm{x})}{\sin (\mathrm{\pi }-\mathrm{x})\cos (\frac{\mathrm{\pi }}{2}+\mathrm{x})}=\frac{-\cos \mathrm{x}\cos \mathrm{x}}{\sin \mathrm{x}\times -\sin \mathrm{x}}\left[\begin{array}{l}\because \cos (\mathrm{\pi }+\mathrm{x})=-\mathrm{cosx}\\ \cos (-\mathrm{x})=\mathrm{cosx}\\ \sin (\mathrm{\pi }-\mathrm{x})=\mathrm{sinx}\\ \cos (\frac{\mathrm{\pi }}{2}+\mathrm{x})=-\mathrm{sinx}\end{array}\right]\\ ={\cot }^2\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.18

$\cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{2}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}\cos (2\mathrm{\pi }\hspace{0.17em}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}[\cot \hspace{0.17em}(\frac{3\mathrm{\pi }}{2}\hspace{0.17em}-\hspace{0.17em}\mathrm{x})\hspace{0.17em}+\hspace{0.17em}\cot \hspace{0.17em}(2\mathrm{\pi }+\mathrm{x})]=1$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\cos (\frac{3\mathrm{\pi }}{2}\hspace{0.17em}+\mathrm{x})\hspace{0.17em}\cos \hspace{0.17em}(2\mathrm{\pi }\hspace{0.17em}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}[\cot (\frac{3\mathrm{\pi }}{2}-\mathrm{x})+\hspace{0.17em}\cot (2\mathrm{\pi }\hspace{0.17em}+\hspace{0.17em}\mathrm{x})]\\ =\sin \mathrm{x}.\cos \hspace{0.17em}\mathrm{x}[\tan \hspace{0.17em}\mathrm{x}+\hspace{0.17em}\cot \mathrm{x}]\\ =\sin \mathrm{x}.\cos \mathrm{x}[\frac{\sin \hspace{0.17em}\mathrm{x}}{\cos \hspace{0.17em}\mathrm{x}}\hspace{0.17em}+\hspace{0.17em}\frac{\cos \hspace{0.17em}\mathrm{x}}{\sin \hspace{0.17em}\mathrm{x}}]\\ =\sin \mathrm{x}.\cos \mathrm{x}\left[\frac{{\sin }^2\mathrm{x}+\hspace{0.17em}{\cos }^2\mathrm{x}}{\sin \mathrm{x}.\cos \hspace{0.17em}\mathrm{x}}\right]\end{array}$ $\begin{array}{l}={\sin }^2\mathrm{x}\hspace{0.17em}+{\cos }^2\mathrm{x}\\ =\hspace{0.17em}1=\hspace{0.17em}\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.19 sin (n + 1) x sin (n + 2) x + cos (n + 1) x cos (n + 2) x = cos x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}:\hspace{0.17em}\text{sin}(\text{n\hspace{0.17em}}+\text{1})\text{x sin\hspace{0.17em}}(\text{n}+\hspace{0.17em}\text{2})\hspace{0.17em}\text{x}\mathrm{}+\text{cos}(\text{n\hspace{0.17em}}+\hspace{0.17em}\text{1})\hspace{0.17em}\text{x cos\hspace{0.17em}}(\text{n\hspace{0.17em}}+\text{2})\text{x}\\ =\text{cos}(\text{n}+\hspace{0.17em}\text{1})\hspace{0.17em}\text{x cos\hspace{0.17em}}(\text{n\hspace{0.17em}}+\hspace{0.17em}\text{2})\hspace{0.17em}\text{x\hspace{0.17em}}+\hspace{0.17em}\text{sin}(\text{n\hspace{0.17em}}+\hspace{0.17em}\text{1})\text{x sin\hspace{0.17em}}(\text{n}+\text{2})\text{x}\\ =\text{cos}\{(\mathrm{n}+\hspace{0.17em}1)\mathrm{x}-\hspace{0.17em}(\mathrm{n}\hspace{0.17em}+\hspace{0.17em}2)\hspace{0.17em}\mathrm{x}\}\\ [\because \mathrm{cosA}\mathrm{cosB}-\mathrm{sinA}\mathrm{sinB}\hspace{0.17em}=\hspace{0.17em}\cos (\mathrm{A}+\hspace{0.17em}\mathrm{B})]\\ =\hspace{0.17em}\text{cos}(\mathrm{nx}+\mathrm{x}\hspace{0.17em}-\mathrm{nx}\hspace{0.17em}-\hspace{0.17em}2\mathrm{x})\\ =\hspace{0.17em}\text{cos\hspace{0.17em}}(-\mathrm{x})[\because \cos (-\mathrm{A})\hspace{0.17em}=\hspace{0.17em}\mathrm{cosA}]\\ =\hspace{0.17em}\cos \mathrm{x}\hspace{0.17em}=\hspace{0.17em}\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.20

$\cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{4}+\mathrm{x})-\cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{4}\hspace{0.17em}-\mathrm{x})=\hspace{0.17em}-\sqrt{2}\sin \mathrm{x}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{4}\hspace{0.17em}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}-\cos \hspace{0.17em}(\frac{3\mathrm{\pi }}{4}-\hspace{0.17em}\mathrm{x})\\ =\hspace{0.17em}-2\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{(\frac{3\mathrm{\pi }}{4}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}+\hspace{0.17em}(\frac{3\mathrm{\pi }}{4}\hspace{0.17em}-\hspace{0.17em}\mathrm{x})}{2}\right)\sin \hspace{0.17em}\left(\frac{(\frac{3\mathrm{\pi }}{4}+\hspace{0.17em}\mathrm{x})\hspace{0.17em}-\hspace{0.17em}(\frac{3\mathrm{\pi }}{4}-\hspace{0.17em}\mathrm{x})}{2}\right)\\ =\hspace{0.17em}-2\sin \left(\frac{2\hspace{0.17em}\left(\frac{3\mathrm{\pi }}{4}\right)}{2}\right)\hspace{0.17em}\sin \left(\frac{2\mathrm{x}}{2}\right)\end{array}$ $\begin{array}{l}=\hspace{0.17em}-2\hspace{0.17em}\sin \hspace{0.17em}(\mathrm{\pi }\hspace{0.17em}-\frac{\mathrm{\pi }}{4})\hspace{0.17em}\sin \mathrm{x}\\ =\hspace{0.17em}-2\sin (\mathrm{\pi }\hspace{0.17em}-\frac{\mathrm{\pi }}{4})\hspace{0.17em}\sin \hspace{0.17em}\mathrm{x}\\ =\hspace{0.17em}-2\sin \hspace{0.17em}\frac{\mathrm{\pi }}{4}\hspace{0.17em}\times \hspace{0.17em}\sin \mathrm{x}\\ =-2\hspace{0.17em}\times \frac{1}{\sqrt{2}}\hspace{0.17em}\times \sin \mathrm{x}\\ =-\sqrt{2}\hspace{0.17em}\sin \hspace{0.17em}\mathrm{x}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.21 Prove that:
sin² 6x – sin² 4x = sin 2x sin 10x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}=\hspace{0.17em}{\text{sin}}^{\text{2}}\text{\hspace{0.17em}6x\hspace{0.17em}}-\hspace{0.17em}{\text{sin\hspace{0.17em}}}^{\text{2}}\text{\hspace{0.17em}4x}\\ =\hspace{0.17em}(\sin 6\mathrm{x}\hspace{0.17em}+\hspace{0.17em}\sin \hspace{0.17em}4\mathrm{x})\hspace{0.17em}(\sin \hspace{0.17em}6\mathrm{x}-\hspace{0.17em}\sin 4\mathrm{x})[\because \hspace{0.17em}{\mathrm{a}}^2\hspace{0.17em}-\hspace{0.17em}{\mathrm{b}}^2\hspace{0.17em}=\hspace{0.17em}(\mathrm{a}\hspace{0.17em}-\hspace{0.17em}\mathrm{b})\hspace{0.17em}(\mathrm{a}\hspace{0.17em}+\hspace{0.17em}\mathrm{b})]\\ =\{2\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{6\mathrm{x}\hspace{0.17em}+4\mathrm{x}}{2}\right)\cos \hspace{0.17em}\left(\frac{6\mathrm{x}\hspace{0.17em}-\hspace{0.17em}4\mathrm{x}}{2}\right)\}\{2\cos \left(\frac{6\mathrm{x}+4\mathrm{x}}{2}\right)\sin \hspace{0.17em}\left(\frac{6\mathrm{x}\hspace{0.17em}-4\mathrm{x}}{2}\right)\}\\ =\hspace{0.17em}\{2\hspace{0.17em}\sin \hspace{0.17em}5\mathrm{x}\cos \mathrm{x}\}\hspace{0.17em}\{2\hspace{0.17em}\cos \hspace{0.17em}5\mathrm{x}\hspace{0.17em}\sin \mathrm{x}\}\\ =\hspace{0.17em}(2\sin \hspace{0.17em}\mathrm{x}\cos \mathrm{x})(2\hspace{0.17em}\sin \hspace{0.17em}5\mathrm{x}\hspace{0.17em}\cos \hspace{0.17em}5\mathrm{x})\\ =\sin \hspace{0.17em}2\mathrm{x}.\sin 10\mathrm{x}[\because \sin 2\mathrm{x}=2\sin \mathrm{x}\cos \mathrm{x}]\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}\mathrm{Hence}\text{proved.}\end{array}$

Q.22 Prove that:
cos² 2x – cos² 6x = sin 4x sin 8x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}={\cos }^{\text{2}}\hspace{0.17em}\text{2x\hspace{0.17em}}-\hspace{0.17em}{\cos }^{\text{2}}\hspace{0.17em}\text{6x}\\ =\hspace{0.17em}(\cos \hspace{0.17em}2\mathrm{x}\hspace{0.17em}+\hspace{0.17em}\cos 6\mathrm{x})(\cos 2\mathrm{x}\hspace{0.17em}-\hspace{0.17em}\cos 6\mathrm{x})\end{array}$ $\begin{array}{l}=\hspace{0.17em}\{2\cos \hspace{0.17em}\left(\frac{2\mathrm{x}\hspace{0.17em}+6\mathrm{x}}{2}\right)\cos \left(\frac{2\mathrm{x}\hspace{0.17em}-6\mathrm{x}}{2}\right)\}\hspace{0.17em}\{-2\hspace{0.17em}\sin \left(\frac{2\mathrm{x}\hspace{0.17em}-6\mathrm{x}}{2}\right)\hspace{0.17em}\sin \hspace{0.17em}\left(\frac{2\mathrm{x}\hspace{0.17em}+6\mathrm{x}}{2}\right)\}\\ =\hspace{0.17em}\{2\cos \hspace{0.17em}4\mathrm{x}\hspace{0.17em}\cos (-2\mathrm{x})\}\hspace{0.17em}\{-2\hspace{0.17em}\sin \hspace{0.17em}(-2\mathrm{x})\hspace{0.17em}\sin 4\mathrm{x}\}\\ =\hspace{0.17em}(2\hspace{0.17em}\sin \hspace{0.17em}2\mathrm{x}\hspace{0.17em}\cos \hspace{0.17em}2\mathrm{x})(2\hspace{0.17em}\sin \hspace{0.17em}4\mathrm{x}\hspace{0.17em}\cos 4\mathrm{x})\\ =\hspace{0.17em}\sin \hspace{0.17em}4\mathrm{x}.\sin \hspace{0.17em}8\mathrm{x}\hspace{0.17em}[\because \hspace{0.17em}\sin 2\mathrm{x}\hspace{0.17em}=\hspace{0.17em}2\sin \hspace{0.17em}\mathrm{x}\cos \mathrm{x}]\\ =\hspace{0.17em}\mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}\mathrm{Hence}\text{\hspace{0.17em}proved.}\end{array}$

Q.23 Prove that:
sin 2x + 2 sin 4x + sin 6x = 4 cos² x sin 4x

Ans.

$\begin{array}{l}\text{sin\hspace{0.17em}2x\hspace{0.17em}}+\text{2 sin 4x}+\text{sin 6x\hspace{0.17em}}=\hspace{0.17em}\text{2 sin 4x\hspace{0.17em}}+\hspace{0.17em}\text{sin 6x}+\hspace{0.17em}\text{sin\hspace{0.17em}2x}\\ =\text{2 sin\hspace{0.17em}4x\hspace{0.17em}}+\hspace{0.17em}2\hspace{0.17em}\sin \left(\frac{6\mathrm{x}\hspace{0.17em}+2\mathrm{x}}{2}\right)\cos \hspace{0.17em}\left(\frac{6\mathrm{x}\hspace{0.17em}-2\mathrm{x}}{2}\right)\\ =\hspace{0.17em}\text{2\hspace{0.17em}sin\hspace{0.17em}4x\hspace{0.17em}}+2\sin \hspace{0.17em}4\mathrm{x}\hspace{0.17em}\cos \hspace{0.17em}2\mathrm{x}\\ =\text{2\hspace{0.17em}sin 4x}(1+\cos \hspace{0.17em}2\mathrm{x})\\ =\text{2\hspace{0.17em}sin 4x\hspace{0.17em}}(2\hspace{0.17em}{\cos }^2\mathrm{x})\\ =\hspace{0.17em}4{\cos }^2\mathrm{x}\text{\hspace{0.17em}sin 4x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.24 Prove that:
cot 4x (sin 5x + sin 3x) = cot x (sin 5x – sin 3x)

Ans.

$\begin{array}{l}\mathrm{We}\text{have, cot 4x\hspace{0.17em}}(\text{sin\hspace{0.17em}5x\hspace{0.17em}}+\hspace{0.17em}\text{sin\hspace{0.17em}3x})=\text{cot\hspace{0.17em}x\hspace{0.17em}}(\text{sin\hspace{0.17em}5x\hspace{0.17em}}–\text{sin\hspace{0.17em}3x})\\ \mathrm{The}\text{given equation can be written as:}\\ \frac{\text{sin 5x}+\text{sin\hspace{0.17em}3x}}{\text{sin\hspace{0.17em}5x}–\text{sin 3x}}=\frac{\text{cot\hspace{0.17em}x}}{\text{cot 4x}}\end{array}$ $\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\frac{\text{sin\hspace{0.17em}5x\hspace{0.17em}}+\hspace{0.17em}\text{sin\hspace{0.17em}3x}}{\text{sin 5x\hspace{0.17em}}–\hspace{0.17em}\text{sin 3x}}=\hspace{0.17em}\frac{2\sin \hspace{0.17em}\left(\frac{5\mathrm{x}\hspace{0.17em}+3\mathrm{x}}{2}\right)\hspace{0.17em}\cos \hspace{0.17em}\left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)}{\text{2 cos\hspace{0.17em}}\frac{5\mathrm{x}\hspace{0.17em}+\hspace{0.17em}3\mathrm{x}}{2}\text{sin}\frac{5\mathrm{x}-\hspace{0.17em}3\mathrm{x}}{2}}\\ =\frac{\sin 4\mathrm{x}\hspace{0.17em}\cos \mathrm{x}}{\cos \hspace{0.17em}4\mathrm{x}\hspace{0.17em}\text{sin\hspace{0.17em}}\mathrm{x}}\\ =\frac{\left(\frac{\cos \hspace{0.17em}\mathrm{x}}{\sin \mathrm{x}}\right)}{\left(\frac{\cos 4\mathrm{x}}{\sin 4\mathrm{x}}\right)}\\ =\hspace{0.17em}\frac{\cot \mathrm{x}}{\cot \hspace{0.17em}4\mathrm{x}}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\hspace{0.17em}\mathrm{Hence}\text{proved.}\end{array}$

Q.25

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{\cos 9\mathrm{x}-\cos 5\mathrm{x}}{\sin 17\mathrm{x}-\sin 3\mathrm{x}}=-\frac{\sin 2\mathrm{x}}{\cos 10\mathrm{x}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\cos 9\mathrm{x}-\cos 5\mathrm{x}}{\sin 17\mathrm{x}-\sin 3\mathrm{x}}=\frac{-2\sin \left(\frac{9\mathrm{x}+5\mathrm{x}}{2}\right)\sin \left(\frac{9\mathrm{x}-5\mathrm{x}}{2}\right)}{2\cos \left(\frac{17\mathrm{x}+3\mathrm{x}}{2}\right)\sin \left(\frac{17\mathrm{x}-3\mathrm{x}}{2}\right)}\\ \left[\begin{array}{l}\because \mathrm{cosA}-\mathrm{cosB}=-2\sin \left(\frac{\mathrm{A}+\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\\ \mathrm{sinA}-\mathrm{sinB}=-2\cos \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\sin \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)\end{array}\right]\end{array}$ $\begin{array}{l}=-\frac{\sin \left(7\mathrm{x}\right)\sin \left(2\mathrm{x}\right)}{\cos \left(10\mathrm{x}\right)\sin \left(7\mathrm{x}\right)}\\ =-\frac{\sin 2\mathrm{x}}{\cos 10\mathrm{x}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{ proved.}\end{array}$

Q.26

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{\sin 5\text{x}+\sin 3\text{x}}{\cos 5\text{x}+\cos 3\text{x}}=\tan 4\text{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\sin 5\mathrm{x}+\sin 3\mathrm{x}}{\cos 5\mathrm{x}+\cos 3\mathrm{x}}=\frac{2\sin \left(\frac{5\mathrm{x}+3\mathrm{x}}{2}\right)\cos \left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)}{2\cos \left(\frac{5\mathrm{x}+3\mathrm{x}}{2}\right)\cos \left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)}\\ =\frac{\sin 4\mathrm{xcosx}}{\cos 4\mathrm{xcosx}}\\ =\tan 4\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.27

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{\mathrm{sinx}-\mathrm{siny}}{\mathrm{cosx}+\mathrm{cosy}}=\tan \left(\frac{\text{x}-\text{y}}{2}\right)\end{array}$

Ans.

$\begin{array}{l}\text{L}.\text{H}.\text{S}.:\\ \frac{\mathrm{sinx}-\mathrm{siny}}{\mathrm{cosx}+\mathrm{cosy}}=\frac{2\cos \left(\frac{\text{x}+\text{y}}{2}\right)\sin \left(\frac{\text{x}-\text{y}}{2}\right)}{2\cos \left(\frac{\text{x}+\text{y}}{2}\right)\cos \left(\frac{\text{x}-\text{y}}{2}\right)}\\ =\frac{\sin \left(\frac{\text{x}-\text{y}}{2}\right)}{\cos \left(\frac{\text{x}-\text{y}}{2}\right)}\\ =\tan \left(\frac{\text{x}-\text{y}}{2}\right)\\ =\text{R}.\text{H}.\text{S}.\mathrm{Hence}\text{\hspace{0.33em}proved.}\end{array}$

Q.28

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \frac{\mathrm{sinx}+\sin 3\text{x}}{\mathrm{cosx}+\cos 3\text{x}}=\tan \text{2x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\mathrm{sinx}+\sin 3\mathrm{x}}{\mathrm{cosx}+\cos 3\mathrm{x}}=\frac{\sin 3\mathrm{x}+\mathrm{sinx}}{\cos 3\mathrm{x}+\mathrm{cosx}}\\ =\frac{2\sin \left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\cos \left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)}{2\cos \left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\cos \left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)}\\ =\frac{\sin 2\mathrm{x}}{\cos 2\mathrm{x}}\\ =\tan 2\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{\hspace{0.17em}proved.}\end{array}$

Q.29

$\begin{array}{l}\mathbf{Prove}\mathbf{\hspace{0.17em}}\mathbf{that}\mathbf{:}\\ \frac{\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{x}\mathbf{}\mathbf{–}\mathbf{\hspace{0.17em}}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{3}\mathbf{x}}{{\mathbf{sin}}^{\mathbf{2}}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{–}\mathbf{}{\mathbf{cos}}^{\mathbf{2}}\mathbf{x}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{\hspace{0.17em}}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\sin \hspace{0.17em}\mathrm{x}\hspace{0.17em}-\sin 3\mathrm{x}}{{\sin }^2\mathrm{x}\hspace{0.17em}-\hspace{0.17em}{\cos }^2\mathrm{x}}=\frac{-(\sin \hspace{0.17em}3\mathrm{x}-\sin \mathrm{x})}{-({\cos }^2\mathrm{x}\hspace{0.17em}-{\sin }^2\mathrm{x})}\\ =\frac{2\hspace{0.17em}\cos \left(\frac{3\mathrm{x}\hspace{0.17em}+\mathrm{x}}{2}\right)\hspace{0.17em}\sin \left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)}{\cos \hspace{0.17em}2\mathrm{x}}\\ =\hspace{0.17em}\frac{2\hspace{0.17em}\cos \hspace{0.17em}2\mathrm{x}\hspace{0.17em}\sin \mathrm{x}}{\cos 2\mathrm{x}}\\ =\hspace{0.17em}2\hspace{0.17em}\sin \mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.30

$\begin{array}{l}\mathbf{Prove}\mathbf{\hspace{0.17em}}\mathbf{that}\mathbf{:}\\ \frac{\mathbf{cos}\mathbf{}\mathbf{4}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{cos}\mathbf{\hspace{0.17em}}\mathbf{3}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{cos}\mathbf{}\mathbf{2}\mathbf{x}}{\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{4}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{\hspace{0.17em}}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{3}\mathbf{x}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{}\mathbf{sin}\mathbf{\hspace{0.17em}}\mathbf{2}\mathbf{x}}\mathbf{\hspace{0.17em}}\mathbf{=}\mathbf{\hspace{0.17em}}\mathbf{cot}\mathbf{}\mathbf{3}\mathbf{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\\ \frac{\cos \hspace{0.17em}4\mathrm{x}\hspace{0.17em}+\cos 3\mathrm{x}+\cos \hspace{0.17em}2\mathrm{x}}{\sin 4\mathrm{x}+\sin \hspace{0.17em}3\mathrm{x}+\sin 2\mathrm{x}}\hspace{0.17em}=\frac{\cos 4\mathrm{x}+\cos 2\mathrm{x}\hspace{0.17em}+\cos 3\mathrm{x}}{\sin 4\mathrm{x}\hspace{0.17em}+\sin 2\mathrm{x}\hspace{0.17em}+\sin 3\mathrm{x}}\\ =\frac{2\cos \hspace{0.17em}\left(\frac{4\mathrm{x}+2\mathrm{x}}{2}\right)\cos \hspace{0.17em}\left(\frac{4\mathrm{x}-2\mathrm{x}}{2}\right)+\cos 3\mathrm{x}}{2\sin \left(\frac{4\mathrm{x}+2\mathrm{x}}{2}\right)\cos \hspace{0.17em}\left(\frac{4\mathrm{x}\hspace{0.17em}-2\mathrm{x}}{2}\right)\hspace{0.17em}+\sin 3\mathrm{x}}\\ =\frac{2\hspace{0.17em}\cos 3\mathrm{x}\cos \mathrm{x}+\hspace{0.17em}\cos \hspace{0.17em}3\mathrm{x}}{2\sin 3\mathrm{x}\cos \mathrm{x}\hspace{0.17em}+\sin 3\mathrm{x}}\end{array}$ $\begin{array}{l}=\frac{\cos 3\mathrm{x}(2\cos \hspace{0.17em}\mathrm{x}+1)}{\sin \hspace{0.17em}3\mathrm{x}(2\cos \mathrm{x}\hspace{0.17em}+1)}\\ \frac{\cos 4\mathrm{x}+\hspace{0.17em}\cos \hspace{0.17em}3\mathrm{x}+\cos 2\mathrm{x}}{\sin 4\mathrm{x}\hspace{0.17em}+\sin 3\mathrm{x}\hspace{0.17em}+\hspace{0.17em}\sin \hspace{0.17em}2\mathrm{x}}=\cot \hspace{0.17em}3\mathrm{x}=\hspace{0.17em}\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{ proved.}\end{array}$

Q.31 Prove that:
cot x cot 2x – cot 2x cot 3x – cot 3x cot x = 1

Ans.

$\begin{array}{l}\because \cot 3\mathrm{x}=\cot (2\mathrm{x}+\mathrm{x})\\ =\frac{\cot 2\mathrm{x}\cot \mathrm{x}-1}{\cot 2\mathrm{x}+\cot \mathrm{x}}\\ \cot 3\mathrm{x}(\cot 2\mathrm{x}+\cot \mathrm{x})=\cot 2\mathrm{x}\cot \mathrm{x}-1\\ \cot 3\mathrm{x}\cot 2\mathrm{x}+\cot 3\mathrm{x}\cot \mathrm{x}=\cot 2\mathrm{x}\cot \mathrm{x}-1\\ 1=\cot \mathrm{x}\cot 2\mathrm{x}-\cot 2\mathrm{x}\cot 3\mathrm{x}-\cot 3\mathrm{x}\cot \mathrm{x}\\ \mathrm{or}\cot \mathrm{x}\cot 2\mathrm{x}-\cot 2\mathrm{x}\cot 3\mathrm{x}-\cot 3\mathrm{x}\cot \mathrm{x}=1.\mathrm{Hence}\text{ proved.}\end{array}$

Q.32

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \tan 4\mathrm{x}=\frac{4\tan \mathrm{x}(1–{\tan }^2\mathrm{x})}{1–6{\tan }^2\mathrm{x}+{\tan }^4\mathrm{x}}\end{array}$

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}:\\ \tan 4\mathrm{x}=\tan 2\left(2\mathrm{x}\right)\\ =\frac{2\tan 2\mathrm{x}}{1-{\tan }^{2}2\mathrm{x}}\end{array}$ $\begin{array}{l}=\frac{2\left(\frac{2\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}\right)}{1-{\left(\frac{2\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}\right)}^2}\\ =\frac{\left(\frac{4\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}\right)}{\left\{\frac{{(1-{\tan }^2\mathrm{x})}^2-4{\tan }^2\mathrm{x}}{{(1-{\tan }^2\mathrm{x})}^2}\right\}}\\ =\frac{4\tan \mathrm{x}}{1-{\tan }^2\mathrm{x}}\times \frac{{(1-{\tan }^2\mathrm{x})}^2}{1-2{\tan }^2\mathrm{x}+{\tan }^4\mathrm{x}-4{\tan }^2\mathrm{x}}\\ =\frac{4\tan \mathrm{x}(1–{\tan }^2\mathrm{x})}{1–6{\tan }^2\mathrm{x}+{\tan }^4\mathrm{x}}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Hence}\text{proved.}\end{array}$

Q.33 Prove that:
cos 4x = 1 – 8sin² x cos² x

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.:\cos 4\mathrm{x}=\cos 2\left(2\mathrm{x}\right)\\ =1-2{\sin }^2\left(2\mathrm{x}\right)[\because \cos 2\mathrm{x}=1-{\sin }^2\mathrm{x}]\\ =1-2{(2\sin \mathrm{x}\cos \mathrm{x})}^2[\because \sin 2\mathrm{x}=2\sin \mathrm{x}\cos \mathrm{x}]\\ =1-2(4{\sin }^2\mathrm{x}{\cos }^2\mathrm{x})\\ =1-8{\sin }^2\mathrm{x}{\cos }^2\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.34 Prove that:
cos 6x = 32 cos⁶ x – 48cos⁴ x + 18 cos² x – 1

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=\text{cos 6x}\\ =\cos 3\left(2\mathrm{x}\right)\\ =4{\cos }^{3}2\mathrm{x}-3\cos 2\mathrm{x}\\ =4{(2{\cos }^2\mathrm{x}-1)}^3-3(2{\cos }^2\mathrm{x}-1)\\ =4(8{\cos }^6\mathrm{x}-12{\cos }^4\mathrm{x}+6{\cos }^2\mathrm{x}-1)-6{\cos }^2\mathrm{x}+3\\ =32{\cos }^6\mathrm{x}-48{\cos }^4\mathrm{x}+24{\cos }^2\mathrm{x}-4-6{\cos }^2\mathrm{x}+3\\ =32{\cos }^6\mathrm{x}-24{\cos }^4\mathrm{x}+18{\cos }^2\mathrm{x}-1\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.35

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{and}\mathrm{general}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equations}:\tan \mathrm{x}=\sqrt{3}\end{array}$

Ans.

$\begin{array}{l}\begin{array}{l}\mathrm{Since},\text{tan}\frac{\text{π}}{3}=\sqrt{3}\text{and tan}\frac{4\text{π}}{3}=\text{tan}\left(\text{π}+\frac{\text{π}}{3}\right)=\tan \frac{\text{π}}{3}=\sqrt{3}\\ \mathrm{Therefore},\text{principal solutions are x}=\frac{\text{π}}{3}\text{and}\frac{4\text{π}}{3}.\\ \text{General solution of tanx}=\sqrt{3}\\ \mathrm{tanx}=\sqrt{3}\\ =\tan \frac{\text{π}}{3}\end{array}\\ \begin{array}{l}\mathrm{tanx}=\tan \left(\mathrm{n\pi }+\frac{\text{π}}{3}\right)\\ \Rightarrow \text{\hspace{0.33em}x}=\mathrm{n\pi }+\frac{\text{π}}{3},\mathrm{where}\text{\hspace{0.33em}n}\in \text{Z}.\end{array}\end{array}$

Q.36

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{and}\mathrm{general}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\sec \mathrm{x}=2\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{secx}=\text{2}\Rightarrow \sec \frac{\mathrm{\pi }}{3}=2\text{and sec}\frac{\text{5}\mathrm{\pi }}{3}=\sec (2\mathrm{\pi }-\frac{\mathrm{\pi }}{3})=\sec \frac{\mathrm{\pi }}{3}=2\\ \mathrm{Therefore},\text{principal solutions are x}=\frac{\mathrm{\pi }}{3}\text{and}\frac{\text{5}\mathrm{\pi }}{3}.\\ \mathrm{G}\text{eneral solution:}\\ \mathrm{secx}=2\\ =\sec \frac{\mathrm{\pi }}{3}\\ =\sec (2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3})\\ \therefore \mathrm{x}=2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{where}\text{ n}\in \mathrm{Z}.\end{array}$

Q.37

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{and}\mathrm{general}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\cot \mathrm{x}=-\sqrt{3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{Since},\text{cot\hspace{0.17em}x}=-\sqrt{3}\\ \Rightarrow \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cot\hspace{0.17em}x}=-\cot \frac{\mathrm{\pi }}{6}\\ =\cot (\mathrm{\pi }-\frac{\mathrm{\pi }}{6})\\ =\cot \left(\frac{5\mathrm{\pi }}{6}\right)\\ \mathrm{and}\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}cot\hspace{0.17em}x}=\cot (2\mathrm{\pi }-\frac{\mathrm{\pi }}{6})\\ =\cot \left(\frac{11\mathrm{\pi }}{6}\right)\\ \mathrm{Therefore},\text{the principal solution are x}=\frac{5\mathrm{\pi }}{6}\text{and}\frac{11\mathrm{\pi }}{6}.\\ \mathrm{General}\text{solution:}\\ \mathrm{cotx}=-\sqrt{3}\\ =\cot \left(\frac{5\mathrm{\pi }}{6}\right)\\ =\cot (\mathrm{n\pi }+\frac{5\mathrm{\pi }}{6})\\ \Rightarrow \mathrm{x}=\mathrm{n\pi }+\frac{5\mathrm{\pi }}{6},\mathrm{where}\text{ n}\in \mathrm{Z}.\end{array}$

Q.38

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{principal}\mathrm{and}\mathrm{general}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\mathrm{cosec}\mathrm{x}=-2\end{array}$

Ans.

$\mathrm{Since},\mathrm{cosec}\mathrm{x}=\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)$ $\begin{array}{l}\therefore \mathrm{cosec}\mathrm{x}=-2\\ =-\mathrm{cosec}\left(\frac{\mathrm{\pi }}{6}\right)\\ =\mathrm{cosec}(\mathrm{\pi }+\frac{\mathrm{\pi }}{6})\\ =\mathrm{cosec}\left(\frac{7\mathrm{\pi }}{6}\right)\\ \mathrm{and}\\ \mathrm{cosec}\mathrm{x}=\mathrm{cosec}(2\mathrm{\pi }-\frac{\mathrm{\pi }}{6})\\ =\mathrm{cosec}\frac{11\mathrm{\pi }}{6}\\ \mathrm{Therefore},\text{the principal solutions are}\frac{7\mathrm{\pi }}{6},\frac{11\mathrm{\pi }}{6}.\\ \mathrm{General}\text{solution:}\\ \text{cosec x}=\mathrm{cosec}\left(\frac{7\mathrm{\pi }}{6}\right)\\ =\mathrm{cosec}(\mathrm{n\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6})\\ \Rightarrow \mathrm{x}=\mathrm{n\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{6},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.39

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\cos 4\mathrm{x}=\cos 2\mathrm{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em} have,}\\ \text{cos 4x}=\text{cos 2x}\end{array}$ $\begin{array}{l}=\cos (2\mathrm{n\pi }\pm 2\mathrm{x})\left[\begin{array}{l}\mathrm{The}\text{general solution of cos}\mathrm{\theta }=\cos \mathrm{\alpha }\\ \Rightarrow \mathrm{\theta }=2\mathrm{n\pi }\pm \mathrm{\alpha }\end{array}\right]\\ \Rightarrow 4\mathrm{x}=2\mathrm{n\pi }\pm 2\mathrm{x}\\ \mathrm{For}(+)\mathrm{ve}\text{sign:}\\ 4\mathrm{x}=2\mathrm{n\pi }+2\mathrm{x}\\ 2\mathrm{x}=2\mathrm{n\pi }\Rightarrow \mathrm{x}=\mathrm{n\pi }\\ \mathrm{For}(-)\mathrm{ve}\text{sign:}\\ 4\mathrm{x}=2\mathrm{n\pi }-2\mathrm{x}\\ 6\mathrm{x}=2\mathrm{n\pi }\Rightarrow \mathrm{x}=\frac{1}{3}\mathrm{n\pi }\\ \mathrm{The}\text{general solutions are x}=\frac{1}{3}\mathrm{n\pi }\text{and}\mathrm{n\pi },\text{where n}\in \mathrm{Z}.\end{array}$

Q.40

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\cos 3\mathrm{x}+\cos \mathrm{x}-\cos 2\mathrm{x}=0\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\text{\hspace{0.17em}have,\hspace{0.17em}}\cos 3\mathrm{x}+\cos \mathrm{x}-\cos 2\mathrm{x}=0\\ 2\cos \left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\cos \left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)-\cos 2\mathrm{x}=0\\ 2\cos 2\mathrm{x}\cos \mathrm{x}-\cos 2\mathrm{x}=0\\ \cos 2\mathrm{x}(2\cos \mathrm{x}-1)=0\\ \Rightarrow \cos 2\mathrm{x}=0\text{}\Rightarrow 2\mathrm{x}=(2\mathrm{n}+1)\frac{\mathrm{\pi }}{2}\\ \Rightarrow \mathrm{x}=(2\mathrm{n}+1)\frac{\mathrm{\pi }}{4},\mathrm{n}\in \mathrm{Z}\end{array}$ $\begin{array}{l}\text{or 2 cos x}-1=0\Rightarrow \cos \mathrm{x}=\frac{1}{2}=\cos \frac{\mathrm{\pi }}{3}\\ \Rightarrow \mathrm{x}=2\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.41

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\sin 2\mathrm{x}+\cos \mathrm{x}=0\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\sin 2\mathrm{x}+\cos \mathrm{x}=0\\ \Rightarrow 2\sin \mathrm{x}\cos \mathrm{x}+\cos \mathrm{x}=0\\ \Rightarrow \cos \mathrm{x}(2\sin \mathrm{x}+1)=0\\ \mathrm{Either}\cos \mathrm{x}=0\Rightarrow \mathrm{x}=(2\mathrm{n}+1)\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}.\\ \mathrm{or}(2\sin \mathrm{x}+1)=0\Rightarrow \sin \mathrm{x}=-\frac{1}{2}\\ \Rightarrow \sin \mathrm{x}=\sin \frac{7\mathrm{\pi }}{2}\\ \Rightarrow \mathrm{x}=\mathrm{n\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}.\\ \mathrm{Thus},\text{the general solution of given equation is:}\\ \mathrm{n\pi }+{(-1)}^{\mathrm{n}}\frac{7\mathrm{\pi }}{2}\text{\hspace{0.17em}\hspace{0.17em}or}(2\mathrm{n}+1)\frac{\mathrm{\pi }}{2},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.42

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:{\sec }^{2}2\text{x}=1-\tan 2\text{x}\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},{\sec }^{2}2\mathrm{x}=1-\tan 2\mathrm{x}\\ \Rightarrow 1+{\tan }^{2}2\mathrm{x}=1-\tan 2\mathrm{x}\end{array}$ $\begin{array}{l}\Rightarrow {\tan }^{2}2\mathrm{x}+\tan 2\mathrm{x}=0\\ \Rightarrow \tan 2\mathrm{x}(\tan 2\mathrm{x}+1)=0\\ \mathrm{Either}\text{tan 2x}=\text{0}\Rightarrow 2\mathrm{x}=\mathrm{n\pi }\\ \Rightarrow \mathrm{x}=\frac{\mathrm{n\pi }}{2},\text{\hspace{0.17em}where n\hspace{0.17em}}\in \mathrm{Z}.\\ \text{or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}\tan 2\mathrm{x}+1=0\Rightarrow \tan 2\mathrm{x}=-1\\ \Rightarrow \tan 2\mathrm{x}=-\tan \frac{\mathrm{\pi }}{4}\\ =\tan (\mathrm{\pi }-\frac{\mathrm{\pi }}{4})\\ \Rightarrow \tan 2\mathrm{x}=\tan \frac{3\mathrm{\pi }}{4}\\ \Rightarrow 2\mathrm{x}=\mathrm{n\pi }+\frac{3\mathrm{\pi }}{4}\\ \Rightarrow \mathrm{x}=\frac{\mathrm{n\pi }}{2}+\frac{3\mathrm{\pi }}{8},\text{where n}\in \mathrm{Z}.\\ \mathrm{The}\text{general solution of the given equation is:}\\ \mathrm{x}=\frac{\mathrm{n\pi }}{2},\frac{\mathrm{n\pi }}{2}+\frac{3\mathrm{\pi }}{8},\text{\hspace{0.17em}where n\hspace{0.17em}}\in \hspace{0.17em}\mathrm{Z}.\end{array}$

Q.43

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{general}\mathrm{solution}\mathrm{for}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}\\ \mathrm{equation}:\sin \mathrm{x}+\sin 3\mathrm{x}+\sin 5\mathrm{x}=0\end{array}$

Ans.

$\begin{array}{l}\mathrm{We}\mathrm{have},\sin \mathrm{x}+\sin 3\mathrm{x}+\sin 5\mathrm{x}=0\\ \Rightarrow \sin 5\mathrm{x}+\sin \mathrm{x}+\sin 3\mathrm{x}=0\\ \Rightarrow 2\sin \left(\frac{5\mathrm{x}+\mathrm{x}}{2}\right)\cos \left(\frac{5\mathrm{x}-\mathrm{x}}{2}\right)+\sin 3\mathrm{x}=0\end{array}$ $\begin{array}{l}\Rightarrow 2\sin 3\mathrm{x}\cos 2\mathrm{x}+\sin 3\mathrm{x}=0\\ \Rightarrow \sin 3\mathrm{x}(2\cos 2\mathrm{x}+1)=0\\ \mathrm{Either}\text{sin3x}=\text{0}\Rightarrow 3\mathrm{x}=\mathrm{n\pi }\\ \Rightarrow \mathrm{x}=\frac{\mathrm{n\pi }}{3},\mathrm{n}\in \mathrm{Z}.\\ \text{or 2 cos 2x}+1=0\Rightarrow \cos 2\mathrm{x}=-\frac{1}{2}\\ \Rightarrow 2\mathrm{x}=2\mathrm{n\pi }\pm \frac{2\mathrm{\pi }}{3}\\ \Rightarrow \mathrm{x}=\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z}.\\ \mathrm{Therefore},\text{ the general solution of the given equations is:}\\ \mathrm{x}=\frac{\mathrm{n\pi }}{3}\text{}\mathrm{or}\mathrm{n\pi }\pm \frac{\mathrm{\pi }}{3},\mathrm{n}\in \mathrm{Z}.\end{array}$

Q.44 In any triangle ABC, if a = 18, b = 24, c = 30, find

1. cos A, cos B, cos C
2. sin A, sin B, sin C

Ans.

$\begin{array}{l}1.\\ \mathrm{Since},\cos \mathrm{A}=\frac{{\text{b}}^2+{\text{c}}^2-{\text{a}}^2}{2\mathrm{bc}}\\ =\frac{{\left(24\right)}^2+{\left(30\right)}^2-{\left(18\right)}^2}{2\left(24\right)\left(30\right)}\\ =\frac{576+900-324}{1440}\\ =\frac{1152}{1440}\\ =\frac{4}{5}\\ \cos \mathrm{B}=\frac{{\text{c}}^2+{\text{a}}^2-{\text{b}}^2}{2\mathrm{ca}}\\ =\frac{{\left(30\right)}^2+{\left(18\right)}^2-{\left(24\right)}^2}{2\left(30\right)\left(18\right)}\\ =\frac{900+324-576}{1080}\\ =\frac{648}{1080}\\ =\frac{3}{5}\\ \cos \mathrm{C}=\frac{{\text{a}}^2+{\text{b}}^2-{\text{c}}^2}{2\mathrm{ab}}\\ =\frac{{\left(18\right)}^2+{\left(24\right)}^2-{\left(30\right)}^2}{2\left(18\right)\left(24\right)}\\ =\frac{324+576-900}{864}\\ =\frac{0}{864}\\ =0\end{array}$

Q.60 Prove that:
(sin 3x + sin x) sin x + (cos 3x – cos x) cos x = 0

Ans.

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=(\sin 3\mathrm{x}+\sin \mathrm{x})\mathrm{sinx}+(\cos 3\mathrm{x}-\cos \mathrm{x})\cos \mathrm{x}\\ =\sin 3\mathrm{x}\sin \mathrm{x}+{\sin }^2\mathrm{x}+\cos 3\mathrm{x}\cos \mathrm{x}-{\cos }^2\mathrm{x}\\ =\cos 3\mathrm{x}\cos \mathrm{x}+\sin 3\mathrm{x}\sin \mathrm{x}-{\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}\\ =\cos (3\mathrm{x}-\mathrm{x})-({\cos }^2\mathrm{x}-{\sin }^2\mathrm{x})\\ [\because \cos (\mathrm{A}-\mathrm{B})=\cos \mathrm{A}\cos \mathrm{B}+\sin \mathrm{A}\cos \mathrm{B}]\\ =\cos 2\mathrm{x}-\cos 2\mathrm{x}[\because {\cos }^2\mathrm{x}-{\sin }^2\mathrm{x}=\cos 2\mathrm{x}]\\ =0\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.61 Prove that:

${(\cos \mathrm{x}+\cos \mathrm{y})}^2+{(\sin \mathrm{x}-\sin \mathrm{y})}^2=4{\cos }^2\frac{\mathrm{x}+\mathrm{y}}{2}$

Ans.

$\begin{array}{l}{\text{L.H.S. : (cos x + cos y)}}^{\text{2}}{\text{+ (sin x-sin y)}}^{\text{2}}\\ \text{=}{\left(\text{2cos}\frac{\text{x + y}}{\text{2}}\text{cos}\frac{\text{x – y}}{\text{2}}\right)}^{\text{2}}\text{+}{\left(\text{2cos}\frac{\text{x + y}}{\text{2}}\text{sin}\frac{\text{x}–\text{y}}{\text{2}}\right)}^{\text{2}}\\ {\text{= 4cos}}^{\text{2}}\left(\frac{\text{x + y}}{\text{2}}\right)\left({\text{cos}}^{\text{2}}\frac{\text{x}–\text{y}}{\text{2}}{\text{+sin}}^{\text{2}}\frac{\text{x}–\text{y}}{\text{2}}\right)\\ {\text{= 4 cos}}^{\text{2}}\left(\frac{\text{x + y}}{\text{2}}\right)\text{× 1}\left[{\text{Q sin}}^{\text{2}}{\text{x + cos}}^{\text{2}}\text{x = 1}\right]\end{array}$ $\begin{array}{l}=4{\cos }^2\left(\frac{\mathrm{x}+\mathrm{y}}{2}\right)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.62 Prove that:

${(\cos \mathrm{x}-\cos \mathrm{y})}^2+{(\sin \mathrm{x}-\sin \mathrm{y})}^2=4{\sin }^2\frac{\mathrm{x}-\mathrm{y}}{2}$

Ans.

$\begin{array}{l}{\text{L.H.S. : (cos x – cos y)}}^{\text{2}}\text{+ (sin x}–{\text{sin y)}}^{\text{2}}\\ \text{=}{\{–\text{2sin}\left(\frac{\text{x + y}}{\text{2}}\right)\text{sin}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\}}^{\text{2}}\text{+}{\left\{\text{2cos}\left(\frac{\text{x + y}}{\text{2}}\right)\text{sin}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\right\}}^{\text{2}}\\ {\text{=4sin}}^{\text{2}}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\left\{{\text{sin}}^{\text{2}}\left(\frac{\text{x + y}}{\text{2}}\right){\text{+cos}}^{\text{2}}\left(\frac{\text{x + y}}{\text{2}}\right)\right\}\\ {\text{= 4 sin}}^{\text{2}}\left(\frac{\text{x}–\text{y}}{\text{2}}\right)\text{× 1}\left[{\text{Q sin}}^{\text{2}}{\text{x + cos}}^{\text{2}}\text{x = 1}\right]\end{array}$ $\begin{array}{l}=4{\sin }^2\left(\frac{\mathrm{x}–\mathrm{y}}{2}\right)\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.63 Prove that:
sin x + sin 3x + sin 5x + sin 7x = 4 cos x cos 2x sin 4x

Ans.

$\begin{array}{l}\text{L.H.S.}=\text{sin x}+\text{sin 3x}+\text{sin 5x}+\text{sin 7x}\\ =\text{sin 7x}+\text{sin x}+\text{sin 5x}+\text{sin 3x}\end{array}$ $\begin{array}{l}=2\sin \left(\frac{7\mathrm{x}+\mathrm{x}}{2}\right)\cos \left(\frac{7\mathrm{x}-\mathrm{x}}{2}\right)+2\sin \left(\frac{5\mathrm{x}+3\mathrm{x}}{2}\right)\cos \left(\frac{5\mathrm{x}-3\mathrm{x}}{2}\right)\\ =2\sin 4\mathrm{x}\cos 3\mathrm{x}+2\sin 4\mathrm{x}\cos \mathrm{x}\\ =2\sin 4\mathrm{x}(\cos 3\mathrm{x}+\cos \mathrm{x})\\ =2\sin 4\mathrm{x}\{2\cos \left(\frac{3\mathrm{x}+\mathrm{x}}{2}\right)\cos \left(\frac{3\mathrm{x}-\mathrm{x}}{2}\right)\}\\ =4\sin 4\mathrm{x}\cos 2\mathrm{x}\cos \mathrm{x}\\ =4\cos \mathrm{x}\cos 2\mathrm{x}\sin 4\mathrm{x}\\ =\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\text{proved.}\end{array}$

Q.64

Prove that:

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

Ans.

$\mathrm{L}.\mathrm{H}.\mathrm{S}.=\frac{(\sin 7\mathrm{x}+\sin 5\mathrm{x})+(\sin 9\mathrm{x}+\sin 3\mathrm{x})}{(\cos 7\mathrm{x}+\cos 5\mathrm{x})+(\cos 9\mathrm{x}+\cos 3\mathrm{x})}$ $\begin{array}{l}=\frac{\left(2\sin \frac{7\mathrm{x}+5\mathrm{x}}{2}\cos \frac{7\mathrm{x}-5\mathrm{x}}{2}\right)+\left(2\sin \frac{9\mathrm{x}+3\mathrm{x}}{2}\cos \frac{9\mathrm{x}-3\mathrm{x}}{2}\right)}{\left(2\cos \frac{7\mathrm{x}+5\mathrm{x}}{2}\cos \frac{7\mathrm{x}-5\mathrm{x}}{2}\right)+\left(2\cos \frac{9\mathrm{x}+3\mathrm{x}}{2}\cos \frac{9\mathrm{x}-3\mathrm{x}}{2}\right)}\\ =\frac{(2\sin 6\mathrm{x}\cos \mathrm{x})+(2\sin 6\mathrm{x}\cos 3\mathrm{x})}{(2\cos 6\mathrm{x}\cos \mathrm{x})+(2\cos 6\mathrm{x}\cos 3\mathrm{x})}\\ =\frac{2\sin 6\mathrm{x}(\cos \mathrm{x}+\cos 3\mathrm{x})}{2\cos 6\mathrm{x}(\cos \mathrm{x}+\cos 3\mathrm{x})}\\ =\tan 6\mathrm{x}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\mathrm{Hence}\mathrm{proved}.\end{array}$

Q.65

$\begin{array}{l}\mathrm{Prove}\mathrm{that}:\\ \sin 3\mathrm{x}+\sin 2\mathrm{x}-\mathrm{sinx}=4\mathrm{sinxcos}\frac{\mathrm{x}}{2}\cos \frac{3\mathrm{x}}{2}\end{array}$

Ans.

$\begin{array}{l}\text{L.H.S. = sin\hspace{0.33em}3x+sin\hspace{0.33em}2x}–\text{sin\hspace{0.33em}x}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= sin\hspace{0.33em}3x}–\text{sin\hspace{0.33em}x + sin\hspace{0.33em}2x}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 2cos}\left(\frac{\text{3x + x}}{\text{2}}\right)\text{sin}\left(\frac{\text{3x}–\text{x}}{\text{2}}\right)\text{+2sin x cos x}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 2cos\hspace{0.33em}2x\hspace{0.33em}sin\hspace{0.33em}x + 2sin x cos x}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 2sin\hspace{0.33em}x}\left(\text{cos\hspace{0.33em}2x + cos\hspace{0.33em}x}\right)\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 2sin\hspace{0.33em}x}\left\{\text{2cos}\left(\frac{\text{2x + x}}{\text{2}}\right)\text{cos}\left(\frac{\text{2x}–\text{x}}{\text{2}}\right)\right\}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 4sin\hspace{0.33em}x\hspace{0.33em}cos}\frac{\text{3x}}{\text{2}}\text{\hspace{0.33em}cos}\frac{\text{x}}{\text{2}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= 4sin\hspace{0.33em}x\hspace{0.33em}cos}\frac{\text{x}}{\text{2}}\text{\hspace{0.33em}cos}\frac{\text{3x}}{\text{2}}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}= R.H.S.\hspace{0.33em}Hence proved.}\end{array}$

Q.66

$\mathrm{Find}\hspace{0.17em}\sin \frac{\mathrm{x}}{2},\cos \frac{\mathrm{x}}{2}\mathrm{and}\tan \frac{\mathrm{x}}{2}\mathrm{in}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}:$ $8.\mathrm{t}\mathrm{an}\mathrm{x}=-\frac{4}{3},\mathrm{x}\mathrm{in}\mathrm{quadrant}\mathrm{II}$ $9.\mathrm{c}\mathrm{os}\mathrm{x}=-\frac{1}{3},\mathrm{x}\mathrm{in}\mathrm{quadrant}\mathrm{III}$ $10.\mathrm{sinx}=\frac{1}{4},\mathrm{x}\text{\hspace{0.33em}in\hspace{0.33em}quadrant\hspace{0.33em}II}$

Ans.

$\begin{array}{l}8.\\ \mathrm{Since},\tan \mathrm{x}=-\frac{4}{3},\mathrm{xinquadrantII}\\ \because {\sec }^2\text{x}=1+{\tan }^2\text{x}\\ =1+{\left(-\frac{4}{3}\right)}^2\\ =1+\frac{16}{9}\\ =\frac{25}{9}\\ \sec \mathrm{x}=\pm \sqrt{\frac{25}{9}}\\ =-\frac{5}{3}\left[\text{x is in II quadrant.}\right]\\ \Rightarrow \cos \mathrm{x}=-\frac{3}{5}\\ 2{\cos }^2\frac{\text{x}}{2}-1=-\frac{3}{5}\\ 2{\cos }^2\frac{\text{x}}{2}=1-\frac{3}{5}\\ =\frac{2}{5}\\ {\cos }^2\frac{\text{x}}{2}=\frac{1}{5}\\ \cos \frac{\text{x}}{2}=\pm \frac{1}{\sqrt{5}}\\ \cos \frac{\text{x}}{2}=\frac{1}{\sqrt{5}}\left[\begin{array}{l}\because \text{x\hspace{0.33em}}\mathrm{lies}\text{\hspace{0.33em}in II quadrant,}\\ \text{So,}\frac{\text{x}}{2}\text{\hspace{0.33em}lies in I quadrant.}\end{array}\right]\\ =\frac{1}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}\\ =\frac{\sqrt{5}}{5}\\ \because \sin \frac{\text{x}}{2}=\sqrt{1-{\cos }^2\frac{\text{x}}{2}}\\ =\pm \sqrt{1-{\left(\frac{1}{\sqrt{5}}\right)}^2}\\ =\pm \sqrt{1-\frac{1}{5}}\\ =\frac{2}{\sqrt{5}}\left[\because \frac{\text{x}}{2}\text{\hspace{0.33em}lies in I quadrant.}\right]\\ =\frac{2}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}\\ =\frac{2\sqrt{5}}{5}\\ \tan \frac{\text{x}}{2}=\frac{\sin \left(\frac{\text{x}}{2}\right)}{\cos \left(\frac{\text{x}}{2}\right)}\\ =\frac{\left(\frac{2}{\sqrt{5}}\right)}{\left(\frac{1}{\sqrt{5}}\right)}=2\\ \mathrm{So},\text{the values of sin}\frac{\text{x}}{2},\text{cos}\frac{\text{x}}{2}\text{and tan}\frac{\text{x}}{2}\text{are}\frac{\sqrt{5}}{5},\frac{2\sqrt{5}}{5}\text{and}\\ \text{2 respectively.}\end{array}$ $\begin{array}{l}9.\\ \mathrm{We}\text{have,}\\ \mathrm{cosx}=-\frac{1}{3}\\ 2{\cos }^2\frac{\text{x}}{2}-1=-\frac{1}{3}\\ 2{\cos }^2\frac{\text{x}}{2}=1-\frac{1}{3}\\ =\frac{2}{3}\\ {\cos }^2\frac{\text{x}}{2}=\frac{1}{3}\\ \cos \frac{\text{x}}{2}=\pm \frac{1}{\sqrt{3}}\\ \cos \frac{\text{x}}{2}=-\frac{1}{\sqrt{3}}\left[\begin{array}{l}\because \text{x\hspace{0.33em}}\mathrm{lies}\text{\hspace{0.33em}in III quadrant,}\\ \text{So,}\frac{\text{x}}{2}\text{\hspace{0.33em}lies in II quadrant.}\end{array}\right]\\ =-\frac{1}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =-\frac{\sqrt{3}}{3}\\ \because \sin \frac{\text{x}}{2}=\sqrt{1-{\cos }^2\frac{\text{x}}{2}}\\ =\pm \sqrt{1-{\left(-\frac{1}{\sqrt{3}}\right)}^2}\\ =\pm \sqrt{1-\frac{1}{3}}\\ =\frac{\sqrt{2}}{\sqrt{3}}\left[\because \frac{\text{x}}{2}\text{\hspace{0.33em}lies in II quadrant.}\right]\\ =\frac{\sqrt{2}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}\\ =\frac{\sqrt{6}}{3}\\ \tan \frac{\text{x}}{2}=\frac{\sin \left(\frac{\text{x}}{2}\right)}{\cos \left(\frac{\text{x}}{2}\right)}\\ =\frac{\left(\frac{\sqrt{2}}{\sqrt{3}}\right)}{\left(-\frac{1}{\sqrt{3}}\right)}=-\sqrt{2}\\ \mathrm{So},\text{the values of sin}\frac{\text{x}}{2},\text{cos}\frac{\text{x}}{2}\text{and tan}\frac{\text{x}}{2}\text{are}\frac{\sqrt{6}}{3},-\frac{\sqrt{6}}{3}\text{and}\\ -\sqrt{\text{2}}\text{respectively.}\end{array}$

$\begin{array}{l}10.\\ \mathrm{We}\mathrm{have},\sin \mathrm{x}=\frac{1}{4},\mathrm{x}\mathrm{in}\mathrm{quadrant}\mathrm{II}\\ \because \cos \mathrm{x}=\sqrt{1-{\sin }^2\text{x}}\\ \cos \mathrm{x}=\pm \sqrt{1-{\left(\frac{1}{4}\right)}^2}\\ =\pm \sqrt{1-\frac{1}{16}}\\ =-\frac{\sqrt{15}}{4}\left[\because \mathrm{x}\mathrm{in}\mathrm{quadrant}\mathrm{II}\right]\\ \mathrm{Since},\cos \mathrm{x}=1-2{\sin }^2\frac{\text{x}}{2}\\ -\frac{\sqrt{15}}{4}=1-2{\sin }^2\frac{\text{x}}{2}\\ 2{\sin }^2\frac{\text{x}}{2}=1+\frac{\sqrt{15}}{4}\\ {\sin }^2\frac{\text{x}}{2}=\frac{1}{2}+\frac{\sqrt{15}}{8}\\ \sin \frac{\text{x}}{2}=\pm \sqrt{\frac{1}{2}+\frac{\sqrt{15}}{8}}\\ =\frac{\sqrt{8+2\sqrt{15}}}{4}\left[\because \frac{\text{x}}{2}\text{lies in I quadrant.}\right]\\ \mathrm{and}\cos \mathrm{x}=2{\cos }^2\frac{\text{x}}{2}-1\\ -\frac{\sqrt{15}}{4}=2{\cos }^2\frac{\text{x}}{2}-1\Rightarrow {\cos }^2\frac{\text{x}}{2}=\frac{1}{2}-\frac{\sqrt{15}}{8}\\ \cos \frac{\text{x}}{2}=\pm \sqrt{\frac{1}{2}-\frac{\sqrt{15}}{8}}\\ =\frac{\sqrt{8-2\sqrt{15}}}{4}\left[\because \frac{\text{x}}{2}\text{lies in I quadrant.}\right]\\ \tan \frac{\text{x}}{2}=\frac{\sin \left(\frac{\text{x}}{2}\right)}{\cos \left(\frac{\text{x}}{2}\right)}\\ =\frac{\frac{\sqrt{8+2\sqrt{15}}}{4}}{\frac{\sqrt{8-2\sqrt{15}}}{4}}\\ =\frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8-2\sqrt{15}}}\times \frac{\sqrt{8+2\sqrt{15}}}{\sqrt{8+2\sqrt{15}}}\\ =\frac{8+2\sqrt{15}}{\sqrt{64-60}}\\ =\frac{8+2\sqrt{15}}{2}\\ \tan \frac{\text{x}}{2}=4+\sqrt{15}\\ \mathrm{So},\text{the values of sin}\frac{\text{x}}{2},\text{cos}\frac{\text{x}}{2}\text{and tan}\frac{\text{x}}{2}\text{are}\frac{\sqrt{8+2\sqrt{15}}}{4},\\ \frac{\sqrt{8-2\sqrt{15}}}{4}\text{and}4+\sqrt{15}\text{respectively.}\end{array}$