Question 15 Marks
If $\sec\theta=\frac{5}{4},$ find the value of $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}.$
AnswerGiven: $\sec\theta=\frac{5}{4}\ \dots(1)$
To find the value of $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}$
Now we know that $\sec\theta=\frac{1}{\cos\theta}$
Therefore,
$\cos\theta=\frac{1}{\cos\theta}$
Therefore from equation (1)
$\cos\theta=\frac{1}{\frac{5}{4}}$
$\cos\theta=\frac{4}{5}\ \dots(2)$
Also, We Know that $\cos^2\theta+\sin^2\theta=1$
Therefore,
$\sin^2\theta=1-\cos^2\theta$
$\sin^2\theta=\sqrt{1-\cos^2\theta}$
Substituting the value of $\cos\theta$ from equation (2)
We get,
$\sin^2\theta=\sqrt{1-\Big(\frac{4}{5}\Big)^2}$
$=\sqrt{1-\frac{4^2}{5^2}}$
$=\sqrt{1-\frac{16}{25}}$
$=\sqrt{\frac{25-16}{25}}$
$=\sqrt{\frac{9}{25}}$
$=\frac{3}{5}$
Therefore,
$\sin\theta=\frac{3}{5}\ \dots(3)$
Also, we know that $\sec^2\theta=1+\tan^2\theta.$
Therefore,
$\tan^2\theta=\sec^2-1$
Therefore
$\tan^2\theta=\Big(\frac{5}{4}\Big)^2-1$
$=\frac{25}{16}-1$
$=\frac{9}{16}$
Therefore,
$\tan\theta=\sqrt{\frac{9}{16}}$
$=\frac{3}{4}$
Therefore,
$\tan\theta=\frac{3}{4}\ \dots(4)$
Also $\cot\theta=\frac{1}{\tan\theta}$
Therefore, from equation (4)
We get,
$\cot\theta=\frac{1}{\frac{3}{4}}$
$\cot\theta=\frac{4}{3}\ \dots(5)$
Substitutiing the value of $\cos\theta,\sin\theta,\cot\theta$ and $\tan\theta$ from equation (2) (3) (4) and (5) respectictively in the expression below
$\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}$
We get,
$\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}=\frac{\frac{3}{5}-2\Big(\frac{4}{5}\Big)}{\frac{3}{4}-\frac{4}{3}}$$$
$=\frac{\frac{3}{5}-\frac{8}{5}}{\frac{(3\times3)-(4\times4)}{4\times3}}$
$=\frac{\frac{3-8}{5}}{\frac{9-16}{4\times3}}$
$=\frac{\frac{-5}{5}}{\frac{-7}{12}}$
$=\frac{12}{7}$
Therefore, $\frac{\sin\theta-2\cos\theta}{\tan\theta-\cot\theta}=\frac{12}{7}$
View full question & answer→Question 25 Marks
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$\sec\theta=\frac{13}{5}$
Answer$\sec\theta=\frac{13}{5}$ $\sec\theta=\frac{\text{hypotenuse}}{\text{adjacent side}}=\frac{13}{5}$ Now consider a right angled $\triangle^\text{le}\text{ABC,}$
By applying pythagoras theorem $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $169=\text{x}^2+25$ $\text{x}^2=169-25=144$ $\text{x}=12$ $\cos\theta =\frac{1}{\sec\theta}=\frac{1}{\frac{13}{5}}=\frac{5}{13}$ $\tan\theta =\frac{\text{opposite side}}{\text{adjacent side}}=\frac{12}{5}$ $\sin\theta =\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{12}{13}$ $\text{cosec }\theta =\frac{1}{\sin\theta}=\frac{1}{\frac{12}{13}}=\frac{13}{12}$ $\sec\theta =\frac{1}{\cos\theta}=\frac{1}{\frac{5}{13}}=\frac{13}{5}$ $\cot\theta =\frac{1}{\tan\theta}=\frac{1}{\frac{12}{5}}=\frac{5}{12}$ View full question & answer→Question 35 Marks
If $3\cot\theta=2,$ find the value of $\frac{4\sin\theta-3\cos\theta}{2\sin\theta+6\cos\theta}.$
AnswerGiven: $3\cot\theta=2$
Therefore,
$\cot\theta=\frac{2}{3}\ \dots(1)$
Now, we know that $\cot\theta=\frac{\cos\theta}{\sin\theta}$
Therefore equuation (i) becomes
$\frac{\cos\theta}{\sin\theta}=\frac{2}{3}\ \dots(2)$
Now by applying Invetendo to equation (ii)
We get,
$\frac{\sin\theta}{\cos\theta}=\frac{3}{2}\ \dots(3)$
Now, multipalying by $\frac{4}{3}$ on botha sides
We get,
$\frac{4}{3}\times\frac{\sin\theta}{\cos\theta}=\frac{4}{3}\times\frac{3}{2}$
Therefore, 3 cancels out on R.H.S and
We get,
$\frac{4\sin\theta}{3\cos\theta}=\frac{2}{1}$
Now by applying dividendo in above equation
We get,
$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}=\frac{2-1}{1}$
$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}=\frac{1}{1}\ \dots(4)$
Now, multiplying by $\frac{2}{6}$ on both sides of equation (3)
We get,
$\frac{2}{6}\times\frac{\sin\theta}{\cos\theta}=\frac{2}{6}\times\frac{3}{2}$
Therefore, 2 cancels out on R.H.S and
We get,
$\frac{2\sin\theta}{6\cos\theta}=\frac{3}{6}$
$\frac{2\sin\theta}{6\cos\theta}=\frac{1}{2}$
Now by applying componendo in above equation
We get,
$\frac{2\cos\theta+6\sin\theta}{6\sin\theta}=\frac{1+2}{2}$
$\frac{2\cos\theta+6\sin\theta}{6\sin\theta}=\frac{3}{2}\ \dots(5)$
Now, by dividing equation (4) by equation (5)
We get,
$\frac{\frac{4\sin\theta-3\cos\theta}{3\sin\theta}}{\frac{2\cos\theta+6\sin\theta}{6\sin\theta}}=\frac{\frac{1}{1}}{\frac{3}{2}}$
Therefore,
$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}\times\frac{6\sin\theta}{2\cos\theta+6\sin\theta}=\frac{1}{1}\times\frac{2}{3}$
$\frac{4\sin\theta-3\cos\theta}{3\sin\theta}\times\frac{2\times(3\sin\theta)}{2\cos\theta+6\sin\theta}=\frac{1}{1}\times\frac{2}{3}$
Therefore, on L.H.S $(3\sin\theta)$ cancels out and we get,
$\frac{2\times(4\sin\theta-3\cos\theta)}{2\cos\theta+6\sin\theta}=\frac{2}{3}$
Now, by taking 2 in the numerator of L.H.S on the R.H.S
We get,
$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{2}{3\times2}$
Therefore, 2 cancels out on R.H.S. and
We get,
$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{1}{3}$
Hence,
$\frac{4\sin\theta-3\cos\theta}{2\cos\theta+6\sin\theta}=\frac{1}{3}$
View full question & answer→Question 45 Marks
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$\cot\theta=\frac{12}{5}$
AnswerGiven: $\cot\theta=\frac{12}{5}\ \dots(1)$ By definition, $\cot\theta=\frac{1}{\tan\theta}$ $\cos \theta=\frac{\text{Base}}{\text{Perpendicular}}\ \dots(2)$ By Comparing (1) and (2) We get, Base = 12 and Perpendicular side = 5
Therefore, By Pythagoras theorem,$ AC^2 = AB^2 + BC^2$
Now we substitute the value of base side (AB) and the perpendicular side (BC) and get hypotenuse (AC) $\text{AC}^2=12^2+5^2$ $\text{AC}^2=144+25$
$\text{AC}^2=169$ $\text{AC}=\sqrt{169}$
$\text{AC}=13$ Hence, Hypotenuse = 13
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse }}$
Therefore, $\sin\theta=\frac{5}{13}$
Now, $\text{cosec }\theta=\frac{1}{\sin\theta}$
Therefore, $\text{cosec }\theta=\frac{\text{Hypotenuse }}{\text{Perpendicular }}$
$\text{cosec }\theta=\frac{13}{5}$
Now, $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
Therefore, $\cos\theta=\frac{12}{13}$
Now, $\sec\theta=\frac{1}{\cos\theta}$
Therefore, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$\sec\theta=\frac{13}{12}$
Now, $\tan\theta=\frac{1}{\cot\theta}$
Therefore, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
$\tan\theta=\frac{5}{12}$ View full question & answer→Question 55 Marks
If $\cos\theta=\frac{3}{5},$ find the value of $\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}.$
AnswerGiven: $\cos\theta=\frac{3}{5}\ ...(1)$
To find the value of $\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}$
Now, we know the following trigonometric identity
$\cos^2\theta+\sin^2\theta=1$
Therefore, by substituting the value of $\cos\theta$ from equation (1),
We get,
$\Big(\frac{3}{5}\Big)^2+\sin^2\theta=1$
Therefore,
$\sin^2\theta=1-\Big(\frac{3}{5}\Big)^2$
$=1-\frac{(3)^2}{(5)^2}$
$=1-\frac{9}{25}$
$\sin^2\theta=\frac{25-9}{25}$
$=\frac{16}{25}$
Therefore by taking square root on both sides
We get,
$\sin\theta=\sqrt{\frac{16}{25}}$
$=\sqrt{\frac{16}{25}}$
$=\frac{4}{5}$
Therefore,
$\sin\theta=\frac{4}{5}\ ...(2)$
Now, we know that
$\tan\theta=\frac{\sin\theta}{\cos\theta}\ ..(3)$
therefore by substituting the value of $\sin\theta$ and $\cos\theta$ from equation (2) and (1) respectively
We get,
$\tan\theta=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac{4}{3}\ ...(4)$
Now, by substituting the value of $\sin\theta$ and $\tan\theta$ from equation (2) and (4) respectively in the expression below,
$\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}$
We get,
$\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}=\frac{\frac{4}{5}-\frac{1}{\frac{4}{3}}}{2\times\frac{4}{3}}$
$=\frac{\frac{4}{5}-\frac{3}{4}}{\frac{2\times4}{3}}$
$=\frac{\frac{4\times4}{5\times4}-\frac{3\times5}{4\times5}}{\frac{8}{3}}$
Therefore,
$\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}=\frac{\frac{16}{20}-\frac{15}{20}}{\frac{8}{3}}$
$=\frac{\frac{1}{20}}{\frac{8}{3}}$
$=\frac{3}{160}$
Therefore, $\frac{\sin\theta-\frac{1}{\tan\theta}}{2\tan\theta}=\frac{3}{160}$
View full question & answer→Question 65 Marks
If $3\cos\theta − 4 \sin \theta = 2 \cos \theta + \sin \theta$ find $\tan\theta.$
AnswerGiven: $3\cos \theta -4\sin \theta =2\cos\theta+\sin\theta$
To find: $\tan\theta$
Now consider the given expression
$3\cos\theta -4\sin\theta=2\cos\theta+\sin\theta$
Now by dividing both sides of the above expression by $\cos\theta$
We get,
$\frac{3\cos\theta-4\sin\theta}{\cos\theta}=\frac{2\cos\theta+\sin\theta}{\cos\theta}$
Now by separating the denominator for each terms
We get,
$\frac{3\cos\theta}{\cos\theta}-\frac{4\sin\theta}{\cos\theta}=\frac{2\cos\theta}{\cos\theta}+\frac{\sin\theta}{\cos\theta}$
Now in the above expression $\cos\theta$ present in both numerator and denominator gets cancelled
Therefore,
$3-\frac{4\sin\theta}{\cos\theta}=2+\frac{\sin\theta}{\cos\theta}$
Now we know that,
$\frac{\sin\theta}{\cos\theta}=\tan\theta$
Therefore by substituting $\frac{\sin\theta}{\cos\theta}=\tan\theta$ in equation (1)
We get,
$3-4\tan\theta=2+\tan\theta$
Now by taking $\tan\theta$ on L.H.S
We get,
$-\tan\theta -4\tan\theta =2-3$
Therefore,
$-5\tan\theta=-1$
$5\tan\theta=1$
$\tan\theta=\frac{1}{5}$
Hence $\tan\theta=\frac{1}{5}$
View full question & answer→Question 75 Marks
If $\sin\theta=\frac{12}{13},$ find the value of $\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}.$
AnswerGiven: $\sin\theta=\frac{12}{13}\dots(1)$
To Find: The value of expression $\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}$
Now, we know that,
$\sin\theta=\frac{\text{Perpendicular side opposite to }\angle\theta}{\text{Hypotenuse}}\dots(2)$
Now when we compare equation (1) and (2)
We get,
Perpendicular side opposite to $\angle\theta=12$
And
Hypotenuse = 13
Therefore, Triangle representing angle $\theta$ is as shown below
Base side BC is unknown and it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
Therefore by substituting the values of known sides
We get,
$13^2=12^2+\text{BC}^2$
Therefore,
$\text{BC}^2=13^2-12^2$
$\text{BC}^2=169-144$
$\text{BC}^2=25$
$\text{BC}=\sqrt{25}$
Therefore,
$\text{BC}=5 \dots(3)$
Now, we know that
$\cos\theta=\frac{\text{Base side adjacent to} \angle\theta}{\text{Hypotenuse}}$
Now from figure (a)
We get,
$\cos\theta=\frac{\text{BC}}{\text{AC}}$
Therefore from figure (a) and equation (3),
$\cos\theta=\frac{5}{13}\dots(4)$
Now we know that,
$\tan\theta=\frac{\sin\theta}{\cos\theta}$
Therefore, substituting the value of $\sin\theta$ and $\cos\theta$ from equation (1) and (4)
We get,
$\tan\theta=\frac{\frac{12}{13}}{\frac{5}{13}}$
$\tan\theta=\frac{12}{13}\times\frac{13}{5}$
Therefore 13 gets cancelled and we get
$\tan\theta=\frac{12}{5}\dots(5)$
Now we substitute the value of $\sin\theta,\cos\theta$ and $\tan\theta$ from equation (1), (4) and (5) respectively in the expression below
$\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}$
Therefore,
We get,
$\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}=\frac{\Big(\frac{12}{13}\Big)^2-\Big(\frac{5}{13}\Big)^2}{2\times\Big(\frac{12}{13}\Big)\times\Big(\frac{5}{13}\Big)}\times\frac{1}{\Big(\frac{12}{5}\Big)^2}$
Therefore by further simplifying we get,
$\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}=\frac{\frac{(12)^2}{(13)^2}-\frac{(5)^2}{(13)^2}}{2\times\Big(\frac{12}{13}\Big)\times\Big(\frac{5}{13}\Big)}\times\frac{1}{\frac{(12)^2}{(5)^2}}$
$=\frac{\frac{144}{169}-\frac{25}{169}}{\frac{2\times12\times5}{13\times13}}\times\frac{25}{144}$
$=\frac{\frac{144-25}{169}}{\frac{120}{169}}\times\frac{25}{144}$
$=\frac{119}{169}\times\frac{169}{120}\times\frac{25}{144}$
Now 169 gets cancelled and $=\frac{25}{120}$ gets reduced to $=\frac{5}{24}$
Therefore
$\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}=\frac{119}{1}\times\frac{1}{24}\times\frac{5}{144}$
$=\frac{119\times5}{24\times144}$
$=\frac{595}{3456}$
Therefore the value of $\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}$ is $\frac{595}{3456}$
That is $\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}=\frac{595}{3456}$ View full question & answer→Question 85 Marks
Evaluate the following:
$4\big(\sin^460^\circ+\cos^430^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ$
AnswerWe have,
$4\big(\sin^460^\circ+\cos^430^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ\dots(1)$
Now,
$\sin60^\circ=\cos30^\circ=\frac{\sqrt{3}}{2} ,\cos45^\circ=\frac{1}{\sqrt{2}},\tan60^\circ =\sqrt{3} ,\tan45^\circ=1 $
So by substituting above values in equation (1)
We get,
$4\big(\sin^460^\circ+\cos^430^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ$
$=4\bigg(\Big(\frac{\sqrt{3}}{2}\Big)^4+\Big(\frac{\sqrt{3}}{2}\Big)^4\bigg)-3\Big(\big(\sqrt{3}\big)^2-1^2\Big)+5\times\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=4\Bigg(\frac{\big(\sqrt{3}\big)^4}{2^4}+\frac{\big(\sqrt{3}\big)^4}{2^4}\Bigg)-3(3-1)+5\times\frac{1^2}{\big(\sqrt{2}\big)^2}$
$=4\Big(\frac{9}{16}+\frac{9}{16}\Big)-3(2)+5\times\frac{1}{2}$
$=4\Big(\frac{9+9}{16}\Big)-6+\frac{5}{2}$
$=4\Big(\frac{18}{16}\Big)-6+\frac{5}{2}$
Now, $=\frac{18}{16}$ gets reduced to $ \frac{9}{8}$
Therefore,
$4\big(\sin^460^\circ+\cos^430^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ$
$=4\Big(\frac{9}{8}\Big)-6+\frac{5}{2}$
$=\frac{36}{8}-6+\frac{5}{2}$
Now, $\frac{36}{8}$ gets reduced to $\frac{9}{2}$
Therefore,
$4\big(\sin^460^\circ+\cos^430^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ$
$=\frac{9}{2}-6+\frac{5}{2}$
Now by taking LCM
We get,
$4\big(\sin^460^\circ+\cos^430^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ$
$=\frac{9}{2}-\frac{6\times2}{1\times2}+\frac{5}{2}$
$=\frac{9}{2}-\frac{12}{2}+\frac{5}{2}$
$=\frac{9-12+5}{2}$
$=\frac{14-12}{2}$
$=\frac{2}{2}$
$=1$
Therefore,
$4\big(\sin^460^\circ+\cos^43^\circ\big)-3\big(\tan^260^\circ-\tan^245^\circ\big)+5\cos^245^\circ=1$
View full question & answer→Question 95 Marks
If $\cos\theta=\frac{12}{13},$ show that $\sin\theta(1-\tan\theta)=\frac{35}{156}.$
AnswerGiven: $\cos\theta=\frac{12}{13}\ \dots(1)$
To show that $\sin\theta(1-\tan\theta)=\frac{35}{156}$
Now, we know that $\cos\theta=\frac{\text{Base side adjacent to}\angle\theta}{\text{Hypotenuse}}\ \dots(2)$
Therefore, by comparing equation (1) and (2) We get, Base side adjacent to $\angle\theta=12$ And Hypotenuse = 13
Therefore from above figure Base side $BC = 12$
Hypotenuse $AC = 13$
Side AB is unknown and it can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
We get, $AC^2= AB^2 + BC^2$
Therefore by substituting the values of known sides
We get, $13^2 = AB^2 + 12^2$
Therefore, $\text{AB}^2=169-144$ $\text{AB}^2=25$
$\text{AB}=\sqrt{25}$ Therefore, $\text{AB}=5\ \dots(3)$
Now, we know that $\sin\theta=\frac{\text{Perpendicular side opposite to}\angle\theta}{\text{Hypotenuse}}$
Now from figure (a) We get, $\sin\theta=\frac{\text{AB}}{\text{AC}}$
Therefore, $\sin\theta=\frac{5}{13}\ \dots(4)$
Now, we know that $\tan\theta=\frac{\text{Perpendicular side opposite to}\angle\theta}{\text{Base side adjacent to }\angle\theta}$
Now from figure (a) We get, $\tan\theta=\frac{\text{AB}}{\text{BC}}$
Therefore, $\tan\theta=\frac{5}{12}\ \dots(5)$
Now L.H.S. of the equation to be proved is as follows
$\text{L.H.S.}=\sin\theta(1-\tan\theta)\ \dots(6)$
Substituting the value of $\sin\theta$ and $\tan\theta$ from equation (4) and (5) respectively We get,
$\text{L.H.S.}=\frac{5}{13}\Big(1-\frac{5}{12}\Big)$
Taking L.C.M inside the bracket We get, $\text{L.H.S.}=\frac{5}{13}\Big(\frac{1\times12}{1\times12}-\frac{5}{12}\Big)$
Therefore, $\text{L.H.S.}=\frac{5}{13}\Big(\frac{12-5}{12}\Big)$
$\text{L.H.S.}=\frac{5}{13}\Big(\frac{7}{12}\Big)$
Now, by opening the bracket and simplifying We get,
$\text{L.H.S.}=\frac{5\times7}{13\times12}$
$\text{L.H.S.}=\frac{35}{156}\ \dots(7)$ From equation (6) and (7), it can be shown that
$\sin\theta(1-\tan\theta)=\frac{35}{156}$ View full question & answer→Question 105 Marks
If $\tan(\text{A}-\text{B})=\frac{1}{\sqrt{3}}$ and $\tan(\text{A}+\text{B})=\sqrt{3},0^\circ<\text{A + B}\leq90^\circ,\text{A}>\text{B}$ find A and B.
AnswerGiven:$\tan(\text{A}-\text{B})=\frac{1}{\sqrt{3}}\ \dots(1)$
$\tan(\text{A}+\text{B})=\sqrt{3}\ \dots(2)$
We know that,$\tan30^\circ=\frac{1}{\sqrt{3}}\ \dots(1)$
$\tan60^\circ-=\sqrt{3}\ \dots(4)$
Now by comparing equation (1) and (3) We get, A - B = 30 ...(5) Now by comparing equation (2) and (4) We get, A + B = 60 ...(6) Now to get the values of A and B, let us solve equation (5) and (6) simultaneously Therefore by adding equation (5) and (6) We get,$\text{A}-\text{B}=90\\\text{A}+\text{B}=0\\\overline{2\text{A}\ \ \ \ \ \ =90}$
Therefore,$2\text{A}=90$
$\Rightarrow\text{A}=\frac{90}{2}$
$\Rightarrow\text{A}=45^\circ$
Hence A = 45° Now by subtracting equation (5) from equation (6) We get,
Therefore, $2\text{B}=30$$\Rightarrow\text{B}=\frac{30}{2}$
$\Rightarrow\text{B}=15^\circ$
Hence B = 15° Therefore the values of A and B are as followsA = 45° and B = 15° View full question & answer→Question 115 Marks
Evaluate the following:
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
AnswerWe have
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ\ \dots(1)$
Now,
$\cot30^\circ=\sqrt{3},\ \cos45^\circ=\frac{1}{\sqrt{2}},\ \sin60^\circ=\frac{\sqrt{3}}{2}$
So by substituting above values in equation (1)
We get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4}{\big(\sqrt{3}\big)^2}+\frac{1}{\Big(\frac{\sqrt{3}}{2}\Big)^2}-\Big(\frac{1}{\sqrt{2}}\Big)^2$
$=\frac{4}{3}+\frac{1}{\frac{\big(\sqrt{3}\big)^2}{2^2}}-\frac{1^2}{\big(\sqrt{2}\big)^2}$
$=\frac{4}{3}+\frac{2^2}{\big(\sqrt{3}\big)^2}-\frac{1}{2}$
$ =\frac{4}{3}+\frac{4}{3}-\frac{1}{2}$
Now LCM of denominator of above expression is 6
Therefore by taking LCM we get,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ$
$=\frac{4\times2}{3\times2}+\frac{4\times2}{3\times2}-\frac{1\times3}{2\times3}$
$ =\frac{8}{6}+\frac{8}{2}-\frac{3}{6}$
$=\frac{8+8-3}{6}$
$ =\frac{16-3}{6}$
$ =\frac{13}{6}$
Hence,
$\frac{4}{\cot^230^\circ}+\frac{1}{\sin^260^\circ}-\cos^245^\circ=\frac{13}{6}$
View full question & answer→Question 125 Marks
In $\triangle\text{PQR},$ right angled at $Q, PQ = 4cm$ and $RQ = 3cm$. Find the values of sin P, sin R, sec P and sec R.
AnswerGiven:
$\triangle\text{PQR}$ is right angled at vertex Q.
$PQ = 4cm$
$RQ = 3cm$
To find:
$\sin\text{P},\sin\text{R},\sec\text{P},\sec\text{R}$
Given $\triangle\text{PQR}$ is as shown below
Hypotenuse side PR is unknown.
Therefore, we find side PR of $\triangle\text{PQR}$ by Pythagoras theorem
By applying Pythagoras theorem to $\triangle\text{PQR}$
We get,
$PR^2 = PQ^2 + RQ^2$
Substituting values of sides from the above figure
$\text{PR}^2=4^4+3^2$
$\text{PR}^2=16+9$
$\text{PR}^2=25$
$\text{PR}=\sqrt{25}$
$\text{PR}=5$
Hence, Hypotenuse = 5
Now by definition,
$\sin\text{P}=\frac{\text{Perpendicular side opposite to }\angle\text{P}}{\text{Hypotenuse}}$
$\sin\text{P}=\frac{\text{RQ}}{\text{PR}}$
Substituting values of sides from the above figure
$\sin\text{P}=\frac{3}{5}$
Now by definition,
$\sin\text{R}=\frac{\text{Perpendicular side opposite to }\angle\text{R}}{\text{Hypotenuse}}$
$\sin\text{R}=\frac{\text{PQ}}{\text{PR}}$
Substituting values of sides from the above figure
$\sin\text{R}=\frac{4}{5}$
By definition,
$\sec\text{P}=\frac{1}{\cos\text{P}}$
$\sec\text{P}=\frac{1}{\frac{\text{Base side adjacent to }\angle\text{P}}{\text{Hypotenuse}}}$
$\sec\text{P}=\frac{\text{Hypotenuse}}{\text{Base side adjacent to }\angle\text{P}}$
Substituting values of sides from the above figure
$\sec\text{P}=\frac{\text{PR}}{\text{PQ}}$
$\sec\text{P}=\frac{5}{4}$
By definition,
$\sec\text{R}=\frac{1}{\cos\text{R}}$
$\sec\text{P}=\frac{1}{\frac{\text{Base side adjacent to }\angle\text{P}}{\text{Hypotenuse}}}$
$\sec\text{R}=\frac{\text{Hypotenuse}}{\text{Base side adjacent to }\angle\text{R}}$
Substituting values of sides from the above figure
$\sec\text{R}=\frac{\text{PR}}{\text{RQ}}$
$\sec\text{R}=\frac{5}{3}$
$\sin\text{P}=\frac{3}{5},\ \sin\text{R}=\frac{4}{5},\ \sec\text{P}=\frac{5}{4}$ and $\sec\text{R}=\frac{5}{3}$ View full question & answer→Question 135 Marks
Evaluate the following:
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
AnswerWe have to find the following expression
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ\dots(1)$
Now,
$\tan30^\circ=\frac{1}{\sqrt3},\ \tan60^\circ=\sqrt3,\ \tan45^\circ=1$
So by substituting above values in equation (1)
We get,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$=\Big(\frac{1}{\sqrt3}\Big)^2+(\sqrt3)^2+(1)^2$
$=\frac{1^2}{(\sqrt3)^2}+(\sqrt3)^2+1$
$=\frac{1}{3}+3+1$
$=\frac{1}{3}+4$
Now by taking LCM
We get,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ$
$=\frac{1}{3}+\frac{4\times3}{1\times3}$
$=\frac{1}{3}+\frac{12}{3}$
$=\frac{1+12}{3}$
$=\frac{13}{3}$
Therefore,
$\tan^230^\circ+\tan^260^\circ+\tan^245^\circ=\frac{13}{3}$
View full question & answer→Question 145 Marks
If $\cos\theta=\frac{5}{13},$ find the value of $\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}.$
AnswerWe have, $\cos\theta=\frac{5}{13}$ In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow(13)^2=\text{AB}^2+(5)^2$ $\Rightarrow169=\text{AB}^2+25$ $\Rightarrow\text{AB}^2=169-25$ $\Rightarrow\text{AB}=12$ $\therefore\sin\theta=\frac{12}{13}$ and $\tan\theta=\frac{12}{5} $ Now, $\frac{\sin^2\theta-\cos^2\theta}{2\sin\theta\cos\theta}\times\frac{1}{\tan^2\theta}=\frac{\Big(\frac{12}{13}\Big)^2-\Big(\frac{5}{13}\Big)^2}{2\times\frac{12}{13}\times\frac{5}{13}}\times\frac{1}{\Big(\frac{12}{5}\Big)^2}$ $=\frac{\frac{144-25}{169}}{\frac{120}{169}}\times\frac{25}{144}$ $=\frac{119}{120}\times\frac{25}{144}$ $=\frac{119\times5}{24\times144}=\frac{595}{3456}$ View full question & answer→Question 155 Marks
If $\sin\theta=\frac{3}{5},$ evaluate $\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}.$
AnswerGiven: $\sin\theta=\frac{3}{5}\ \dots(1)$
$\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}$
To find the value of
Now, we know the following trigonometric identity
$\cos^2+\sin^2\theta=1$
Therefore, by substituting the value of $\sin\theta$ from equation (1),
We get,
$\cos^2\theta+\Big(\frac{3}{5}\Big)^2=1$
Therefore,
$\cos^2\theta=1-\Big(\frac{3}{5}\Big)^2$
$=1-\frac{(3)^2}{(5)^2}$
$=1-\frac{9}{25}$
Now by taking L.C.M
We get,
$\cos^2\theta=\frac{25-9}{25}$
$=\frac{16}{25}$
Therefore by taking square root on both sides
We get,
$\cos\theta=\sqrt{\frac{16}{25}}$
$=\frac{\sqrt{16}}{\sqrt{25}}$
$=\frac{4}{5}$
Therefore,
$\cos\theta=\frac{4}{5}\ \dots(2)$
Now, we know that
$\tan\theta=\frac{\sin\theta}{\cos\theta}$
Therefore by substituting the value of $\sin\theta$ and $\cos\theta$ from equation (1) and (2) respectively
We get,
$\tan\theta=\frac{\frac{3}{5}}{\frac{4}{5}}$
$\tan\theta=\frac{3}{5}\times\frac{5}{4}$
$=\frac{3}{4}$
$\tan\theta=\frac{3}{4}\ \dots(3)$
Also, we know that
$\cot\theta=\frac{1}{\tan\theta}$
Therefore from equation (4),
We get,
$\cot\theta=\frac{1}{\frac{3}{4}}$
$=\frac{4}{3}$
Therefore,
$\cot\theta=\frac{4}{3}\ \dots(4)$
Now, by substituting the value of $\cos\theta,\tan\theta$ and $\cot\theta$ from equation (2), (3) and (4) respectively in the expression below
$\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}$
We get,
$=\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{\frac{4}{5}-\frac{1}{\frac{3}{4}}}{2\times\frac{4}{3}}$
$=\frac{\frac{4}{5}-\frac{4}{3}}{\frac{2\times4}{3}}$
$=\frac{\frac{4\times3}{5\times3}-\frac{4\times5}{3\times5}}{\frac{8}{3}}$
$=\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{\frac{12}{15}-\frac{20}{15}}{\frac{8}{3}}$
$=\frac{\frac{-8}{15}}{\frac{8}{3}}$
$=\frac{-8}{15}\times\frac{3}{8}$
$=\frac{-1}{5}$
Therefore, $\frac{\cos\theta-\frac{1}{\tan\theta}}{2\cot\theta}=\frac{-1}{5}$
View full question & answer→Question 165 Marks
If $\sec\text{A}=\frac{5}{4},$ verify that $\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}.$
AnswerGiven: $\sec\text{A}=\frac{5}{4}\dots(1)$ To verify: $=\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}\dots(2)$ Now we know that $\sec\text{A}=\frac{1}{\cos\text{A}}$ Therefore $\cos\text{A}=\frac{1}{\sec\text{A}}$ Now, by substituting the value of sec A from equation (1) We get, $\cos\text{A}=\frac{1}{\frac{5}{4}}$ $=\frac{4}{5}$ Therefore, $\cos\text{A}=\frac{4}{5}\dots(3)$ Now, we know the following trigonometric identity $\cos^2\text{A}+\sin^2\text{A}=1$ Therefore, $\sin^2\text{A}=1-\cos^2\text{A}$ Now by substituting the value of cos A from equation (3) We get, $\sin^2\text{A}=1-\Big(\frac{4}{5}\Big)^2$ $=1-\frac{\big(4\big)^2}{\big(5\big)^2}$ $=1-\frac{16}{25}$ Now by taking L.C.M We get, $\sin^2\text{A}=\frac{25-16}{25}$ $=\frac{9}{25}$ Now, by taking square root on both sides We get, $\sin\text{A}=\sqrt{\frac{9}{25}}$ $=\frac{\sqrt{9}}{\sqrt{25}}$ $=\frac{3}{5}$ Therefore, $\sin\text{A}=\frac{3}{5}\dots(4)$ Now, we know that $\tan\text{A}=\frac{\sin\text{A}}{\cos\text{A}}$ Now by substituting the value of sin A and cos A from equation (3) and (4) respectively We get, $\tan\text{A}=\frac{\frac{3}{5}}{\frac{4}{5}}$ $=\frac{3}{5}\times\frac{5}{4}$ $=\frac{3}{4}$ Therefore $\tan\text{A}=\frac{3}{4}\dots(5)$ Now from the expression of equation (2) $\text{L.H.S}=\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}$Now by substituting the value of cos A and sin A from equation (3) and (4)
We get, $\text{L}.\text{H}.\text{S}=\frac{3\Big(\frac{3}{5}\Big)-4\Big(\frac{3}{5}\Big)^3}{4\Big(\frac{4}{5}\Big)^3-3\Big(\frac{4}{5}\Big)}$ Therefore, $\text{L}.\text{H}.\text{S}=\frac{\frac{9}{5}-4\Big(\frac{27}{125}\Big)}{4\Big(\frac{64}{125}\Big)-\frac{12}{5}}$ $=\frac{\frac{9}{5}-\frac{108}{125}}{\frac{256}{125}-\frac{12}{5}}$ Now by taking L.C.M of both numerator and denominator We get, $\text{L}.\text{H}.\text{S}=\frac{\frac{9\times25}{5\times25}-\frac{108}{125}}{\frac{256}{125}-\frac{12\times25}{5\times25}}$ $=\frac{\frac{225}{125}-\frac{108}{125}}{\frac{256}{125}-\frac{300}{125}}$ $=\frac{\frac{225-108}{125}}{\frac{256-300}{125}}$ $=\frac{\frac{117}{125}}{\frac{-44}{125}}$ $=\frac{-117}{44}$ $=\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{-117}{44}\dots(6)$ Now from the expression of equation (2) $\text{R}.\text{H}.\text{S}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}$ Now by substituting the value of tan A from equation (5) We get, $\text{R}.\text{H}.\text{S}=\frac{3\Big(\frac{3}{4}\Big)-\Big(\frac{3}{4}\Big)^3}{1-3\Big(\frac{3}{4}\Big)^2}$ $=\frac{\frac{9}{4}-\frac{27}{64}}{1-\frac{3\times9}{16}}$ Now by taking L.C.M We get, $\text{R}.\text{H}.\text{S}=\frac{\frac{9\times16}{4\times16}-\frac{27}{64}}{\frac{16-27}{16}}$ $=\frac{\frac{144}{64}-\frac{27}{64}}{\frac{-11}{16}}$ $=\frac{\frac{144-27}{64}}{\frac{-11}{16}}$ $=\frac{\frac{117}{64}}{\frac{-11}{16}}$ $=\frac{117}{64}\times\frac{16}{-11}$ $16\times4=64$ Now, Therefore, $\text{R}.\text{H}.\text{S}=\frac{117}{4}\times\frac{1}{-11}$ $=\frac{117\times1}{4\times-11}$ $=\frac{117}{-44}$ $=\frac{117}{-44}$ Therefore, $\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}=\frac{-117}{44}\dots(7)$ Now by comparing equation (6) and (7) $\frac{3\sin\text{A}-4\sin^3\text{A}}{4\cos^3\text{A}-3\cos\text{A}}=\frac{3\tan\text{A}-\tan^3\text{A}}{1-3\tan^2\text{A}}$ We get, L.H.S. = R.H.S.
View full question & answer→Question 175 Marks
If $\sin (\text{A} − \text{B}) = \sin \text{A} \cos \text{B} − \cos \text{A} \sin \text{B}$ and $\cos (\text{A − B}) = \cos\text{A} \cos\text{B} + \sin \text{A} \sin \text{B},$ find the values of sin 15° and cos 15°.
AnswerGiven:
$\sin(\text {A}-\text{B})=\sin\text {A}\cos\text {B}-\cos\text{A}\sin\text {B}\ \dots(1)$
$\cos(\text {A}-\text{B})=\cos\text {A}\cos\text {B}+\sin\text{A}\sin\text {B}\ \dots(2)$
To find: The values of sin 15° and cos 15°
In this problem we need to find sin 15 and cos 15°
Hence to get 15° angle we need to choose the value of A and B such that (A - B) = 15°
So if we choose A = 45° and B = 30°
Then we get, (A - B) = 15°
Therefore by substituting A = 45° and B = 30° in equation (1)
We get,
$\sin(45^\circ-30^\circ)=\sin45^\circ\cos30^\circ-\cos45^ \circ\sin30^\circ$
Therefore,
$\sin(15^\circ)= \sin45^\circ\cos30^\circ-\cos45^\circ \sin30^\circ\ \dots(3)$
Now we know that,
$\sin45^\circ=\cos45^ \circ=\frac{1}{\sqrt{2}},\ \sin30^ \circ=\frac{1}{2},\ \cos30^\circ= \frac{\sqrt{3}}{2}$
Now by substituting above values in equation (3)
We get,
$\sin(15^\circ)=\bigg (\frac{1}{\sqrt{2}}\bigg)\times\bigg (\frac{\sqrt{3}}{2}\bigg)-\bigg(\frac {1}{\sqrt{2}}\bigg)\times\bigg(\frac{1} {2}\bigg)$
$=\frac{\sqrt{3}} {2\sqrt{2}}-\frac{1}{2\sqrt{2}}$
$=\frac{\sqrt{3}-1} {2\sqrt{2}}$
Therefore,
$\sin(15^\circ)=\frac {\sqrt{3}-1}{2\sqrt{2}}\ ....(4)$
Now by substituting A = 45° and B = 30° in equation (2)
We get,
$\cos(45^\circ-30^ \circ)=\cos45^\circ\cos30^\circ+\sin45^ \circ\sin30^\circ$
Therefore,
$\cos(15^\circ)= \cos45^\circ\cos30^\circ+\sin45^\circ \sin30^\circ\ \dots(5)$
Now we know that,
$\sin45^\circ=\cos45^ \circ=\frac{1}{\sqrt{2}},\ \sin30^ \circ=\frac{1}{2},\ \cos30^\circ=\frac {\sqrt{3}}{2}$
Now by substituting above values in equation (5)
We get,
$\cos(15^\circ)=\bigg (\frac{1}{\sqrt{2}}\bigg)\times\bigg (\frac{\sqrt{3}}{2}\bigg)+\bigg(\frac {1}{\sqrt{2}}\bigg)\times\bigg(\frac{1} {2}\bigg)$
$=\frac{\sqrt{3}} {2\sqrt{2}}+\frac{1}{2\sqrt{2}}$
$=\frac{\sqrt{3}+1} {2\sqrt{2}}$
Therefore,
$\cos(15^\circ)=\frac {\sqrt{3}+1}{2\sqrt{2}}\ \dots(6)$
Therefore from equation (4) and (6)
$\sin(15^\circ)=\frac {\sqrt{3}-1}{2\sqrt{2}}$
$\cos(15^\circ)=\frac {\sqrt{3}+1}{2\sqrt{2}}$
View full question & answer→Question 185 Marks
In the given figure, find tan P and cot R. Is tan P = cot R?
AnswerThe given figure is below:
To Find: tan P, cot R In the given right angled? $PQR$, length of side $QR$ is unknown.
Therefore, to find length of side QR we use Pythagoras Theorem
Hence, by applying Pythagoras theorem in $\triangle\text{PQR},$
We get, $PR^2 = PQ^2 + QR^2$
Now, we substitute the length of given side $PR$ and $PQ$ in the above equation
$13^2=12^2+\text{QR}^2$
$\text{QR}^2=13^2-12^2$
$\text{QR}^2=169-144$
$\text{QR}^2=25$ $\text{QR}^2=\sqrt{25}$
$\text{QR}^2=5$ By definition, we know that $\tan\text{P}=\frac{\text{Perpendicular side opposite to }\angle \text{P}}{\text{Base side adjacent to }\angle \text{P}}$
$\tan\text{P}=\frac{\text{QR}}{\text{PQ}}$
$\tan\text{P}=\frac{5}{12}\ \dots(1)$
Also, by definition, we know that $\cot\text{P}=\frac{\text{Base side adjacent to }\angle \text{R}}{\text{Perpendicular side opposite to }\angle \text{R}}$
$\cot\text{R}=\frac{\text{QR}}{\text{PQ}}$
$\cot\text{R}=\frac{5}{12}\ \dots(2)$
Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal
Therefore, L.H.S of both the equation are also equal
$\tan\text{P}=\cot\text{R}$ Yes, $\tan\text{P}=\cot\text{R}=\frac{5}{12}$ View full question & answer→Question 195 Marks
Evaluate the following:
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
AnswerWe have,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}\ \dots(1)$
Now,
$\sin30^\circ=\frac{1}{2},\ \sin90^\circ=\cos0^\circ=1,$ $\tan30^\circ=\frac{1}{\sqrt{3}},\tan60^\circ=\sqrt{3}$
So by substituting above values in equation (1)
We get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{\frac{1}{2}-1+2\times1}{\frac{1}{\sqrt{3}}\times\sqrt{3}}$
Now, $\sqrt{3}$ present in the denominator of above expression gets cancelled and we get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{\frac{1}{2}-1+2}{1}$
$=\frac{1}{2}-1+2$
Now by taking LCM in the above expression we get,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}$
$=\frac{1}{2}-\frac{1\times2}{1\times2}+\frac{2\times2}{1\times2}$
$=\frac{1}{2}-\frac{2}{2}+\frac{4}{2}$
$=\frac{1-2+4}{2}$
$=\frac{5-2}{2}$
$=\frac{3}{2}$
Therefore,
$\frac{\sin30^\circ-\sin90^\circ+2\cos0^\circ}{\tan30^\circ\tan60^\circ}=\frac{3}{2}$
View full question & answer→