Question 11 Mark
Write ‘True’ or ‘False’ and justify your answer.
$\frac{\tan47^\circ}{\cot43^\circ}=1$
AnswerTrue.47º and 43º are complementry angles.
$\therefore\ \frac{\tan47^\circ}{\cot43^\circ}=\frac{\tan47^\circ}{\cot(90^\circ-47^\circ)}=\frac{\tan47^\circ}{\tan47^\circ}=1$
Hence, the given expression is true.
View full question & answer→Question 21 Mark
State whether the following are true or false. Justify your answer.$\sin \theta=\frac{4}{3} $ for some angle $\theta.$
AnswerFalse as $\sin \theta$ cannot be > 1.
View full question & answer→Question 31 Mark
Write ‘True’ or ‘False’ and justify your answer.
The value of the expression (sin80º - cos80º) is negative.
AnswerFalse.80º is near to 90º, sin90º = 1 and cos90º = 0
So, the given expression sin80º - cos80º > 0
So, the value of the given expression is posittive. So, the given statement is false.
View full question & answer→Question 41 Mark
Write 'True' or 'False' and justify your answer in the following:
The value of $\sin\theta+\cos\theta$ is always greater than 1.
AnswerFalse.
The value of $(\sin\theta+\cos\theta)\text{ for }\theta=0^\circ$ is 1.
View full question & answer→Question 51 Mark
Write 'True' or 'False' and justify your answer in the following:
The value of the expression $\sin80^\circ-\cos80^\circ$ is negative.
AnswerFalse.
$\sin80^\circ-\cos80^\circ$
$=\sin80^\circ-\sin(90^\circ-80^\circ)$
$=\sin80^\circ-\sin10^\circ\text{is }+\text{ve} $
$\because\ \sin\theta$ increase as $\theta$ increase.
View full question & answer→Question 61 Mark
State whether the following are true or false. Justify your answer.The value of $\sin \theta$ increases as $\theta$ increases.
AnswerValue of $\sin q$ as $\theta$ increases.True.
View full question & answer→Question 71 Mark
Write ‘True’ or ‘False’ and justify your answer.
The value of $2\sin\theta\text{can be}\Big(\text{a}+\frac{1}{\text{a}}\Big),$ where a is a positive number, and $\text{a}\neq1.$
AnswerFalse.Consider 'a' and$\frac{'1'}{\text{a}}$ as positive numbers and $\text{a}\neq0$
Arithumetic mean (AM) of a and $\frac{1}{\text{a}}=\frac{\Big(\text{a}+\frac{1}{\text{a}}\Big)}{2}$
Geometric mean (GM) of a and $\frac{1}{\text{a}}=\sqrt{\text{a}\times\frac{1}{\text{a}}}=1$
$\because\ \text{AM}>\text{GM}$
$\therefore\ \frac{\Big(\text{a}+\frac{1}{\text{a}}\Big)}{2}>1$
$\Rightarrow\ \Big(\text{a}+\frac{1}{\text{a}}\Big)>2$
Let $\text{a}+\frac{1}{\text{a}}=2\sin\theta$
$\Rightarrow\ 2\sin\theta>2$
$\Rightarrow\ \sin\theta>1$
Which can never be possible.
Hence, our consideration that $\text{a}+\frac{1}{\text{a}}=2\sin\theta$ is false.
Alternate Answer
a is positive and $\text{a}\neq1$ i.e., a can be above 0 all real and values except 1.
Let 0 < a < 1 then $\frac{1}{\text{a}}$ will be more then 1 so $\text{a}+\frac{1}{\text{a}}>2$ for any value of a.
Let $\text{a}=0.2\Rightarrow\text{a}+\frac{1}{\text{a}}=0.2+\frac{1}{0.2}=5.2$
$\text{a}=0.9\Rightarrow\text{a}+\frac{1}{\text{a}}=0.9+\frac{1}{0.9}=0.9+1.111=2.011$
Put $\text{a}+\frac{1}{\text{a}}=2\sin\theta$
$\therefore\ 2\sin\theta>2$
$\Rightarrow\ \sin\theta>1$
Which is impossible so $2\sin\theta\neq\text{a}+\frac{1}{\text{a}}$
Hence, the given statement is false. If we take any value of a more than one, then the value of $\text{a}+\frac{1}{\text{a}}$ is always greater then 2 which repeats the result.
View full question & answer→Question 81 Mark
State whether the following are true or false. Justify your answer.$\sin \theta=\cos \theta$ for all values of $\theta$.
Answer$\sin \theta=\cos \theta$ for all values of $\theta$.False.
View full question & answer→Question 91 Mark
Write 'True' or 'False' and justify your answer in the following:
$\cos\theta=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}$, where a and b are two lab distinct numbers such that ab > 0.
AnswerFalse.
Given, a and b are two distinct numbers such that ab > 0.
Using, AM > GM
[since, AM and GM of two number a and b are $\frac{\text{a}+\text{b}}{2}$ and $\sqrt{\text{ab}},$ respectively]
$\Rightarrow\ \frac{\text{a}^2+\text{b}^2}{2}>\sqrt{\text{a}^2\times\text{b}^2}$
$\Rightarrow\ \text{a}^2+\text{b}^2>2\text{ab}$
$\Rightarrow\ \frac{\text{a}^2+\text{b}^2}{2\text{ab}}>1\ \Big[\because\ \cos\theta=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}\Big]$
$\Rightarrow\ \cos\theta>1\ \big[\because\ -1\leq\cos\theta\leq1\big]$
Which is not possible.
Hence, $\cos\theta\neq\frac{\text{a}^2+\text{b}^2}{2\text{ab}}$
View full question & answer→Question 101 Mark
State whether the following are true or false. Justify your answer.$\sec \text{A}=\frac{12}{5} $ for some value of angle A.
AnswerTrue as $\sec \text{A}$ is always greater than 1.
View full question & answer→Question 111 Mark
Write 'True' or 'False' and justify your answer in the following:
The value of $\sin\theta$ is $\text{x}+\frac{1}{\text{x}},$ where 'x' is a positive real number.
AnswerFalse. We know that $\Big(\sqrt{\text{x}}-\frac{1}{\sqrt{\text{x}}}\Big)^2\geq0$ or $\Big(\text{x}+\frac{1}{\text{x}}\Big)\geq2,$ but $\sin\theta$ is not greater than 1. Alternatively, there exists the following three posibilities: Case 1: If $\text{x}<1,\text{then}\Big(\text{x}+\frac{1}{\text{x}}\Big)<1$Case 2:
If $\text{x}=1,\text{then}\Big(\text{x}+\frac{1}{\text{x}}\Big)=1$Case 3:
If $\text{x}>1,\text{then}\Big(\text{x}+\frac{1}{\text{x}}\Big)>1$ However, $\sin\theta$ cannot be greater than 1.
View full question & answer→Question 121 Mark
Write 'True' or 'False' and justify your answer.
The value of the expression $\left(\cos ^2 23^{\circ}-\sin ^2 67^{\circ}\right)$ is positive.
AnswerFalse. $23^{\circ}$ and $67^{\circ}$ are complementry angles so.
$\cos ^2 23^{\circ}-\sin ^2 67^{\circ}=\cos ^2 23^{\circ}-\sin ^2\left(90^{\circ}-23^{\circ}\right)$
$=\cos ^2 23^{\circ}-\cos ^2 23^{\circ}$
$=0$
So, the value of the given expression is not positive. Hence, the given statement is false.
View full question & answer→Question 131 Mark
Write 'True' or 'False' and justify your answer in the following:
The value of $\cos^2{23}-\sin^2{67}$ is positive.
AnswerFalse.
$\cos^2{23}-\sin^2{67}$
$=\sin^2(90-23)-\sin^2{67}$
$=\sin^2{67}-\sin^2{67}=0$
View full question & answer→Question 141 Mark
Write ‘True’ or ‘False’ and justify your answer.
$(\tan\theta+2)(2\tan\theta+1)=5\tan\theta+\sec^2\theta$
AnswerFalse.LHS$=(\tan\theta+2)(2\tan\theta+1)$
$=\tan\theta(2\tan\theta+1)+2(2\tan\theta+1)$
$=2\tan^2\theta+\tan\theta+4\tan\theta+2$
$=2\tan^2\theta+5\tan\theta+2$
$=2(\tan^2\theta+1)+5\tan\theta$
$=2\sec^2\theta+5\tan\theta\neq$ RHS
Hence, the given statement is false.
View full question & answer→Question 151 Mark
Write 'True' or 'False' and justify your answer.
If $\cos A+\cos ^2 A=1$, then $\sin ^2 A+\sin ^4 A=1$
AnswerTrue.True. $\cos A +\cos ^2 A=1$ [Given]
$\Rightarrow cosA = 1 - cos^2A$
$\Rightarrow cosA = sin^2A$
$\Rightarrow cos^2A = sin^4A$
$Now, LHS = sin^2A + sin^4A$
$= cosA + cos^2A$
$= 1 = RHS$
Hence, the given statement is true.
View full question & answer→Question 161 Mark
State whether the following are true or false. Justify your answer.$\cos \text{A} $ is the abbreviation used for the cosecant of angle A.
AnswerFalse as $\cos \text{A}$ is the abbreviation of cosine A.
View full question & answer→Question 171 Mark
State whether the following are true or false. Justify your answer.$\cot \text{A}$ is not defined for $\text{A}=0^\circ.$
Answer$\cot \text{A}$ is not defined for $\text{A}=0^\circ$True
View full question & answer→Question 181 Mark
State whether the following are true or false. Justify your answer.The value of $\cos \theta$ increases as $\theta$ increases.
AnswerValue of $\cos \theta \uparrow$ as $\theta$ increases.False.
View full question & answer→Question 191 Mark
State whether the following are true or false. Justify your answer.$\sin (\text{A}+\text{B})=\sin \text{A}+\sin\text{B}.$
Answer$\sin (\text{A}+\text{B})=\sin \text{A}+\sin \text{B}$False, suppose $\text{A}=30^\circ, \text{B}=60^\circ$ because $\sin(\text{A}+\text{B})=\sin 90^\circ=1$ $\sin \text{A}+\sin \text{B}=\frac{1}{2}+\frac{\sqrt{3}}{2}\neq1$
View full question & answer→Question 201 Mark
State whether the following are true or false. Justify your answer.The value of $\tan \text{A}$ is always less than 1.
AnswerFalse because sides of a right triangle may have any length, so $\tan \text{A}$ may have any value.
View full question & answer→Question 211 Mark
Write ‘True’ or ‘False’ and justify your answer.
$\cos\theta=\frac{\text{a}^2+\text{b}^2}{2\text{ab}},$ where a and b are two distinct numbers such that ab > 0.
AnswerFalse.Consider two numbers $a ^2$ and $b ^2$ then
Arithmetic mean (AM) of $a^2$ and $b^2=\frac{a^2+b^2}{2}$
Geometric mean (GM) of $a ^2$ and $b^2=\sqrt{a^2 \times b^2}$
$\Rightarrow\ \text{GM}=\text{ab}$
$\because\ \text{AM}>\text{GM}$
$\therefore\ \frac{\text{a}^2+\text{b}^2}{2}>\text{ab}$
$\Rightarrow\ \frac{\text{a}^2+\text{b}^2}{2\text{ab}}>1$
$\Rightarrow\ \cos\theta>1\ \Big(\text{Given}\cos\theta=\frac{\text{a}^2+\text{b}^2}{2\text{ab}}\Big)$
But, the value of $\cos\theta$ can never be greater than 1.
So,the given expression is false.
View full question & answer→Question 221 Mark
Write ‘True’ or ‘False’ and justify your answer.
$\sqrt{(1-\cos^2\theta)\sec^2\theta}=\tan\theta$
AnswerTrue.LHS $=\sqrt{(1-\cos^2\theta)\sec^2\theta}=\sqrt{\sin^2\theta.\frac{1}{\cos^2\theta}}$
$=\sqrt{\frac{\sin^2\theta}{\cos^2\theta}}=\tan\theta=$ RHS
Hence, the given expression is true.
View full question & answer→Question 231 Mark
State whether the following are true or false. Justify your answer.$\cot \text{A}$ is the product of $\cot$ and A.
AnswerFalse as $\cot \text{A}$ is not the product of ‘cot’ and A. ‘cot’ is separated from A has no meaning.
View full question & answer→