Question 11 Mark
Find the general solution for the equation: sinx + sin 3x + sin 5x = 0
Answersin x + sin3x + sin 5x = 0
$ \Rightarrow $ (sin 5x + sin x) + sin 3x = 0
$ \Rightarrow 2\sin \left( {\frac{{5x + x}}{2}} \right)\cos \left( {\frac{{5x - x}}{2}} \right) + \sin 3x$ = 0
$ \Rightarrow $ 2 sin 3x cos 2x + sin 3x = 0
$ \Rightarrow $ sin 3x (2 cos 2x + 1) = 0
$ \Rightarrow $ either sin 3x = 0 or 2 cos 2x + 1= 0
$ \Rightarrow 3x = n\pi $ or $\cos 2x = - \frac{1}{2} = \cos \frac{{2\pi }}{3},n \in Z$
$ \Rightarrow x = \frac{{n\pi }}{3}$ or $2x = 2n\;\pi \pm \frac{{2\pi }}{3},n \in Z$
$ \Rightarrow x = \frac{{n\pi }}{3}$ or $x = n\pi \pm n \in Z$
View full question & answer→Question 21 Mark
Find the general solution for $\sec ^2 2 x=1-\tan 2 x$ equation.
Answer$\sec ^2 2 x=1-\tan 2 x$
$ \Rightarrow 1 + {\tan ^2}2x = 1 - \tan 2x$
$ \Rightarrow {\tan ^2}2x + \tan 2x = 0$
$ \Rightarrow \tan 2x(\tan 2x + 1) = 0$
$ \Rightarrow $ either tan 2x = 0 or tan 2x + 1 = 0
$ \Rightarrow 2x = n\pi $ or $\tan 2x = - 1 = - \tan \frac{\pi }{4} = \tan \left( { - \frac{\pi }{4}} \right),n \in Z$
$ \Rightarrow x = \frac{{n\pi }}{2}$ or $x = \frac{{n\pi }}{2} + \frac{{3\pi }}{8},n \in Z$
View full question & answer→Question 31 Mark
Find the general solution for the equation: sin 2x + cos x = 0
Answersin 2x + cos x = 0
$ \Rightarrow $ 2 sin x cos x + cos x = 0
$ \Rightarrow $ cos x (2 sin x + 1) = 0
$ \Rightarrow $ Either cos x = 0 or 2 sin x + 1 = 0
$ \Rightarrow x = (2n + 1)\frac{\pi }{2}$ or $\sin x = - \frac{1}{2} = - \sin \frac{\pi }{6} = \sin \left( { - \frac{\pi }{6}} \right),n \in Z$
$ \Rightarrow x = (2n + 1)\frac{\pi }{2}$ or $x = n\pi {( - 1)^n}\left( { - \frac{\pi }{6}} \right)$
$x = (2n + 1)\frac{\pi }{2}$ or $x = n\;\pi + {( - 1)^{n + 1}}\left( {\frac{\pi }{6}} \right)$
or $x = n\;\pi {( - 1)^n}\frac{{7\pi }}{6}$$\left[ {\because \sin \left( {\pi + \frac{\pi }{6}} \right) = - \sin \frac{\pi }{6}} \right]$
$ \Rightarrow x = (2n + 1)\frac{\pi }{2}$ or $x = n\;\pi + {( - 1)^n}\frac{{7\pi }}{6},n \in Z$
View full question & answer→Question 41 Mark
Find the general solution for the equation: cos 3x + cos x – cos 2x = 0
AnswerGiven: cos 3x + cos x - cox 2x = 0
$\Rightarrow$ $2 \cos \frac{(3 x+x)}{2} \cos \frac{3 x-x}{2}-\cos 2 x=0\left(\because \cos A+\cos B=2 \cos \frac{a+b}{2} \cos \frac{A-B}{2}\right)$
$\Rightarrow$ 2 cos 2 x cos x- cos 2 x = 0
$\Rightarrow$ cos 2 x(2 cos x-1) = 0
$\Rightarrow$ cos 2 x = 0 or 2 cos x-1 = 0
$\Rightarrow$ cos 2 x = 0 or cos x = 1/2
We know that the general solution for cos x = 0 is x = (2 n + 1) $\pi$/ 2, where n $\in$ Z and Z is set of integers.
Therefore, 2 x = (2n + 1) $\frac{\pi}{2}$ or cos x = cos $\frac{\pi}{3}$, where n $\in$ Z and Z is set of integers
$\Rightarrow$ x = (2n +1) $\frac{\pi}{4}$ or x = 2 n $\pi \pm \frac{\pi}{3}$, where n $\in$ Z and z is set of integers
View full question & answer→Question 51 Mark
Find the general solution for the equation: cos 4x = cos 2x
AnswerHere cos 4x = cos 2x
$ \Rightarrow 4x = 2n\pi \pm 2x,n \in Z$ [ $\because$ If cos x = cos y $ \Rightarrow x = 2n\;\pi \pm y$]
$ \Rightarrow 4x - 2x = 2n\;\pi $ or $4x + 2x = 2\;\pi ,n \in Z$
$ \Rightarrow 2x = 2n\;\pi $ or $6x = 2n\;\pi ,n \in Z$
$ \Rightarrow x = 2\pi $ or $x = \frac{{n\pi }}{3},n \in Z$
Hence general solutions are $n\pi $ or $\frac{{n\pi }}{3},n \in Z$
View full question & answer→Question 61 Mark
Find the principal and general solution of the equation: cosec x = -2
AnswerHere cosec x = - 2 $ \Rightarrow \sin x = - \frac{1}{2}$, which is negative, so x lies in third or fourth quadrant.
$\therefore \sin x=-\frac{1}{2}=-\sin 30^{\circ}=\sin \left(180^{\circ}+30^{\circ}\right) \text { or } \sin \left(360^{\circ}-30^{\circ}\right)$
$ =\sin 210^{\circ} \text { or } \sin 330^{\circ}$
$ = \sin \frac{{7\pi }}{6}$ or $\sin \frac{{11\pi }}{6}$
Hence the principal solutions are $\frac{{7\pi }}{6},\frac{{11\pi }}{6}$
Now $\sin x = - \sin \frac{\pi }{6}$
$ \Rightarrow x = n\pi + {( - 1)^n}\left( {\frac{{7\pi }}{6}} \right)$ where $n \in Z$
View full question & answer→Question 71 Mark
Find the principal and general solution of the equation: cot x = - $\sqrt 3$
AnswerGiven: cot x = $-\sqrt{3}$
We know that, $\cot \frac{\pi}{6}=\sqrt{3} \text { and } \cot \left(\pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$
Also $\cot \left(2 \pi-\frac{\pi}{6}\right)=-\cot \frac{\pi}{6}=-\sqrt{3}$
$\Rightarrow \cot \frac{5 \pi}{6}=-\sqrt{3} \text { and } \cot \frac{11 \pi}{6}=-\sqrt{3}$
$\therefore$ the principal solutions of the equat̥ion are $x=\frac{5 \pi}{6} \text { and } \frac{11 \pi}{6}$
Now when, $\cot x=\cot \frac{5 \pi}{6}$
$\Rightarrow \tan x=\tan \frac{5 \pi}{6}\left(\because \cot x=\frac{1}{\tan x}\right)$
$\Rightarrow \mathrm{x}=\mathrm{n} \pi+\frac{5 \pi}{6}$, where n$\in$ z and z is set of integers.
Hence the general solution of the equation $\mathrm{x}=\mathrm{n} \pi+\frac{5 \pi}{6}$ where n $\in$ Z and z is set of integers.
View full question & answer→Question 81 Mark
Find the principal and general solution of the equation: sec x = 2
AnswerHere sec x = 2 $ \Rightarrow \cos x = \frac{1}{2}$, which is positive, so x lies in first or fourth quadrant.
$\cos \;x = \frac{1}{2} = \cos 60$ or cos ($360^{\circ}-60^{\circ}$)
= cos $60^{\circ}$ or cos $300^{\circ} = \cos \frac{\pi }{3}$ or $\cos \frac{{5\pi }}{3}$
Hence the principal solutions are $\frac{\pi }{3},\frac{{5\pi }}{3}$.
Now $\cos = \cos \frac{\pi }{3} \Rightarrow x = 2\pi \pm \frac{\pi }{3}$ where $n \in Z$
View full question & answer→Question 91 Mark
Find the principal and general solution of the equation: $\tan x = \sqrt 3 $.
AnswerHere $\tan x = \sqrt 3 $, which is positive, so x lies in first or third quadrant.
$\therefore \tan x = \sqrt 3 = \tan 60^\circ $ or tan (180 + 60°)
= tan 60° or tan 240° = $\tan \frac{\pi }{3}$ or $\tan \frac{{4\pi }}{3}$
Hence the principal solutions are $\frac{\pi }{3},\frac{{4\pi }}{3}$
Now tan x =$\tan \frac{\pi }{3} \Rightarrow x = n\;\pi + \frac{\pi }{3}$ where $n \in Z$. which is the general solution
View full question & answer→Question 101 Mark
Prove that: $\frac { \cos ( \pi + x ) \cos ( - x ) } { \sin ( \pi - x ) \cos \left( \frac { \pi } { 2 } + x \right) } = \cot ^ { 2 } x$ .
AnswerLHS = $\frac { \cos ( \pi + x ) \cdot \cos ( - x ) } { \sin ( \pi - x ) \cdot \cos \left( \frac { \pi } { 2 } + x \right) }$
= $\frac { - \cos x \cdot \cos x } { \sin x \cdot ( - \sin x ) }$
[$\because$ cos $( \pi + \theta )$ $= - cos\theta\ and\ cos (-\theta ) = cos\theta ]$
[$sin$ ($( \pi - \theta )$ $= sin$ $\theta$ and $cos$ $\left( \frac { \pi } { 2 } + \theta \right)$ $= - sin$ $\theta$]
= $\frac { \cos ^ { 2 } x } { \sin ^ { 2 } x }$$ = cot^2 x =$ RHS
Hence proved.
View full question & answer→Question 111 Mark
Prove that: $\frac { \tan \left( \frac { \pi } { 4 } + x \right) } { \tan \left( \frac { \pi } { 4 } - x \right) } = \left( \frac { 1 + \tan x } { 1 - \tan x } \right) ^ { 2 }$
AnswerLHS = $\frac { \tan \left( \frac { \pi } { 4 } + x \right) } { \tan \left( \frac { \pi } { 4 } - x \right) }$
= $\frac { \left( \frac { \tan \frac { \pi } { 4 } + \tan x } { 1 - \tan \frac { \pi } { 4 } \cdot \tan x } \right) } { \left( \frac { \tan \frac { \pi } { 4 } - \tan x } { 1 + \tan \frac { \pi } { 4 } \cdot \tan x } \right) }$ [$\because$ $tan (x + y) =$ $\frac { \tan x + \tan y } { 1 - \tan x \cdot \tan y }$ and $tan (x - y) =$ $\frac { \tan x - \tan y } { 1 + \tan x \cdot \tan y }$]
= $\frac { \left( \frac { 1 + \tan x } { 1 - \tan x } \right) } { \left( \frac { 1 - \tan x } { 1 + \tan x } \right) } = \left( \frac { 1 + \tan x } { 1 - \tan x } \right) ^ { 2 }$ = RHS
$\therefore$ LHS = RHS
Hence proved.
View full question & answer→Question 121 Mark
Prove that $\cos \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - y} \right) - \sin \left( {\frac{\pi }{4} - x} \right)$$\sin \left( {\frac{\pi }{4} - y} \right) = \sin (x + y)$
AnswerWe have L.H.S=. $\cos \left( {\frac{\pi }{4} - x} \right)\cos \left( {\frac{\pi }{4} - y} \right)\sin \left( {\frac{\pi }{4} - x} \right)$$\sin \left( {\frac{x}{4} - y} \right)$
$ = \cos \left[ {\frac{\pi }{4} - x + \frac{\pi }{4} - y} \right]$[ $\because $cos (A+ B) = cos A cos B - sin A sin B]
$ = \cos \left[ {\frac{\pi }{2} - (x + y)} \right]$= sin (x + y) = R.H.S
View full question & answer→Question 131 Mark
Prove that cot x cot2x - cot2x cot3x - cot3x cotx = 1
AnswerWe have cot 3x = cot (2x + x)
$ = \frac{{\cot 2x\cot x - 1}}{{\cot 2x + \cot x}}$
$\therefore $ cot 3x (cot 2x + cot x ) = cot 2x.cot x - 1
$ \Rightarrow \cot 3x \cdot \cot 2x$+ cot 3x.cot x = cot 2x.cot x - 1
$ \Rightarrow $ cot 3x cot 2x + cot 3x cot x - cot 2x cot x +1 = 0
$\therefore \cot x\cot 2x - \cot 2x\cot 3x$$ - \cot 3x \cdot \cot x = 1$
View full question & answer→Question 141 Mark
Prove that: $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
AnswerTo prove: $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}=\tan 2 x$
Taking L.H.S, we have
L.H.S = $\frac{\sin x+\sin 3 x}{\cos x+\cos 3 x}$
We know,
cos A + cos B = 2 cos $\frac{(A+B)}{2} \cos \frac{(A-B)}{2}$& sin A + sin B = 2 sin$\frac{(A+B)}{2} \cos \frac{(A-B)}{2}$
LHS = $\frac{2 \sin \frac{(x+3 x)}{2} \cos \frac{(x-3 x)}{2}}{2 \cos \frac{(x+3 x)}{2} \cos \frac{(x-3 x)}{2}}$ = $\frac{\sin 2 x \cos (-x)}{\cos 2 x \cos (-x)}=\frac{\sin 2 x}{\cos 2 x}$
LHS = tan 2x
$\therefore$ LHS = RHS
Hence, proved.
View full question & answer→Question 151 Mark
Prove that: $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
AnswerTo prove: $\frac{\sin x-\sin y}{\cos x+\cos y}=\tan \frac{x-y}{2}$
Taking L.H.S., we have
LHS = $\frac{\sin x-\sin y}{\cos x+\cos y}$
We know,
cos A + cos B = 2 cos $\frac{(A+B)}{2} \cos \frac{(A-B)}{2}$& sin A - sin B = 2 cos $\frac{(A+B)}{2} \sin \frac{(A-B)}{2}$
LHS = $\frac{2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}}{2 \cos \frac{x+y}{2} \cos \frac{x-y}{2}}=\frac{\sin \frac{x-y}{2}}{\cos \frac{x-y}{2}}$
LHS = $\tan \frac{x-y}{2}$
$\therefore$ LHS = RHS
Hence, proved.
View full question & answer→Question 161 Mark
Prove that: $\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}=\tan 4 x$.
AnswerLHS $=\frac{\sin 5 x+\sin 3 x}{\cos 5 x+\cos 3 x}$ $=\frac{2 \sin \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}{2 \cos \left(\frac{5 x+3 x}{2}\right) \cos \left(\frac{5 x-3 x}{2}\right)}$
$=\frac{2 \sin 4 x \cos x}{2 \cos 4 x \cos x}$ = tan 4x = RHS
Hence proved.
View full question & answer→Question 171 Mark
Prove that: $\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}=-\frac{\sin 2 x}{\cos 10 x}$.
AnswerLHS $=\frac{\cos 9 x-\cos 5 x}{\sin 17 x-\sin 3 x}$
$=\frac{-2 \sin \left(\frac{9 x+5 x}{2}\right) \sin \left(\frac{9 x-5 x}{2}\right)}{2 \sin \left(\frac{17 x-3 x}{2}\right) \cos \left(\frac{17 x+3 x}{2}\right)}$
$=\frac{-2 \sin 7 x \sin 2 x}{2 \sin 7 x \cos 10 x}$ $=\frac{-\sin 2 x}{\cos 10 x}$ = RHS
View full question & answer→Question 181 Mark
Prove that: $\sin ^2 6 x-\sin ^2 4 x=\sin 2 x \sin 10 x$
AnswerWe have L.H.S. $ = {\sin ^2}6x - {\sin ^2}4x$
$ = \sin (6x + 4x) \cdot \sin (6x - 4x)$
$[\because {\sin ^2}A - {\sin ^2}B = \sin (A + B)\sin (A - B)]$
$ = \sin 10x \cdot \sin 2x = R.H.S.$
View full question & answer→Question 191 Mark
Prove that: sin (n + 1) x sin (n + 2) x + cos (n +1) x cos (n + 2) x = cos x
AnswerWe have L.H.S. = sin (n +1) x sin (n +2) x + cos (n +1 ) x cos (n +2) x
= cos [n + 1)x - (n + 2)x]
[$\because $ cos (A - B) = cos A cos B + sin A sin B]
=cos[nx + x - nx - 2x]
= cos(-x) = cos x = R.H.S. [ $\because \cos ( - \theta ) = \cos \theta $]
View full question & answer→Question 201 Mark
Find the values of the trigonometric function: $\sin \left( { - \frac{{11\pi }}{3}} \right)$
Answer$\sin \left( { - \frac{{11\pi }}{3}} \right) = \sin \left( {\frac{{ - 11 \times 180^\circ }}{3}} \right) = \sin ( - 660^\circ )$
$\sin ( - 660^\circ ) = \sin ( - 2 \times 360^\circ + 60^\circ ) = \sin 60^\circ = \frac{{\sqrt 3 }}{2}$
View full question & answer→Question 211 Mark
Find the values of the trigonometric function: $\tan \frac{{19\pi }}{3}$
Answer$\tan \frac{{19\pi }}{3} = \tan \frac{{19}}{3} \times 180 = \tan 1140^\circ $
tan 1140° =$\tan (3 \times 360^\circ + 60^\circ ) = \tan 60^\circ = \sqrt 3 $
View full question & answer→Question 221 Mark
Find the values of the trigonometric function: cosec $(-1410)^\circ$
Answercosec $(-1410)^\circ = \cos ec( - 4 \times 360^\circ + 30^\circ )$ = cosec 30° = 2
View full question & answer→Question 231 Mark
Find the values of the trigonometric function: sin 765°
Answer$\sin 765^\circ = \sin (2 \times 360^\circ + 45^\circ ) = \sin 45^\circ = \frac{1}{{\sqrt 2 }}$
View full question & answer→Question 241 Mark
Find the values of the trigonometric function: $\cot \left( {\frac{{ - 15\pi }}{4}} \right)$
Answer$\cot \left( {\frac{{ - 15\pi }}{4}} \right) = \cot \left( {\frac{{ - 15 \times 180^\circ }}{4}} \right) = \cot ( - 675^\circ )$
$\cot ( - 675^\circ ) = \cot ( - 2 \times 360^\circ + 45^\circ )$ = cot 45° = 1
View full question & answer→Question 251 Mark
Find the degree measure to the radian measures $\left( {use\;\pi \frac{{22}}{7}} \right)\frac{{7\pi }}{6}$
Answer${\left( {\frac{{7\pi }}{6}} \right)^C} = {\left( {\frac{{7\pi }}{6} \times \frac{{180}}{\pi }} \right)^o} = 210^\circ $
View full question & answer→Question 261 Mark
Find the degree measure to the radian measure $\left( {use\;\pi \frac{{22}}{7}} \right)\frac{{5\pi }}{3}$
Answer${\left( {\frac{{5\pi }}{3}} \right)^C} = {\left( {\frac{{5\pi }}{3} \times \frac{{180}}{\pi }} \right)^o} = 300^\circ $
View full question & answer→Question 271 Mark
Find the degree measurement of $-4$ radians.
AnswerGiven that: -4 radians
We know that 1 radian = $\frac{180}{\pi}$
$\therefore$ $-4=-4 \times \frac{180}{\pi}=-4 \times \frac{180 \times 7}{22}$ = - 229°
View full question & answer→Question 281 Mark
Find the degree measure to the radian measure $\left( {use\;\pi \frac{{22}}{7}} \right)\frac{{11}}{{16}}$
AnswerTo convert radian into degree measures, we multiply by $\frac{{180}}{\pi }$
$\therefore {\left( {\frac{{11}}{16}} \right)^C} = {\left( {\frac{{11}}{16} \times \frac{{180}}{\pi }} \right)^o} = {\left( {\frac{{11}}{{16}} \times \frac{{180 \times 7}}{{22}}} \right)^o}$$ = {\left( {\frac{{315}}{8}} \right)^o} = {\left( {39\frac{3}{8}} \right)^o}$
$ = 39^\circ \left( {\frac{3}{8} \times 60} \right)$'$ = 39^\circ \left( {\frac{{45}}{2}} \right)$'$ = 39^\circ \left( {22\frac{1}{2}} \right)$'
$ = 39^\circ 22'\left( {\frac{1}{2} \times 60} \right)$"
$=39^{\circ} 22^{\prime} 30^{\prime \prime}$
View full question & answer→Question 291 Mark
Find the radian measure to the degree measure: 520°
AnswerTo convert degree measures into radians, we multiply by $\frac{\pi }{{180}}$
$520^\circ = {\left( {520 \times \frac{\pi }{{180}}} \right)^C} = {\left( {\frac{{26\pi }}{9}} \right)^C}$
View full question & answer→Question 301 Mark
Find the radian measure to the degree measure: 240°
AnswerTo convert degree measures into radians, we multiply by $\frac{\pi }{{180}}$
$240^\circ = {\left( {240 \times \frac{\pi }{{180}}} \right)} = {\left( {\frac{{4\pi }}{3}} \right)}$
View full question & answer→Question 311 Mark
Find the radian measure to the degree measure: $-47^{\circ} 30^{\prime}$
AnswerTo convert degree measures into radians, we multiply by $\frac{\pi }{{180}}$
$-47^{\circ} 30^{\prime}$
$ = - {\left( {47\frac{{30}}{{60}}} \right)^°} = - {\left( {\frac{{95}}{2}} \right)^°}$
$ = - {\left( {\frac{{95}}{2} \times \frac{\pi }{{180}}} \right)^C} = - {\left( {\frac{{19\pi }}{{72}}} \right)^C}$
View full question & answer→Question 321 Mark
Find the radian measure to the degree measure: 25°
AnswerTo convert degree measures into radians, we multiply by $\frac{\pi }{{180}}$
$\therefore 25^\circ = {\left( {25 \times \frac{\pi }{{180}}} \right)^C} = {\left( {\frac{{5\pi }}{{36}}} \right)^C}$
View full question & answer→Question 331 Mark
Find the value of cos (–1710°).
AnswerLet y = cos (–1710°)
We know that values of cos x repeats after an interval of 2$\pi$ or 360°.
Therefore, we have
y = cos (–1710°) = cos (–1710° + 5 $\times$ 360°) = cos (–1710° + 1800°) = cos 90° = 0
View full question & answer→Question 341 Mark
Find the value of sin $\begin{equation} \frac{31 \pi}{3} \end{equation}$
AnswerLet y = sin $\begin{equation} \frac{31 \pi}{3} \end{equation}$
We know that values of sin x repeat after an interval of 2$\pi$. Therefore
$\begin{equation} \sin \frac{31 \pi}{3}=\sin \left(10 \pi+\frac{\pi}{3}\right)=\sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \end{equation}$
View full question & answer→Question 351 Mark
If cot x $= -\frac{5}{12}, x$ lies in second quadrant, find the values of other five trigonometric functions.
AnswerGiven: cot x = $= -\frac{5}{12}$, x lies in second quadrant
Since cot x = $\begin{equation} -\frac{5}{12} \end{equation}$, we have
tan x = $\begin{equation} -\frac{12}{5} \end{equation}$
Now $sec^2 x = 1 + tan^2 x = 1 + \begin{equation} \frac{144}{25}=\frac{169}{25} \end{equation}$
Hence sec x = $\begin{equation} =\pm\frac{13}{5} \end{equation}$
Since x lies in second quadrant, sec x will be negative. Therefore
sec x = $\begin{equation} -\frac{13}{5} \end{equation}$,
Which also gives
cos x = $\begin{equation} -\frac{5}{13} \end{equation}$
Further, we have
sin x = tan x cos x = $\begin{equation} \left(-\frac{12}{5}\right) \times\left(-\frac{5}{13}\right)=\frac{12}{13} \end{equation}$
and cosec x = $\begin{equation} \frac{1}{\sin x}=\frac{13}{12} \end{equation}$
View full question & answer→Question 361 Mark
If cos x = $- \frac {3}{5}$, x lies in the third quadrant, find the values of other five trigonometric functions.
AnswerGiven: cos x = $- \frac {3}{5}$, x lies in the third quadrant,
Since cos x = $- \frac {3}{5}$, we have sec x = $- \frac {5}{3}$
Now $\sin ^2 x+\cos ^2 x=1$, i.e., $\sin ^2 x=1-\cos ^2 x$ or $\sin ^2 x=1-\frac{9}{25}=\frac{16}{25}$
Hence sin x = $\begin{equation} \pm \frac{4}{5} \end{equation}$
Since x lies in third quadrant, sin x is negative. Therefore
sin x = $\begin{equation} - \frac{4}{5} \end{equation}$
which also gives
cosec x = $\begin{equation} - \frac{5}{4} \end{equation}$
Further, we have
$\begin{equation} \tan x=\frac{\sin x}{\cos x}=\frac{4}{3} \text { and } \cot x=\frac{\cos x}{\sin x}=\frac{3}{4} \end{equation}$
View full question & answer→Question 371 Mark
If the arcs of the same lengths in two circles subtend angles $65^{\circ}$ and $110^{\circ}$ at the centre, then find the ratio of their radii.
AnswerSuppose $r_1$ and $r_2$ be the radii of two circles.
Given, $\theta_1$ $= 65^o$
= $\frac { \pi } { 180 } \times 65 = \frac { 13 \pi } { 36 }$ rad
[$\because$ radian measure = $\frac { \pi } { 180 } \times$ degree measure]
and $\theta_2$ $= 110^o$
= $\frac { \pi } { 180 } \times 110 = \frac { 22 \pi } { 36 }$ rad
Let $l$ be the length of each of the arc. Then, $l = r _ { 1 } \theta _ { 1 } = r _ { 2 } \theta _ { 2 }$
i.e., $\frac { r _ { 1 } } { r _ { 2 } } = \frac { 22 } { 13 }$
$\therefore$$r_1 : r_2 = 22 : 13$
View full question & answer→Question 381 Mark
The minute hand of the watch is 1.5 cm long. How far does its tip move in 40 minutes? (Use $\pi$ = 3.14).
AnswerIn 60 minutes, the minute hand of a watch completes one revolution.
Therefore, in 40 minutes, the minute hand turns through $\frac{2}{3}$ of a revolution.
$\therefore$ $\begin{equation} \theta=\frac{2}{3} \times 360^{\circ} \end{equation}$ or $\begin{equation} \frac{4 \pi}{3} \end{equation}$ radian
Hence, the required distance traveled is given by:
x $\begin{equation} =r \ \theta=1.5 \times \frac{4 \pi}{3} \mathrm{cm}=2 \pi \mathrm{cm}=2 \times 3.14 \mathrm{cm} \end{equation}$ = 6.28 cm
View full question & answer→Question 391 Mark
Find the radius of the circle in which a central angle of 60° intercepts an arc of length 37.4 cm (use $\begin{equation} \pi=\frac{22}{7} \end{equation}$).
AnswerHere length, $l$ = 37.4 cm and $\begin{equation} \theta=60^{\circ}=\frac{60 \pi}{180}\ \mathrm{radian}=\frac{\pi}{3} \end{equation}$
Hence, by $\begin{equation} r=\frac{l}{\theta} \end{equation}$, we have
$\begin{equation} r=\frac{37.4 \times 3}{\pi}=\frac{37.4 \times 3 \times 7}{22} \end{equation}$ = 35.7 cm
View full question & answer→Question 401 Mark
Convert 6 radian into degree measure.
AnswerWe know, degree measure = $\frac { 180 } { \pi }$ $\times$ radian measure
$\therefore$ Required degree measure
= $\frac { 180 } { \frac { 22 } { 7 } } \times 6$ [$\because \pi = \frac { 22 } { 7 }$]
= $\frac { 1080 \times 7 } { 22 }$ degree
$= 343$ $\frac { 7 } { 11 }$degree
$= 343^o +$ $\frac { 7 } { 11 }$ $\times$ $60\ min$ [$\because 1^o = 60'$]
$= 343^o + 38 $$\frac { 2 } { 11 }$ min
$= 343^o + 38' +$ $\frac { 2 } { 11 }$ min
$= 343^o +38' +$ $\frac { 2 } { 11 }$ $\times$ 60"
$= 343^o + 38' + 10.9"$
$=343^{\circ} 38^{\prime} 11^{\prime \prime} \text { (approx) }$
$\text { Hence, } 6 \mathrm{rad}=343^{\circ} 38^{\prime} 11^{\prime \prime} \text { (approx) }$
View full question & answer→Question 411 Mark
Convert 40° 20′ into radian measure.
AnswerWe know that 180° = $\pi$ radians
Hence, 40° 20′ = $\begin{equation} 40 \frac{1}{3} \text { degree }=\frac{\pi}{180} \times \frac{121}{3} \text { radian }=\frac{121 \pi}{540} \text { radian } \end{equation}$
Therefore, we have
40° 20′ = $\begin{equation} \frac{121 \pi}{540} \end{equation}$ radian
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