Question 12 Marks
Evaluate the following:
$\frac{\sec70^\circ}{\text{cosec }20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}$
AnswerWe have to find: $\frac{\sec70^\circ}{\text{cosec }20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}$$$ Since $\frac{\sec70^\circ}{\text{cosec }20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}$ and $\sec(90^\circ-\theta)=\text{cosec }\theta$ So, $\frac{\sec70^\circ}{\text{cosec }20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}=\frac{\sec(90^\circ-20^\circ)}{\text{cosec 20}^\circ}+\frac{\sin(90^\circ-31^\circ)}{\cos31^\circ}$ $=\frac{\text{cosec }20^\circ}{\text{cosec }20^\circ}+\frac{\cos31^\circ}{\cos31^\circ}$ $= 1 + 1$ $= 2$value of $\frac{\sec70^\circ}{\text{cosec }20^\circ}+\frac{\sin59^\circ}{\cos31^\circ}\text{ is 2}$
View full question & answer→Question 22 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.$\cos78^\circ+\sec78^\circ$
Answer$\cos78^\circ=\cos(90^\circ-12^\circ)=\sin12^\circ$
$\sec78^\circ=\sec(90^\circ-12^\circ)=\text{cosec }12^\circ$
$\Rightarrow\sin12^\circ+\text{cosec }12^\circ$
View full question & answer→Question 32 Marks
Evaluate the following:
$\text{cosec 31}^\circ-\sec59^\circ$
AnswerWe have to find: $\text{cosec 31}^\circ-\sec59^\circ$$$Since $\text{cosec}(90^\circ-\theta)=\sec\theta$
So, $=\text{cosec 31}^\circ-\sec59^\circ$ $=\text{cosec}(90^\circ-59^\circ)-\sec59^\circ$ $\sec59^\circ-\sec59^\circ$ $= 0$So value of $=\text{cosec 31}^\circ-\sec59^\circ\text{ is 0}$
View full question & answer→Question 42 Marks
Find the value of x in the following:
$\cos2\text{x}=\cos60^\circ\cos30^\circ+\sin60^\circ\sin30%\circ$
Answer$\cos2\text{x}=\frac{1}{2}\cdot\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\cdot\frac{1}{2}$$\bigg[\because\cos60^\circ=\sin30^\circ=\frac{1}{2}\sin60^\circ=\cos30^\circ=\frac{\sqrt{3}}{2}\bigg]$
$\cos2\text{x}=2\cdot\frac{\sqrt{3}}{4}$
$\Rightarrow\cos2\text{x}=\frac{\sqrt{3}}{2}$
$\cos2\text{x}=\cos30^\circ$
$2\text{x}=30^\circ$
$\text{x}=15^\circ$
View full question & answer→Question 52 Marks
Evaluate:
$=\frac{\sin50^\circ}{\cos40^\circ}+\frac{\text{cosec }40^\circ}{\sec50^\circ}-4 \cos50^\circ\ \text{cosec }40^\circ$
AnswerWe will use the properties of complementary angles.
$=\frac{\sin50^\circ}{\cos40^\circ}+\frac{\text{cosec }40^\circ}{\sec50^\circ}-4 \cos50^\circ\text{cosec }40^\circ$
$ =\frac{\sin50^\circ}{\sin50^\circ}+\frac{\text{cosec }40^\circ}{\text{cosec }40^\circ}-4\frac{\cos50^\circ}{\cos50^\circ}$
$=1+1-4$
$=-2$
View full question & answer→Question 62 Marks
Evaluate the following:
$\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1$
AnswerWe have to find: $\frac{\tan35^\circ}{\cot55^\circ}+\frac{\cot78^\circ}{\tan12^\circ}-1$$$ Since $\tan(90^\circ-\theta)=\cot\theta$ and $\cot(90^\circ-\theta)=\tan\theta$ $\frac{\tan(90-55)}{\cot55}+\frac{\cot(90-12)}{\tan12}-1$$\frac{\cot55^\circ}{\cot55^\circ}+\frac{\tan12^\circ}{\tan12^\circ}-1$
$1+1-1=1$
View full question & answer→Question 72 Marks
If $\cot\theta=\frac{7}{8},$ evaluate:
$\cot^2\theta$
AnswerGiven: $\cot\theta=\frac{7}{8}$
to evoluate: $\cot^\theta$
$\cot\theta=\frac{7}{8}$
Squaring on both sides,
We get,
$(\cot\theta)^2=\Big(\frac{7}{8}\Big)^2$
$\cot^2\theta=\frac{7^2}{8^2}$
$\cot^2=\frac{49}{64}$
$\cot^2\theta=\frac{49}{64}$
View full question & answer→Question 82 Marks
If A, B, C are the interior angles of a triangle ABC, prove that.
$\tan\Big(\frac{\text{C+A}}{2}\Big)=\cot\frac{\text{ B}}{2}$
Answer$\tan\Big(\frac{\text{C+A}}{2}\Big)=\cot\frac{\text{ B}}{2}$
In $\triangle\text{ABC},$
$\text{A + B + C}=180^\circ$
$\text{A + C}=180^\circ-\text{B}\dots(1)$
$\text{L.H.S}=\tan\Big(\frac{\text{C+A}}{2}\Big)$
$=\tan\Big(\frac{180^\circ-\text{B}}{2}\Big)$ [From (1)]
$=\tan\Big(90^\circ-\frac{\text{B}}{2}\Big)$
$=\cot\Big(\frac{\text{B}}{2}\Big)$
$=\text{R.H.S}$
View full question & answer→Question 92 Marks
Evaluate the following:
$\sin35^\circ\sin55^\circ-\cos35^\circ\cos55^\circ$
AnswerWe find: $\sin35^\circ\sin55^\circ-\cos35^\circ\cos55^\circ$Since $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
$\sin35^\circ\sin55^\circ-\cos35^\circ\cos55^\circ$ $=\sin(90^\circ-55^\circ)\sin55^\circ-\cos(90^\circ-55^\circ)\cos55^\circ$ $=\cos55^\circ\sin55^\circ-\sin55^\circ\cos55^\circ$ $=1-1$$=0$
So value of $\sin35^\circ\sin55^\circ-\cos35^\circ\cos55^\circ\text{ is }0$
View full question & answer→Question 102 Marks
Write the value of $\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ?$
AnswerWe have to find: $\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
$=\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
$=\tan\big(90^\circ-80^\circ\big)\tan\big(90^\circ-75^\circ\big)\tan75^\circ\tan80^\circ$
$=\cot80^\circ\cot75^\circ\tan75^\circ\tan80^\circ$
$=\big(\cot75^\circ\tan75^\circ\big)\big(\cot80^\circ\tan80^\circ\big)$ $\big[\tan\big(90^\circ-\theta\big)=\cot\theta\big]$
$=1\times1$
$=1$ $\big[\cot\theta.\tan\theta=1\big]$
Hence the value of $\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ\text{ is }1$
View full question & answer→Question 112 Marks
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$\cos\theta=\frac{12}{15}$
AnswerGiven: $\cos\theta=\frac{12}{15}\ \dots(1)$ By definition, $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}\ \dots(2)$
By Comparing (1) and (2) We get, Base $= 12$ and Hypotenuse $= 15$
Therefore, By Pythagoras theorem, $AC^2 = AB^2 + BC^2$
Now we substitute the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC) $15^2=12^2+\text{BC}^2$
$\text{BC}^2=15^2-12^2$ $\text{BC}^2=225-144$
$\text{BC}^2=81$ $\text{BC}=\sqrt{81}$
$\text{BC}=9$
Hence, Perpendicular side = $9$ Now,
$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ Therefore.
$\sin\theta=\frac{9}{15}$
$\text{cosec }\theta=\frac{1}{\sin\theta}$ Now, Therefore,
$\text{cosec }\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}}$
$\text{cosec }\theta=\frac{15}{9}$
Now, $\sec\theta=\frac{1}{\cos\theta}$ Therefore,
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$\sec\theta=\frac{15}{12}$
Now, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$ Therefore,
$\tan\theta=\frac{9}{12}$
Now, $\cot\theta=\frac{1}{\tan\theta}$ Therefore,
$\cot\theta=\frac{\text{Base}}{\text{Perpendicular}}$
$\cot\theta=\frac{12}{9}$ View full question & answer→Question 122 Marks
If A and B are acute angles such that:
$\tan\text{A}=\frac{1}{2},\ \tan\text{B}=\frac{1}{3}$ and $\tan(\text{A + B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}},$ find A + B.
Answer$\tan\text{A}=\frac{1}{2}\ \tan\text{B}=\frac{1}{3}$
$\tan(\text{A}+\text{B})=\frac{\tan\text{A}+\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$ $=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\cdot\frac{1}{3}}=\frac{\frac{5}{6}}{1-\frac{1}{6}}=1$
$\tan(\text{A}+\text{B})=\tan45^\circ$
$\therefore\text{A}+\text{B}=45^\circ$
View full question & answer→Question 132 Marks
Prove that:$\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ=0$
AnswerWe have, $\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ.\text{cosec }20^\circ=0$
So we will calculate left hand side,
$\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cosec }20^\circ}{\sec70^\circ}-2\cos70^\circ\text{cosec }20^\circ$$=\frac{\sin70^\circ}{\cos20^\circ}+\frac{\text{cos}70^\circ}{\sin20^\circ}-2\cos70^\circ\cdot\text{cosec }(90^\circ-70^\circ)$
$=\frac{\sin(90^\circ-20^\circ)}{\cos20^\circ}+\frac{\cos(90^\circ-20^\circ)}{\sin20^\circ}-2\cos70^\circ\text.\sec70^\circ$
$=\frac{\cos20^\circ}{\cos20^\circ}+\frac{\sin20^\circ}{\sin20^\circ}-2\times1$
$=1+1-2$
$=2-2$
$=0$
Hence proved.
View full question & answer→Question 142 Marks
If A = 30° and B = 60°, verify that.
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
Answer$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
$\text{A}=30^\circ\ \text{B}=60^\circ$
$\cos(90^\circ)=\cos30^\circ\cos60^\circ-\sin30^\circ\sin60^\circ$
$=\cos(90^\circ)=\frac{1}{2}\cdot\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}\cdot\frac{1}{2}$
$0=0$
$\text{L.H.S} = \text{R.H.S}$
View full question & answer→Question 152 Marks
If A, B, C are the interior angles of a $\triangle\text{ABC},$ show that:$\sin\frac{\text{B+C}}{2}=\cos\frac{\text{A}}{2}$
AnswerWe have to prove: $\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Since we know that in triangle ABC
$\text{A + B + C}=180^\circ$
$\Rightarrow\ \text{B + C}=180^\circ-\text{A}$
Dividing by 2 on both sides, we get
$\Rightarrow\ \frac{\text{B+C}}{2}=90^\circ-\frac{\text{A}}{2}$
$\Rightarrow\ \sin \frac{\text{B+C}}{2}=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$\Rightarrow\ \sin\frac{\text{B+C}}{2}=\cos\frac{\text{A}}{2}$
View full question & answer→Question 162 Marks
Evaluate the following:
$\sec50^\circ\sin40^\circ+\cos40^\circ\text{cosec }50^\circ$
AnswerWe have to find $\sec50^\circ\sin40^\circ+\cos40^\circ\text{cosec }50^\circ$
Since $\cos(90^\circ-\theta)=\sin\theta,\sec(90^\circ-\theta)=\text{cosec }\theta$ and $\sin\theta.\text{cosec }\theta=1$
So, $\sec50^\circ\sin40^\circ+\cos40^\circ\text{cosec }50^\circ$
$=\sec(90^\circ-40^\circ)\sin40^\circ+\cos(90^\circ-50^\circ)\text{cosec }50^\circ$
$=1+1$
$=2$
So value of $\sec50^\circ\sin40^\circ+\cos40^\circ\text{cosec }50^\circ\text{ is }2$
View full question & answer→Question 172 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.$\sec76^\circ+\text{cosec }52^\circ$
Answer$\sec76^\circ=\sec(90^\circ-14^\circ)=\text{cosec }14^\circ$
$\text{cosec }52^\circ=\text{cosec }(90^\circ-88^\circ)=\sec38^\circ$
$\Rightarrow\text{cosec }14^\circ+\sec38^\circ$
View full question & answer→Question 182 Marks
If $\sin3\theta=\cos(\theta-6^\circ),$ where $3\theta$ and $\theta-6^\circ$ are acute angles, find the value of $\theta.$
Answer$\sin3\theta=\cos(\theta-6^\circ)$
$\Rightarrow\cos(90^\circ-3\theta)=\cos(\theta-6^\circ)$
$\Rightarrow90^\circ-3\theta=\theta-6^\circ$
$\Rightarrow3\theta+\theta=90^\circ+6^\circ$
$\Rightarrow4\theta=96^\circ$
$\Rightarrow\theta=\frac{96^\circ}{4}$
$\Rightarrow\theta=24^\circ$
View full question & answer→Question 192 Marks
If $\cos2\theta=\sin4\theta,$ where $2\theta$ and $4\theta$ are acute angles, find the value of $\theta.$
AnswerWe know that $ \sin(90-\theta)=\cos\theta$
$\sin4\theta=\cos2\theta$
$\sin4\theta=\sin(90-2\theta)$
$4\theta=90-2\theta$
$6\theta=90$
$\theta=\frac{90}{6}$
$\theta=15^\circ$
View full question & answer→Question 202 Marks
If $\sec2\text{A}=\text{cosec(A}-42^\circ),$ where 2A is an acute angles, find the value of A.
AnswerWe know that $(\sec(90-\theta))=\text{cosec }\theta$
$\sec2\text{A}=\sec(90-(\text{A}-42))$
$ \sec2\text{A}=\sec(90-\text{A}+42)$
$\sec2\text{A}=\sec(132-\text{A})$
Now equating both the angles we get
$2\text{A}=132-\text{A}$
$3\text{A}=132$
$\text{A}=44$
View full question & answer→Question 212 Marks
Evaluate the following:
$\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$
Answer$\sin60^\circ\cos30^\circ+\cos60^\circ\sin30^\circ$
$=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}+\frac{1}{2}\times\frac{1}{2}$
$=\frac{3}{4}+\frac{1}{4}$
$=\frac{3+1}{4}$
$=\frac{4}{4}$
$=1$
View full question & answer→Question 222 Marks
Evaluate the following:
$(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)$
AnswerWe have to find: $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)$$$Since $\sin(90^\circ-\theta)=\cos\theta$
So, $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)=(\sin72^\circ)^2-(\cos18^\circ)^2$ $=[\sin(90^\circ-18^\circ)]^2-(\cos18^\circ)^2$ $=(\cos18^\circ)^2-(\cos18^\circ)^2$ $=\cos^218^\circ-\cos^218^\circ$ $=0$So value of $(\sin72^\circ+\cos18^\circ)(\sin72^\circ-\cos18^\circ)\text{ is }0$
View full question & answer→Question 232 Marks
If A = B = 60°, verify that.
$\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$
Answer$\Rightarrow\tan(\text{A}-\text{B})=\frac{\tan\text{A}-\tan\text{B}}{1-\tan\text{A}\tan\text{B}}$ put $\text{A}=\text{B}=60^\circ$ $\Rightarrow\tan(60^\circ-60^\circ)=\frac{\tan60^\circ-\tan60^\circ}{1-\tan60^\circ\tan60^\circ}$ $\tan0^\circ=\frac{0}{1-\tan^260^\circ}$$\Rightarrow0=0$
$\Rightarrow\text{L.H.S} = \text{R.H.S}$
View full question & answer→Question 242 Marks
Express $\cos75^\circ+\cot75^\circ$ in terms of angles between 0° and 30°.
AnswerGive that: $\cos75^\circ+\cot75^\circ$
$=\cos75^\circ+\cot75^\circ$
$=\cos(90^\circ-15^\circ)+\cot(90^\circ-15^\circ)$
$=\sin15^\circ+\tan15^\circ$
Hence the correct answer is $\sin15^\circ+\tan15^\circ$ $$
View full question & answer→Question 252 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.$\tan65^\circ+\cot49^\circ$
Answer$\tan65^\circ+\cot49^\circ$
$\tan65^\circ=\tan(90^\circ-25^\circ)=\cot25^\circ$
$\cot49^\circ=\cot(90^\circ-41)=\tan(41^\circ)$
$\Rightarrow\cot25^\circ+\tan41^\circ$
View full question & answer→Question 262 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.
$\cot85^\circ+\cos75^\circ$
Answer$\cot85^\circ=\cot(90^\circ-5^\circ)=\tan5^\circ$
$\cos75^\circ=\cos(90^\circ-15^\circ)=\sin15^\circ$
$=\tan5^\circ+\sin15^\circ$
View full question & answer→Question 272 Marks
Find the value of x in the following:
$2\sin3\text{x}=\sqrt{3}$
Answer$2\sin3\text{x}=\sqrt{3}$
$\Rightarrow\sin3\text{x}=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin3\text{x}=\sin60^\circ$
$\Rightarrow3\text{x}=60^\circ$
$\Rightarrow\text{x}=\frac{60}{3}=20^\circ$
Thus, $\text{x}=20^\circ$
View full question & answer→Question 282 Marks
Evaluate the following:
$\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2$
AnswerWe have to find: $\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2$
Since $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\Big(\frac{\sin(90^\circ-41^\circ)}{\cos41^\circ}\Big)^2+\Big(\frac{\cos(90^\circ-49^\circ)}{\sin49^\circ}\Big)^2=\Big(\frac{\cos41^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\sin49^\circ}{\sin49^\circ}\Big)^2$
$= 1 + 1$
$= 2$
So value of $=\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2\text{ is }2$
View full question & answer→Question 292 Marks
Evaluate the following:
$\frac{\sec11^\circ}{\text{cosec }79^\circ}$
Answer$\frac{\sec11^\circ}{\text{cosec }79^\circ}\Rightarrow\frac{\sec(90^\circ-79^\circ)}{\text{cosec }79^\circ}=\frac{\text{cosec }79^\circ}{\text{cosec }79^\circ}$ $\big[\because\ \sec(90-\theta).\text{cosec }\theta\big]$
$=1$
View full question & answer→Question 302 Marks
Find the value of x in the following:
$2\sin\frac{\text{x}}{2}=1$
AnswerWe have,
$2\sin\frac{\text{x}}{2}=1$
$\Rightarrow\sin\frac{\text{x}}{2}=\frac{1}{2}$
$\sin30^\circ=\frac{1}{2}$
Since,
Therefore,
$\sin\frac{\text{x}}{2}=\frac{1}{2}$
$\Rightarrow\frac{\text{x}}{2}=30^\circ$
$\Rightarrow\text{x}=2\times30^\circ$
$\Rightarrow\text{x}=60^\circ$
Therefore,
$\text{x}=60^\circ$
View full question & answer→Question 312 Marks
Evaluate the following:
$\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)$
AnswerWe have to find: $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)$$$
Since $\cot(90^\circ-\theta)=\tan\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)=\frac{(\cot90^\circ-50^\circ)}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos(90^\circ-55^\circ)}{\sin55^\circ}\Big)$
$=\frac{\tan50^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\sin55^\circ}{\sin55^\circ}\Big)$
$=1-\frac{1}{2}$
$=\frac{1}{2}$
So value of $\frac{\cot40^\circ}{\tan50^\circ}-\frac{1}{2}\Big(\frac{\cos35^\circ}{\sin55^\circ}\Big)\text{ is }\frac{1}{2}$
View full question & answer→Question 322 Marks
If A, B, C are the interior angles of a $\triangle\text{ABC},$ show that:
$\cos\frac{\text{B+C}}{2}=\sin\frac{\text{A}}{2}$
Answer$\cos\frac{\text{B+C}}{2}=\sin\frac{\text{A}}{2}$
$=\text{L.H.S}$
$=\cos\frac{\text{B+C}}{2}$
$=\cos\frac{180^\circ-\text{A}}{2}$
$=\cos\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$=\sin\frac{\text{A}}{2}$
$=\text{R.H.S}$
View full question & answer→Question 332 Marks
Prove the following:
$\sin\theta\sin(90^\circ-\theta)-\cos\theta\cos(90^\circ-\theta)=0$
AnswerWe have to prove: $\sin\theta.\sin(90^\circ-\theta)-\cos\theta.\cos(90^\circ-\theta)=0$
Left hand side
$=\sin\theta.\sin(90^\circ-\theta)-\cos\theta.\cos(90^\circ-\theta)$
$=\sin\theta.\cos\theta-\cos\theta.\sin\theta$
$=\sin\theta(\cos\theta-\cos\theta)$
$=0$
= Right hand side
Proved.
View full question & answer→Question 342 Marks
In a $\triangle\text{ABC},$ right angled at B, AB = 24cm, BC = 7cm. Determine:
$\sin \text{C}, \cos \text{C}$
Answer$\triangle\text{ABC}$ is right angled at B AB = 24cm, BC = 7cm. Let 'x' be the hypotenuse, By applying Pythagoras $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\text{x}^2=24^2+7^2$ $\text{x}^2=576+49$ $\text{x}^2=625$ $\text{x}=25$$\sin\text{C}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}$
$\cos\text{C}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}$
View full question & answer→Question 352 Marks
If A = B = 60°, verify that.
$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$
Answer$\sin(\text{A}-\text{B})=\sin\text{A}\cos\text{B}-\cos\text{A}\sin\text{B}$
$\Rightarrow\sin(60^\circ-60^\circ)=\sin60^\circ\cos60^\circ-\cos60^\circ\sin^\circ60^\circ$
$\Rightarrow\sin0^\circ=\frac{\sqrt{3}}{2}\times\frac{1}{2}-\frac{1}{2}\times\frac{\sqrt{3}}{2}$
$\Rightarrow0=\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}$
$\Rightarrow0=0$
$\Rightarrow\text{L.H.S} = \text{R.H.S}$
View full question & answer→Question 362 Marks
If A, B, C are the interior angles of a triangle ABC, prove that.
$\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{ A}}{2}$
Answer$\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\Big(\frac{\text{ A}}{2}\Big)$
$\text{L.H.S}=\sin\Big(\frac{\text{B+C}}{2}\Big)$
$=\sin\Big(\frac{180^\circ-\text{A}}{2}\Big)$
$=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
$=\cos\frac{\text{A}}{2}$
$=\text{R.H.S}$
View full question & answer→Question 372 Marks
If $\theta=30^\circ,$ verify that.
$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
Answer$\tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}$
Put $\theta=30^\circ$
$\Rightarrow\tan2\times30^\circ=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$\Rightarrow\tan60^\circ=\frac{2\times\frac{1}{\sqrt{3}}}{1-\big(\frac{1}{\sqrt{3}}\big)^2}$
$\Rightarrow\sqrt{3}=\frac{2}{\sqrt{3}}\times\frac{1}{1-\frac{1}{3}}$
$\Rightarrow\sqrt{3}=\frac{2}{\sqrt{3}}\times\frac{3}{2}$
$\Rightarrow\sqrt{3}=\sqrt{3}$
$\Rightarrow\text{L.H.S}=\text{R.H.S}$
View full question & answer→Question 382 Marks
State whether the following are true or false. Justify your answer.
cos A is the abbreviation used for the cosecant of angle A.
Answercos A is the abbreviation used for cosecant of angle A.
The given statement is false.
$\therefore$ cos A is abbreviation used for cos of angle A but not for cosecant of angle A.
View full question & answer→Question 392 Marks
Write the acute angle $\theta$ satisfying $\sqrt{3}\sin\theta=\cos\theta.$
AnswerWe have: $\sqrt{3}\sin\theta=\cos\theta$
$\Rightarrow\sqrt{3}\sin\theta=\cos\theta$
$\Rightarrow\frac{\sin\theta}{\cos\theta}=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\theta=\tan30^\circ$
$\Rightarrow\theta=30^\circ$
Hence the acute angle is 30°
View full question & answer→Question 402 Marks
Evaluate:
$=\tan7^\circ\tan23^\circ\tan60^\circ\tan67^\circ\tan83^\circ$
AnswerWe will use the properties of complementary angles.
$=\tan7^\circ\tan23^\circ\tan60^\circ\tan67^\circ\tan83^\circ$
$ =\cos83^\circ\cot67^\circ\tan60^\circ\tan67^\circ\tan83^\circ$
$=\sqrt{3}$
View full question & answer→Question 412 Marks
Evaluate the following:
$\frac{\sin21^\circ}{\cos69^\circ}$
Answer$\frac{\sin21^\circ}{\cos69^\circ}=\frac{\sin(90-69^\circ)}{\cos69^\circ}$
$=\frac{\cos69^\circ}{\cos69^\circ}$
$=1$
View full question & answer→Question 422 Marks
Evaluate the following:
$\frac{\tan10^\circ}{\cot80^\circ}$
Answer$\frac{\tan10^\circ}{\cot80^\circ}\Rightarrow\frac{\tan(90^\circ-80^\circ)}{\cot80^\circ}=\frac{\cot80^\circ}{\cot80^\circ}$ $\big[\because \sin(90^\circ-\theta)=\cot\theta\big]$
$=1$
View full question & answer→Question 432 Marks
If $\sin \theta = \cos (\theta − 45^\circ),$ where $\theta$ and $\theta-45^\circ$ are acute angles, find the degree measure of $\theta.$
AnswerWe have,
$\sin\theta=\cos(\theta-45^\circ)$
$\Rightarrow\cos(90^\circ-\theta)=\cos(\theta-45^\circ)$
$\Rightarrow90^\circ-\theta=\theta-45^\circ$
$\Rightarrow\theta+\theta=90^\circ+45^\circ$
$\Rightarrow2\theta=135^\circ$
$\Rightarrow\theta=\frac{135^\circ}{2}$
$\Rightarrow \theta=67.5^\circ$
View full question & answer→Question 442 Marks
Evaluate the following:
$\cos^230^\circ+\cos^245^\circ+\cos^260^\circ+\cos^290^\circ$
Answer$\cos^230^\circ+\cos^245^\circ+\cos^260^\circ+\cos^290^\circ=\Big(\frac{\sqrt3}{2}\Big)^2+\Big(\frac{1}{\sqrt2}\Big)^2+\Big(\frac{1}{2}\Big)^2+(0)^2$
$=\frac{3}{4}+\frac{1}{2}+\frac{1}{4}+0$
$=\frac{3+2+1}{4}$
$=\frac{6}{4}=\frac{3}{2}$
View full question & answer→Question 452 Marks
Evaluate:
$ \frac{\cos58^\circ}{\sin32^\circ}+\frac{\sin22^\circ}{\cos68^\circ}-\frac{\cos38^\circ\text{cosec }52^\circ}{\tan18^\circ\tan35^\circ\tan60^\circ\tan72^\circ\tan65^\circ}$
Answer$ \cos58^\circ=\cos(90^\circ-32^\circ)=\sin32^\circ$
$ \sin22^\circ=\sin(90^\circ-68^\circ)=\cos68^\circ$
$ \cos38^\circ=\cos(90-52)=\sin52^\circ$
$ \tan18^\circ=\cot72\tan35^\circ=\cot55^\circ$
$ \Rightarrow \frac{\sin32^\circ}{\sin32^\circ}+\frac{\cos68^\circ}{\cos68^\circ}-\frac{\sin52 \text{ cosec 52}}{\tan72.\cot72\tan55\cot55.\tan60}$
$ =1+1-\frac{1}{\sqrt{3}}=\frac{2\sqrt{3}-1}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}}=\frac{6-\sqrt{3}}{3}$
View full question & answer→Question 462 Marks
Evaluate the following:
$\frac{\sin20^\circ}{\cos70^\circ}$
Answer$\Rightarrow\frac{\sin(90^\circ-70^\circ)}{\cos70^\circ}\Rightarrow\frac{\cos70^\circ}{\cos70^\circ}$ $\big[\because\sin(90^\circ-\theta)=\cos\theta\big]$
$\Rightarrow\frac{\cos70^\circ}{\cos70^\circ}=1$
View full question & answer→Question 472 Marks
Evaluate the following:
$\cos48^\circ-\sin42^\circ$
AnswerWe have to find: $\cos48^\circ-\sin42^\circ$ Since $\cos(90^\circ-\theta)=\sin\theta$So, $\cos48^\circ-\sin42^\circ=\cos(90^\circ-42^\circ)-\sin42^\circ$
$=\sin42^\circ-\sin42^\circ$ $= 0$ So value of $\cos48^\circ-\sin42^\circ\text{ is }0$
View full question & answer→Question 482 Marks
Evaluate the following:
$\text{cosec}^330^\circ\cos60^\circ\tan^345^\circ\sin^290^\circ\sec^245^\circ\cot30^\circ$
Answer$\text{cosec}^330^\circ\cos60^\circ\tan^345^\circ\sin^290^\circ\sec^245^\circ\cot30^\circ$
$=(2)^3\times\frac{1}{2}\times(1)^3\times(1)^2\times(\sqrt{2})^2\times\sqrt{3}$
$=8\times\frac{1}{2}\times1\times1\times2\times\sqrt{3}$
$=8\sqrt{3}$
View full question & answer→Question 492 Marks
Evaluate:
$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$
Answer$4(\sin^430^\circ+\cos^460^\circ)-\frac{2}{3}(\sin^260^\circ-\cos^245^\circ)+\frac{1}{2} \ \tan^260^\circ$
$= 4\bigg\{\Big(\frac{1}{2}\Big)^4+\Big(\frac{1}{2}\Big)^4\bigg\}-\frac{2}{3}\bigg\{\Big(\frac{\sqrt{3}}{2}\Big)^2-\Big(\frac{1}{\sqrt{2}}\Big)^2\bigg\}+\frac{1}{2}\times\big(\sqrt{3}\big)^2$
$ =4\Big(\frac{1}{16}+\frac{1}{16}\Big)-\frac{2}{3}\Big(\frac{3}{4}-\frac{1}{2}\Big)+\frac{3}{2}$
$ =4\Big(\frac{2}{16}\Big)-\frac{2}{3}\Big(\frac{3-2}{4}\Big)+\frac{3}{2}$
$ =\frac{8}{16}-\frac{2}{3}\times\frac{1}{4}+\frac{3}{2}$
$ =\frac{1}{2}-\frac{1}{6}+\frac{3}{2}$
$ = \frac{3-1+9}{6}$
$=\frac{11}{6}$
View full question & answer→Question 502 Marks
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$\cos\theta=\frac{7}{25}$
Answer$\cos\theta=\frac{7}{25}.$ We know that $\cos\theta=\frac{\text{adjacent side}}{\text{hypoteunse}}$ Now consider a right angled $ \triangle^\text{le}\text{ABC,}$
Let x be the opposite's side. By applying pythagorn's theorem $\text{AC}^2=\text{AB}^2+\text{BC}^2$ $(25)^2=\text{x}^2+7^2$ $625-49=\text{x}^2$ $576=\sqrt{576}=24$ $\sin \theta=\frac{\text{opposite side}}{\text{hypotenuse}}=\frac{24}{25}$ $\tan \theta=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{24}{7}$ $\text{cosec }\theta=\frac{1}{\sin\theta}=\frac{1}{\frac{24}{25}}=\frac{25}{24}$ $\sec\theta=\frac{1}{\cos\theta}=\frac{1}{\frac{7}{25}}=\frac{25}{7}$ $\cot\theta=\frac{1}{\tan\theta}=\frac{1}{\frac{24}{7}}=\frac{7}{24}$ View full question & answer→Question 512 Marks
Evaluate:
$=\frac{\sin18^\circ}{\cos72^\circ}+\sqrt{3}\ \{\tan10^\circ\tan30^\circ\tan40^\circ\tan50^\circ\tan80^\circ\}$
AnswerWe will use the properties of complementary angles.
$=\frac{\sin18^\circ}{\cos72^\circ}+\sqrt{3}\ (\tan10\tan30\tan40\tan50\tan80)$
$ =\frac{\sin18^\circ}{\sin18^\circ}+\sqrt{3}\ (\cot80\tan30\cot50\tan50\tan80)$
$ =1+\sqrt{3}\Big(\frac{1}{\sqrt{3}}\Big)$
$ =2$
View full question & answer→Question 522 Marks
Prove that:$\sin48^\circ\sec42^\circ+\cos48^\circ\text{cosec }42^\circ=2$
Answer$\text{L.H.S}=\sin48^\circ\sec42^\circ+\cos48^\circ\text{cosec }42^\circ$
$=\sin48\sec(90^\circ-48^\circ)+\cos48^\circ\text{cosec }(90^\circ-48^\circ)$
$=\sin48^\circ\cdot\text{cosec }48^\circ+\cos48^\circ\cdot\sec48^\circ$
$=\sin48^\circ\times\frac{1}{\sin48^\circ}+\cos48^\circ\times\frac{1}{\cos48^\circ}$
$=1+1$
$=2$
$=\text{R.H.S}$
View full question & answer→Question 532 Marks
Evaluate the following:
$\frac{\sin19^\circ}{\cos71^\circ}$
Answer$\Rightarrow\frac{\sin(90^\circ-71^\circ)}{\cos71^\circ}=\frac{\cos71^\circ}{\cos71^\circ}$
$=1$
View full question & answer→Question 542 Marks
If $\theta=30^\circ,$ verify that.
$\cos3\theta=4\cos^3\theta-3\cos\theta$
AnswerGiven:
$\theta=30^\circ\dots(1)$
To verify:
$\cos3\theta=4\cos^3\theta-3\cos\theta\dots(2)$
Now consider left hand side of the expression in equation (2)
Therefore
$\cos3\theta=\cos3\times30$
$=\cos90$
$=0$
Now consider right hand side of the expression to be verified in equation (2)
Therefore,
$4\cos^3\theta-3\cos\theta=4\cos^330-3\cos30$
$=4\bigg(\frac{\sqrt{3}}{2}\bigg)^3-3\times\frac{\sqrt{3}}{2}$
$=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}$
$=0$
Hence it is verified that,
$\cos3\theta=4\cos^3\theta-3\cos\theta$
View full question & answer→Question 552 Marks
State whether the following are true or false. Justify your answer.
cot A is the product of cot and A.
Answercot A is a trigonometric ratio which means cotangent of angle A.
Hence, cot A is the product of cot and A is False.
View full question & answer→Question 562 Marks
In a $\triangle\text{ABC}$ right angled at B, $\angle\text{A}=\angle\text{C.}$ Find the values of.
$\sin\text{A}\cos\text{C}+\cos\text{A}\sin\text{C}$
AnswerIn $\triangle\text{le}\text{ ABC}\ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$ $\angle\text{A}+90^\circ+\angle\text{A}=180^\circ$ $2\angle\text{A}=90^\circ$ $\angle\text{A}=45^\circ$$\therefore\ \angle\text{A}=45^\circ$
$\sin45^\circ\cos45^ \circ+\cos45^\circ\sin45^\circ$ $\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt {2}}+\frac{1}{\sqrt{2}}\cdot\frac{1}{\sqrt {2}}=\frac{1}{2}+\frac{1}{2}=1$
View full question & answer→Question 572 Marks
Prove that:
$\frac{\cos80^\circ}{\sin10^\circ}+\cos59^\circ\text{cosec }31^\circ=2$
Answer$\cos80^\circ=\cos(90^\circ-10^\circ)=\sin10^\circ$
$\cos59^\circ=\cos(90^\circ-31^\circ)=\sin31^\circ$
$\Rightarrow\frac{\sin10^\circ}{\sin10^\circ}+\sin31^\circ\text{cosec }31^\circ$
$=1+1=2$ $[\because\sin\theta\text{ cosec }\theta=1]$
Hence proved.
View full question & answer→Question 582 Marks
Evaluate the following:
$\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
AnswerWe have to find $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$
Since $\tan(90^\circ-\theta)=\cot\theta$
So, $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ$ $=\tan(90^\circ-42^\circ)\tan(90^\circ-67^\circ)\tan42^\circ\tan67^\circ$
$=\cot42^\circ\cot67^\circ\tan42^\circ\tan67^\circ$
$=(\tan67^\circ\cot67^\circ)(\tan42^\circ\cot42^\circ)$
$=1\times1$
$=1$
So value of $\tan48^\circ\tan23^\circ\tan42^\circ\tan67^\circ\text{ is }1$
View full question & answer→Question 592 Marks
In a rectangle ABCD, AB = 20cm, $\angle\text{BAC} = 60^\circ,$ calculate side BC and diagonals AC and BD.
AnswerWe have drawn the following figure
Since ABCD is a rectangle Therefore, $\angle\text{ABC}=\angle\text{BCD}=90^\circ$ Now, consider $\triangle\text{ABC}$ We know that sum of all the angles of any triangle is 180° Therefore, $\angle\text{BAC}+\angle\text{ABC}+\angle\text{ACB}=180^\circ\dots(1)$ Now by substituting the values of known angles $\angle\text{BAC}$ and $\angle\text{ABC}$ in equation (1) We get, $60^\circ+90^\circ+\angle\text{ACB}=180^\circ$ $\Rightarrow150^\circ+\angle\text{ACB}=180^\circ$ $\Rightarrow\angle\text{ACB}=180^\circ-150^\circ$ $\Rightarrow\angle\text{ACB}=30^\circ$ Now in $\triangle\text{ABC}$ We know that, $\cos\text{A}=\cos60^\circ$ $\Rightarrow\frac{\text{AB}}{\text{AC}}=\cos60^\circ$ Now we have, $\cos60^\circ=\frac{1}{2}$ AB = 20cm and Therefore by substituting above values in equation (2) We get, $\cos\text{A}=\cos60^\circ$ $\Rightarrow\frac{20}{\text{AC}}=\frac{1}{2}$ Now by cross multiplying we get, $20\times2=1\times\text{AC}$ $\Rightarrow40=\text{AC}$ $\Rightarrow\text{AC}=40$ Therefore, $\text{AC}=40\text{cm}\ \dots(3)$ Now in $\triangle\text{ABC}$ We know that, $\sin\text{A}=\sin60^\circ$ $\Rightarrow\frac{\text{BC}}{\text{AC}}=\sin60^\circ$ Now we have from equation (3), $\sin60^\circ=\frac{\sqrt{3}}{2}$ AC=40cm and Therefore by substituting above values in equation (4) We get, $\sin\text{A}=\sin60^\circ$ $\Rightarrow\frac{\text{BC}}{40}=\frac{\sqrt{3}}{2}$ Now by cross multiplying we get, $\text{BC}\times2=\sqrt{3}\times40$ $\Rightarrow\text{BC}=\frac{\sqrt{3}\times40}{2}$ $\Rightarrow\text{BC}=20\sqrt{3}\text{m}$ Therefore, $\Rightarrow\text{BC}=20\sqrt{3}\text{m}\ \dots(5)$ Since ABCD is a rectangle Therefore, $\text{AB}=\text{CD}=20\text{cm}\ \dots(6)$ And $\text{BC}=\text{AD}=20\sqrt{3}\text{cm}\ \dots(7)$ Now in $\triangle\text{BCD}$ We know that, $\tan\text{B}=\frac{\text{CD}}{\text{BC}}$ Now by substituting the values of sides from equation (6) and (7) We get, $\Rightarrow\tan\text{B}=\frac{20}{20\sqrt{3}}$ $\Rightarrow\tan\text{B}=\frac{1}{\sqrt{3}}$ Since $\tan30^\circ=\frac{1}{\sqrt{3}}$ Therefore, $\angle\text{B}=30^\circ$ That is in $\triangle\text{BCD}$ $\angle\text{DBC}=30^\circ\ \dots(8)$ NOW in $\triangle\text{BCD}$ We know that, $\cos\text{B}=\frac{BC}{BD}$ From equation (7) and (8) $\Rightarrow\cos30^\circ=\frac{20\sqrt{3}}{\text{BD}}$ Since $\Rightarrow\cos30^\circ=\frac{\sqrt{3}}{2}$ Therefore, $\frac{\sqrt{3}}{2}=\frac{20\sqrt{3}}{\text{BD}}$ Now by cross multiplying we get, $\sqrt{3}\times\text{BD}=20\sqrt{3}\times2$ $\Rightarrow\sqrt{3}\times\text{BD}=40\sqrt{3}$ $\Rightarrow\text{BD}=\frac{40\sqrt{3}}{\sqrt{3}}$ $\Rightarrow\text{BD}=40$ Therefore, $\text{BD}=40\text{cm}\ \dots(9)$ Hence from equation (3), (5) and (9) $\text{AC}=40\text{cm},\ \text{BC}=20\sqrt{3}\text{cm},\ \text{BD}=40\text{cm}$ View full question & answer→Question 602 Marks
Write the maximum and minimum values of $\cos\theta.$
AnswerThe maximum value of $\cos\theta$ is 1 and the minimum value of $\cos\theta$ is -1 because value of $\cos\theta$ lies between -1 and 1.
View full question & answer→Question 612 Marks
If $\sec4\text{A}=\text{cosec(A}-20^\circ),$ where 4A is an acute angles, find the value of A.
AnswerGiven: $\sec4\text{A}=\text{cosec}(\text{A}-20^\circ)$ and 4A is an acute angle
We have to find $ \theta$
Now
$\sec4\text{A}=\text{cosec}(\text{A}-20^\circ)$
$ \sec4\text{A}=\sec\big\{90^\circ-(\text{A}-20^\circ)\big\}$
$ \sec4\text{A}=\sec(90^\circ-\text{A}+20^\circ)$
$ \sec4\text{A}=\sec(110^\circ-\text{A})$
$ 5\text{A}=110^\circ$
$ \text{A}=22^\circ$
Hence the value of A is 22º
View full question & answer→Question 622 Marks
If A + B = 90° and $\tan\text{A}=\frac{3}{4},$ what is cot B?
AnswerWe have,
$\text{A}+\text{B}=90^\circ$
So, $\text{A}=90^\circ-\text{B}$
And, $\tan\text{A}=\frac{3}{4}$
$\Rightarrow\tan\text{A}=\tan(90^\circ-\text{B})$
$\Rightarrow\cot\text{B}=\frac{3}{4}$
View full question & answer→Question 632 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.$\text{cosec }54^\circ+\sin72^\circ$
Answer$\text{cosec }54^\circ=\text{cosec }(90^\circ-36^\circ)=\sec36^\circ$
$\sin72^\circ=\sin(90^\circ-18^\circ)=\cos18^\circ$
$\Rightarrow \sec36^\circ+\cos18^\circ$
View full question & answer→Question 642 Marks
State whether the following are true or false. Justify your answer.
$\sin\theta=\frac{4}{3}$ for some angle $\theta.$
Answer$\sin\theta=\frac{4}{3}$ for some angle $\theta$
Given statement is false
Since value of $\sin\theta$ is less than (or) equal to one.
Here value of $\sin\theta$ exceeds one, so given statement is false.
View full question & answer→Question 652 Marks
If $2\theta + 45^\circ$ and $30^\circ-\theta$ are acute angles, find the degree measure of $\theta$ satisfying $\sin (2\theta + 45^\circ) = \cos (30^\circ − \theta).$
AnswerHere $2\theta+45^\circ$ and $30-\theta^\circ$ are acute angles:
We know that $ (90-\theta)=\cos\theta$
$\sin(2\theta+45^\circ)=\sin(90-(30-\theta))$
$ \sin(2\theta+45^\circ)=\sin(90-30+\theta)$
$\sin(2\theta+45^\circ)=\sin(60+\theta)$
On equting sin of angle of we get
$ 2\theta+45=60+\theta$
$2\theta-\theta=60-45$
$\theta=15^\circ$
View full question & answer→Question 662 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.$\sin59^\circ+\cos56^\circ$
Answer$\sin59^\circ=\sin(90^\circ-59^\circ)=\cos31^\circ$
$\cos56^\circ=\cos(65^\circ-34^\circ)=\sin34^\circ$
$\Rightarrow\cos31^\circ+\sin34^\circ$
View full question & answer→Question 672 Marks
Evaluate:
$=\frac{3\cos55^\circ}{7\sin35^\circ}-\frac{4(\cos70^\circ\text{cosec}20^\circ)}{7(\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}$
AnswerWe will use the properties of complementary angles. $=\frac{3\cos55^\circ}{7\sin35^\circ}-\frac{4(\cos70^\circ\text{cosec}20^\circ)}{7(\tan5^\circ\tan25^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}$ $ =\frac{3\cos55^\circ}{7\cos55^\circ}-\frac{4(\cos70^\circ\sec70^\circ)}{7(\cot85^\circ\cot65^\circ\tan45^\circ\tan65^\circ\tan85^\circ)}$ $ =\frac{3}{7}-\frac{4}{7}$$=-\frac{1}{7}$
View full question & answer→Question 682 Marks
If $\tan\text{A}=\frac{3}{4}$ and A + B = 90°, then what is the value of cot B?
AnswerGiven that:
$\text{A}+\text{B}=90^\circ$
$\tan\text{A}=\frac{3}{4}$
$\text{A}+\text{B}=90^\circ$
$\Rightarrow\text{B}=90^\circ-\text{A}$
$\Rightarrow\cot\text{B}=\cot(90^\circ-\text{A})$
$\Rightarrow\cot\text{B}=\tan\text{A}$
$\Rightarrow\cot\text{B}=\frac{3}{4}\big[\cot\big(90^\circ-\text{A}\big)=\tan\text{A}\big]$
Hence the value of cot B is $\frac{3}{4}$
View full question & answer→Question 692 Marks
Find the value of x in the following:
$\sqrt{3}\sin\text{x}=\cos\text{x}$
Answer$\sqrt{3}\sin\text{x}=\cos\text{x}$
$\frac{\sin\text{x}}{\cos\text{x}}=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\text{x}=\frac{1}{\sqrt{3}}$
$\Rightarrow\tan\text{x}=\tan30^\circ$
$\Rightarrow\text{x}=30^\circ$
View full question & answer→Question 702 Marks
Express the following in terms of trigonometric ratios of angles lying between 0° and 45°.
$\sin67^\circ+\cos75^\circ$
Answer$\sin67^\circ=\sin(90^\circ-23^\circ)=\cos23^\circ$
$\cos75^\circ=\cos(90^\circ-15^\circ)=\sin15^\circ$
$=\cos23^\circ+\sin15^\circ$
View full question & answer→Question 712 Marks
State whether the following are true or false. Justify your answer.
The value of tan A is always less than 1.
Answer$\tan\text{A}\angle1$
Value of tan A at 45° i.e., tan 45 = 1
As value of A increases to 90°
Tan A becomes infinite
So, given statement is false.
View full question & answer→Question 722 Marks
Evaluate the following:
$\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2$
AnswerWe have to find: $\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2$$$
Since $\sin(90^\circ-\theta)=\cos\theta$ and $\cos(90^\circ-\theta)=\sin\theta$
So, $\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2=\bigg[\frac{\sin(90^\circ-63^\circ)}{\cos63^\circ}\bigg]^2-\bigg[\frac{\cos(90^\circ-27^\circ)}{\sin27^\circ}\bigg]^2$
$=\Big(\frac{\cos63^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\sin27^\circ}{\sin27^\circ}\Big)^2$
$= 1 - 1$
$= 0$
So value of $\Big(\frac{\sin27^\circ}{\cos63^\circ}\Big)^2-\Big(\frac{\cos63^\circ}{\sin27^\circ}\Big)^2\text{ is }0$
View full question & answer→Question 732 Marks
If $\sin 3\text{A} = \cos (\text{A} - 26^\circ),$ where 3A is an acute angles, find the value of A.
Answer$\cos\theta=\sin(90^\circ-\theta)$
$\Rightarrow\cos(\text{A}-26)=\sin(90^\circ-(\text{A}-26^\circ))$
$\Rightarrow\sin3\text{A}=\sin(90^\circ-(\text{A}-26))$
Equating anglees on both sides,
$3\text{A}=90^\circ-\text{A}+26^\circ$
$4\text{A}=116^\circ\Rightarrow\text{A}=\frac{116}{4}=29^\circ$
$\therefore\ \text{A}=29^\circ$
View full question & answer→Question 742 Marks
Prove that:$\tan20^\circ\tan35^\circ\tan45^\circ\tan55^\circ\tan70^\circ=1$
Answer$\tan20^\circ=\tan(90^\circ-70^\circ)=\cot70^\circ$
$\tan35^\circ=\tan(90^\circ-55^\circ)=\cot55^\circ$
$\tan45^\circ=1$
$\Rightarrow\cot70^\circ\tan70^\circ\times\cot55^\circ\tan55^\circ\tan45^\circ$
$\Rightarrow1\times1\times1=1$
Hence proved.
View full question & answer→Question 752 Marks
What is the maximum value of $\frac{1}{\text{cosec }\theta}?$
AnswerThe maximum value of $\frac{1}{\text{cosec }\theta}$ is 1 because the maximum value of $\sin\theta$ is 1 that is
$\frac{1}{\text{cosec }\theta}=\sin\theta$
$\frac{1}{\text{cosec }\theta}=1$
View full question & answer→Question 762 Marks
If $\cos\theta=\frac{2}{3},$ find the value of $\frac{\sec\theta-1}{\sec\theta+1}.$
Answer$\cos\theta=\frac{2}{3}\Rightarrow\sec\theta=\frac{3}{2}$
$\therefore\frac{\sec\theta-1}{\sec\theta+1}=\frac{\frac{3}{2}-1}{\frac{3}{2}+1}$
$=\frac{\frac{1}{2}}{\frac{5}{2}}=\frac{1}{2}\times\frac{2}{5}=\frac{1}{5}$
View full question & answer→Question 772 Marks
$\frac{2\sin68} {\cos22}-\frac{2\cot15^\circ}{5\tan75^ \circ}-\frac{3\tan45^\circ\tan20^\circ \tan40^\circ\tan50^\circ\tan70^\circ} {5}$
Answer$\sin68^\circ=\sin(90 -22)=\cos22$
$\cot15^\circ=\cot(90 -75)=\tan75 $
$2\cdot\frac{\cos22} {\cos22}-\frac{2\tan75^\circ}{5\tan75^ \circ}-\frac{3\tan45^\circ\tan20^\circ \tan40^\circ\cot40^\circ\cot20^\circ} {5}\ $
$=2-\frac{2}{5}-\frac {3}{5}=2-1=1$
View full question & answer→Question 782 Marks
Evaluate:
$=\tan35^\circ \tan40 \tan50^\circ \tan55^\circ$
Answer$ \tan35^\circ=\tan(90^\circ-55^\circ)=\cot55^\circ$
$ \tan40^\circ=\tan(90^\circ-50^\circ)=\cot+50^\circ$
$=\cot55^\circ\cdot\cot50^\circ\cdot\tan50^\circ\cdot\tan55^\circ$
$=\cot55^\circ\cdot\tan55^\circ\cdot\cot50^\circ\cdot\tan50^\circ$
$=1$
View full question & answer→Question 792 Marks
Evaluate:
$\text{cosec }(65^\circ+\theta)-\sec(25^\circ-\theta)-\tan(55^\circ-\theta)+\cot(35^\circ+\theta)$
Answer$\text{cosec}(65^\circ+\theta)-\sec(25^\circ-\theta)-\tan(55^\circ-\theta)+\cot(35^\circ+\theta)$
$ = \text{cosec}\big\{90^\circ-(25^\circ-\theta)\big\}-\sec(25^\circ-\theta)-\tan(55^\circ-\theta)+\cot\big\{90^\circ-(55^\circ-\theta)\big\}$
$ =\sec(25^\circ-\theta)-\sec(25^\circ-\theta)-\tan(55^\circ-\theta)+\tan(55^\circ-\theta)$
$= 0$
View full question & answer→