MCQ 11 Mark
The value of $\tan45^\circ\times\cot45^\circ$ is :
- A$0$
- ✓$1$
- C$2$
- D$\frac{1}{2}$
Answer
View full question & answer→Correct option: B.
$1$
$\tan45^\circ\times\cot45^\circ$
$=1\times1=1$
$=1\times1=1$
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MCQ 21 Mark
If $\tan\theta = \frac{\text{a}}{\text{b}}$ then $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}=$
- A$\frac{\text{a}-\text{b}}{\text{a + b}}$
- B$\frac{\text{b}-\text{b}}{\text{b+a}}$
- ✓$\frac{\text{b + a}}{\text{b}-\text{a}}$
- DNone of these
Answer
View full question & answer→Correct option: C.
$\frac{\text{b + a}}{\text{b}-\text{a}}$
Given : $\tan\theta = \frac{\text{a}}{\text{b}}$
Dividing all terms of $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$ by $\cos\theta$
$=\frac{1+\tan\theta}{1-\tan\theta} =\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}} = \frac{\text{b + a}}{\text{b}-\text{a}}$
Dividing all terms of $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$ by $\cos\theta$
$=\frac{1+\tan\theta}{1-\tan\theta} =\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}} = \frac{\text{b + a}}{\text{b}-\text{a}}$
MCQ 31 Mark
Directions : In the following questions, a statement of assertion $(A)$ is followed by a statement of reason $(R).$ Mark the correct choice as :
Assertion : In a right $\triangle\text{ABC}$, right angled at $B,$ if $ \tan\text{A}=1$, then $ 2\sin\text{A}$. $ \cos\text{A}=1$
Reason : $ \text{cosec}\text { A}$ is the abbreviation used for cosecant of angle $A$.
Assertion : In a right $\triangle\text{ABC}$, right angled at $B,$ if $ \tan\text{A}=1$, then $ 2\sin\text{A}$. $ \cos\text{A}=1$
Reason : $ \text{cosec}\text { A}$ is the abbreviation used for cosecant of angle $A$.
- ABoth assertion $(A)$ and reason $(R)$ are true and reason $(R)$ is the correct explanation of assertion $(A).$
- ✓Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A).$
- CAssertion $(A)$ is true but reason $(R)$ is false.
- DAssertion $(A)$ is false but reason $(R)$ is true.
Answer
View full question & answer→Correct option: B.
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A).$
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A).$
MCQ 41 Mark
The value of $\frac{\cos^320^\circ-\cos^370^\circ}{\sin^370^\circ-\sin^320^\circ}$ is :
- A$\frac{1}{2}$
- B$\frac{1}{\sqrt{2}}$
- ✓$1$
- D$2$
Answer
View full question & answer→Correct option: C.
$1$
We have to evaluate the value.
The formula to be used,
$\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2+\text{b}^2-\text{ab})$
$\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{b}^2+\text{ab})$
So,
$=\frac{\cos^320^\circ-\cos^370^\circ}{\sin^370^\circ-\sin^320^\circ}$
$=\frac{(\cos20^\circ-\cos70)(\cos^220^\circ+\cos^270+\cos20^\circ\cos70^\circ)}{(\sin70^\circ-\sin20^\circ)(\sin^270^\circ+\sin^220^\circ+\sin70^\circ\sin20^\circ)}$
Now using the properties of complementary angles,
$=\frac{(\sin70^\circ-\sin20)(\sin^270^\circ+\cos^270+\cos20^\circ\cos70^\circ)}{(\sin70^\circ-\sin20^\circ)(\sin^270^\circ+\cos^270^\circ+\sin70^\circ\sin20^\circ)}$
$=\frac{1+\cos20^\circ\cos70^\circ}{1+\sin70^\circ\sin20^\circ}$
$=\frac{1+\cos20^\circ\cos70^\circ}{1+\cos20^\circ\cos70^\circ}$
$=1$
Hence the correct option is $(c)$
The formula to be used,
$\text{a}^3+\text{b}^3=(\text{a}+\text{b})(\text{a}^2+\text{b}^2-\text{ab})$
$\text{a}^3-\text{b}^3=(\text{a}-\text{b})(\text{a}^2+\text{b}^2+\text{ab})$
So,
$=\frac{\cos^320^\circ-\cos^370^\circ}{\sin^370^\circ-\sin^320^\circ}$
$=\frac{(\cos20^\circ-\cos70)(\cos^220^\circ+\cos^270+\cos20^\circ\cos70^\circ)}{(\sin70^\circ-\sin20^\circ)(\sin^270^\circ+\sin^220^\circ+\sin70^\circ\sin20^\circ)}$
Now using the properties of complementary angles,
$=\frac{(\sin70^\circ-\sin20)(\sin^270^\circ+\cos^270+\cos20^\circ\cos70^\circ)}{(\sin70^\circ-\sin20^\circ)(\sin^270^\circ+\cos^270^\circ+\sin70^\circ\sin20^\circ)}$
$=\frac{1+\cos20^\circ\cos70^\circ}{1+\sin70^\circ\sin20^\circ}$
$=\frac{1+\cos20^\circ\cos70^\circ}{1+\cos20^\circ\cos70^\circ}$
$=1$
Hence the correct option is $(c)$
MCQ 51 Mark
$\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ=?$
- A$\sqrt{3}$
- B$\frac{1}{\sqrt{3}}$
- C$-1$
- ✓$1$
Answer
View full question & answer→Correct option: D.
$1$
$\tan10^\circ\tan15^\circ\tan75^\circ\tan80^\circ$
$=\tan10^\circ\tan15^\circ\tan(90^\circ-15^\circ)\tan(90^\circ-10^\circ)$
$=\tan10^\circ\tan15^\circ\cot15^\circ\cot10^\circ$
$=(\tan10^\circ\cot10^\circ)(\tan15^\circ\cot15^\circ)$
$=1\times1$
$=1$
$=\tan10^\circ\tan15^\circ\tan(90^\circ-15^\circ)\tan(90^\circ-10^\circ)$
$=\tan10^\circ\tan15^\circ\cot15^\circ\cot10^\circ$
$=(\tan10^\circ\cot10^\circ)(\tan15^\circ\cot15^\circ)$
$=1\times1$
$=1$
MCQ 61 Mark
Choose the correct answer from the given four options. The value of $(\tan1^\circ\tan2^\circ\tan3^\circ...\tan89^\circ)$ is :
- A$0$
- ✓$1$
- C$2$
- D$\frac{1}{2}$
Answer
View full question & answer→Correct option: B.
$1$
$\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan89^\circ$
$=\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan44^\circ.\tan45^\circ.\tan46^\circ...\tan87^\circ-\tan88^\circ\tan89^\circ$
$=\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan44^\circ.(1)$
$-\tan(90^\circ-44^\circ)...\tan(90^\circ-3^\circ)$
$\tan(90^\circ-2^\circ)-\tan(90^\circ-1^\circ)(\therefore\tan45^\circ=1)$
$=\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan44^\circ.(1).$
$\cot44^\circ......\cot3^\circ-\cot2^\circ-\cot1^\circ$
$[\because\tan(90^\circ-\theta)=\cot\theta]$
$=\tan1^\circ.\tan2^\circ.\tan3^\circ...\tan44^\circ(1).$
$\frac{1}{\tan44^\circ}...\frac{1}{\tan30^\circ}.\frac{1}{\tan2^\circ}.\frac{1}{\tan1^\circ}$
$\Big[\because\cot\theta=\frac{1}{\tan\theta}\Big]$
$=1$
$=\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan44^\circ.\tan45^\circ.\tan46^\circ...\tan87^\circ-\tan88^\circ\tan89^\circ$
$=\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan44^\circ.(1)$
$-\tan(90^\circ-44^\circ)...\tan(90^\circ-3^\circ)$
$\tan(90^\circ-2^\circ)-\tan(90^\circ-1^\circ)(\therefore\tan45^\circ=1)$
$=\tan1^\circ-\tan2^\circ-\tan3^\circ...\tan44^\circ.(1).$
$\cot44^\circ......\cot3^\circ-\cot2^\circ-\cot1^\circ$
$[\because\tan(90^\circ-\theta)=\cot\theta]$
$=\tan1^\circ.\tan2^\circ.\tan3^\circ...\tan44^\circ(1).$
$\frac{1}{\tan44^\circ}...\frac{1}{\tan30^\circ}.\frac{1}{\tan2^\circ}.\frac{1}{\tan1^\circ}$
$\Big[\because\cot\theta=\frac{1}{\tan\theta}\Big]$
$=1$
MCQ 71 Mark
Choose the correct option and justify your choice : $\frac{2\tan30^\circ}{1-\tan^230^\circ}= $
- A$\cos 60^\circ$
- B$\sin 60^\circ$
- ✓$\tan 60^\circ$
- D$\sin 30^\circ$
Answer
View full question & answer→Correct option: C.
$\tan 60^\circ$
$\frac{2\tan30^\circ}{1-\tan^230^\circ}=\frac{2\times\frac{1}{\sqrt{3}}}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{\frac{2}{\sqrt{3}}}{\frac{3-1}{3}} $
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} $
$=\frac{2}{\sqrt{3}}\times\frac{3}{2}$
$=\frac{3}{\sqrt{3}} $
$=\frac{3}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} $
$=\frac{3\sqrt{3}}{3}$
$=\sqrt{3}=\tan60^\circ$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}} $
$=\frac{2}{\sqrt{3}}\times\frac{3}{2}$
$=\frac{3}{\sqrt{3}} $
$=\frac{3}{\sqrt{3}}\times\frac{\sqrt{3}}{\sqrt{3}} $
$=\frac{3\sqrt{3}}{3}$
$=\sqrt{3}=\tan60^\circ$
MCQ 81 Mark
$\tan5^\circ\times\tan30^\circ\times4\tan85^\circ$ is equal to :
- ✓$\frac{4}{\sqrt{3}}$
- B$4\sqrt{3}$
- C$1$
- D$4$
Answer
View full question & answer→Correct option: A.
$\frac{4}{\sqrt{3}}$
We have to find $\tan5^\circ\times\tan30^\circ\times4\tan85^\circ$
We know that
$\tan(90^\circ-\theta)=\cot\theta$
$\tan\theta\cot\theta=1$
$\tan30^\circ=\frac{1}{\sqrt{3}}$
So,
$\tan5^\circ\times\tan30^\circ\times4\tan85^\circ$
$=\tan(90^\circ-85^\circ)\times\tan30^\circ\times4\tan85^\circ$
$=\cot85^\circ\times\tan30^\circ\times4\tan85^\circ$
$=4\cot85^\circ\times\tan85^\circ\tan30^\circ$
$=4\times1\times\frac{1}{\sqrt{3}}$
$=\frac{4}{\sqrt{3}}$
Hence the correct option is $(a)$
We know that
$\tan(90^\circ-\theta)=\cot\theta$
$\tan\theta\cot\theta=1$
$\tan30^\circ=\frac{1}{\sqrt{3}}$
So,
$\tan5^\circ\times\tan30^\circ\times4\tan85^\circ$
$=\tan(90^\circ-85^\circ)\times\tan30^\circ\times4\tan85^\circ$
$=\cot85^\circ\times\tan30^\circ\times4\tan85^\circ$
$=4\cot85^\circ\times\tan85^\circ\tan30^\circ$
$=4\times1\times\frac{1}{\sqrt{3}}$
$=\frac{4}{\sqrt{3}}$
Hence the correct option is $(a)$
MCQ 91 Mark
The value of $\tan1^\circ\tan2^\circ\tan3^\circ.....\tan89^\circ$ is :
- ✓$1$
- B$-1$
- C$0$
- DNone of these
Answer
View full question & answer→Correct option: A.
$1$
Here we have to find : $\tan1^\circ\tan2^\circ\tan3^\circ.....\tan89^\circ$
$\tan1^\circ\tan2^\circ\tan3^\circ.....\tan89^\circ$
$=\tan(90^\circ-89^\circ)\tan(90^\circ-88^\circ)\tan(90^\circ-87^\circ) ...\tan87^\circ\tan88^\circ\tan89^\circ$
$=\cot89^\circ\cot88^\circ\cot87^\circ...\tan87^\circ\tan88^\circ\tan89^\circ$
$=(\cot89^\circ-\tan89^\circ)(\cot88^\circ\tan88^\circ)$
$\ \ \ \ (\cot87^\circ\tan87^\circ)...(\cot44^\circ\tan44^\circ)\tan45^\circ$
$=1\times1\times1...1\times1\ [\text{since}\cot\theta\tan\theta=1]$
$=1$
Hence the correct option is $(a)$
$\tan1^\circ\tan2^\circ\tan3^\circ.....\tan89^\circ$
$=\tan(90^\circ-89^\circ)\tan(90^\circ-88^\circ)\tan(90^\circ-87^\circ) ...\tan87^\circ\tan88^\circ\tan89^\circ$
$=\cot89^\circ\cot88^\circ\cot87^\circ...\tan87^\circ\tan88^\circ\tan89^\circ$
$=(\cot89^\circ-\tan89^\circ)(\cot88^\circ\tan88^\circ)$
$\ \ \ \ (\cot87^\circ\tan87^\circ)...(\cot44^\circ\tan44^\circ)\tan45^\circ$
$=1\times1\times1...1\times1\ [\text{since}\cot\theta\tan\theta=1]$
$=1$
Hence the correct option is $(a)$
MCQ 101 Mark
If $\text{x}\tan45^\circ\cos60^\circ=\sin60^\circ\cot60^\circ,$ then x is equal to:
- ✓$1$
- B$\sqrt{3}$
- C$\frac{1}{2}$
- D$\frac{1}{\sqrt{2}}$
Answer
View full question & answer→Correct option: A.
$1$
(A) 1
MCQ 111 Mark
$\frac{2\tan30^\circ}{1-\tan^230^\circ}$ is equal to :
- A$\cos60^\circ$
- B$\sin60^\circ$
- ✓$\tan60^\circ$
- D$\sin30^\circ$
Answer
View full question & answer→Correct option: C.
$\tan60^\circ$
We are asked to find the value of the following
$\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
We know that $\begin{bmatrix}\tan30^\circ=\frac{1}{\sqrt{3}}\\\tan60^\circ=\sqrt{3}\end{bmatrix}$
$=\frac{3}{\sqrt3}$
$=\frac{3}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\sqrt3$
$=\tan60^\circ$
Hence the correct option is $(c)$
$\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\tan30^\circ}{1-\tan^230^\circ}$
$=\frac{2\times\frac{1}{\sqrt{3}}}{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}$
$=\frac{\frac{2}{\sqrt{3}}}{1-\frac{1}{3}}$
$=\frac{\frac{2}{\sqrt{3}}}{\frac{2}{3}}$
We know that $\begin{bmatrix}\tan30^\circ=\frac{1}{\sqrt{3}}\\\tan60^\circ=\sqrt{3}\end{bmatrix}$
$=\frac{3}{\sqrt3}$
$=\frac{3}{\sqrt3}\times\frac{\sqrt3}{\sqrt3}$
$=\sqrt3$
$=\tan60^\circ$
Hence the correct option is $(c)$
MCQ 121 Mark
If $\cot\theta = \frac{7}{8}$ then the value of $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ is :
- A$\frac{8}{7}$
- ✓$\frac{49}{64}$
- C$\frac{7}{8}$
- D$\frac{64}{49}$
Answer
View full question & answer→Correct option: B.
$\frac{49}{64}$
Given : $\cot\theta = \frac{7}{8}$
Now, $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$= \frac{1-\sin^{2}\theta}{1-\cos^{2}\theta}$
$= \big(\frac{7}{8}\big)^{2}$
$=\frac{49}{64}$
Now, $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$= \frac{1-\sin^{2}\theta}{1-\cos^{2}\theta}$
$= \big(\frac{7}{8}\big)^{2}$
$=\frac{49}{64}$
MCQ 131 Mark
If $\theta$ is an acute angle such that $\tan^2\theta=\frac{8}{7},$ then the value of $\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$ is :
- ✓$\frac{7}{8}$
- B$\frac{8}{7}$
- C$\frac{7}{4}$
- D$\frac{64}{49}$
Answer
View full question & answer→Correct option: A.
$\frac{7}{8}$
Given that : $\tan^2\theta=\frac{8}{7}$ and $\theta$ is an acute angle
We have to find the following expression
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
Since
$\tan^2\theta=\frac{8}{7}$
$\tan^2\theta=\sqrt\frac{8}{7}$
$\tan\theta=\frac{\sqrt8}{\sqrt7}$
Since $\tan\theta=\frac{\text{Perpedicular}}{\text{Base}}$
$\Rightarrow{\text{perpedicular}}=\sqrt{8}$
$\Rightarrow\text{Base}=\sqrt{7}$
$\Rightarrow\text{Hypotenuse}=\sqrt{8+7}$
$\Rightarrow\text{Hypotenuse}=\sqrt{15}$
We know that $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ and $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
We find :
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$=\frac{\Big(1+\frac{\sqrt{8}}{\sqrt{15}}\Big)\Big(1-\frac{\sqrt{8}}{\sqrt{15}}\Big)}{\Big(1+\frac{\sqrt{7}}{\sqrt{15}}\Big)\Big(1-\frac{\sqrt{7}}{\sqrt{15}}\Big)}$
$=\frac{\Big(1-\frac{8}{15}\Big)}{\Big(1-\frac{7}{15}\Big)}$
$=\frac{\frac{7}{15}}{\frac{8}{15}}$
$=\frac{7}{8}$
Hence the correct option is $(a)$
We have to find the following expression
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
Since
$\tan^2\theta=\frac{8}{7}$
$\tan^2\theta=\sqrt\frac{8}{7}$
$\tan\theta=\frac{\sqrt8}{\sqrt7}$
Since $\tan\theta=\frac{\text{Perpedicular}}{\text{Base}}$
$\Rightarrow{\text{perpedicular}}=\sqrt{8}$
$\Rightarrow\text{Base}=\sqrt{7}$
$\Rightarrow\text{Hypotenuse}=\sqrt{8+7}$
$\Rightarrow\text{Hypotenuse}=\sqrt{15}$
We know that $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$ and $\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
We find :
$\frac{(1+\sin\theta)(1-\sin\theta)}{(1+\cos\theta)(1-\cos\theta)}$
$=\frac{\Big(1+\frac{\sqrt{8}}{\sqrt{15}}\Big)\Big(1-\frac{\sqrt{8}}{\sqrt{15}}\Big)}{\Big(1+\frac{\sqrt{7}}{\sqrt{15}}\Big)\Big(1-\frac{\sqrt{7}}{\sqrt{15}}\Big)}$
$=\frac{\Big(1-\frac{8}{15}\Big)}{\Big(1-\frac{7}{15}\Big)}$
$=\frac{\frac{7}{15}}{\frac{8}{15}}$
$=\frac{7}{8}$
Hence the correct option is $(a)$
MCQ 141 Mark
If $\text{x}\sin(90^\circ-\theta)\cot(90^\circ-\theta)=\cos(90^\circ-\theta),$ then $x =$
- A$0$
- ✓$1$
- C$-1$
- D$2$
Answer
View full question & answer→Correct option: B.
$1$
We have : $\text{x}\sin\text({90}^\circ-\theta)\cot(90^\circ-\theta)=\cos(90^\circ-\theta)$
Here we have to find the value of $x$
$\begin{bmatrix}\sin(90^\circ-\theta)=\cos\theta \\\cos(90^\circ-\theta)=\sin\theta\\\cot(90^\circ-\theta)=\tan\theta \end{bmatrix}$
We know that
$\Rightarrow\text{x}\sin(90^\circ-\theta)\cot(90^\circ-\theta)=\cos(90^\circ-\theta)$
$\Rightarrow\text{x}\cos\theta\tan\theta=\sin\theta$
$\Rightarrow\text{x}\cos\theta\times\frac{\sin\theta}{\cos\theta}=\sin\theta$
$\Rightarrow\text{x}=1$
Hence the correct option is $(b)$
Here we have to find the value of $x$
$\begin{bmatrix}\sin(90^\circ-\theta)=\cos\theta \\\cos(90^\circ-\theta)=\sin\theta\\\cot(90^\circ-\theta)=\tan\theta \end{bmatrix}$
We know that
$\Rightarrow\text{x}\sin(90^\circ-\theta)\cot(90^\circ-\theta)=\cos(90^\circ-\theta)$
$\Rightarrow\text{x}\cos\theta\tan\theta=\sin\theta$
$\Rightarrow\text{x}\cos\theta\times\frac{\sin\theta}{\cos\theta}=\sin\theta$
$\Rightarrow\text{x}=1$
Hence the correct option is $(b)$
MCQ 151 Mark
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$ is equal to :
- A$\sec^2\text{A}$
- B$-1$
- C$\cot^2\text{A}$
- ✓$\tan^2\text{A}$
Answer
View full question & answer→Correct option: D.
$\tan^2\text{A}$
$\frac{1+\tan^2\text{A}}{1+\cot^2\text{A}}$
$=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}$
$=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$=\tan^2\text{A}$
$=\frac{\sec^2\text{A}}{\text{cosec}^2\text{A}}$
$=\frac{\sin^2\text{A}}{\cos^2\text{A}}$
$=\tan^2\text{A}$
MCQ 161 Mark
The value of $\frac{\tan55^\circ}{\cot35^\circ}+\cot1^\circ\cot2^\circ\cot3^\circ....\cot90^\circ,$ is :
- A$-2$
- B$2$
- ✓$1$
- D$0$
Answer
View full question & answer→Correct option: C.
$1$
We have to find the value of the following expression
$\frac{\tan55^\circ}{\cot35^\circ}+\cot1^\circ\cot2^\circ\cot3^\circ....\cot90^\circ$
$=\frac{\tan55^\circ}{\cot35^\circ}+\cot1^\circ\cot2^\circ\cot3^\circ ....\cot90^\circ$
$=\frac{\tan(90^\circ-35^\circ)}{\cot35^\circ}+\cot(90^\circ-89^\circ)\cot(90^\circ-88^\circ)$
$\ \ \ \ \cot(90^\circ-87^\circ) ....\cot87^\circ\cot88^\circ\cot89^\circ ....\cot90^\circ$
$=\frac{\cot35^\circ}{\cot35^\circ}+\tan89^\circ\tan88^\circ\tan87^\circ ....\cot87^\circ\cot88^\circ\cot89^\circ....\cot90^\circ$
$=1+1\times1\times1\ ....\times\ 0$
$=1$
As $\cot90^\circ=0$
Hence the correct option is $ (c)$
$\frac{\tan55^\circ}{\cot35^\circ}+\cot1^\circ\cot2^\circ\cot3^\circ....\cot90^\circ$
$=\frac{\tan55^\circ}{\cot35^\circ}+\cot1^\circ\cot2^\circ\cot3^\circ ....\cot90^\circ$
$=\frac{\tan(90^\circ-35^\circ)}{\cot35^\circ}+\cot(90^\circ-89^\circ)\cot(90^\circ-88^\circ)$
$\ \ \ \ \cot(90^\circ-87^\circ) ....\cot87^\circ\cot88^\circ\cot89^\circ ....\cot90^\circ$
$=\frac{\cot35^\circ}{\cot35^\circ}+\tan89^\circ\tan88^\circ\tan87^\circ ....\cot87^\circ\cot88^\circ\cot89^\circ....\cot90^\circ$
$=1+1\times1\times1\ ....\times\ 0$
$=1$
As $\cot90^\circ=0$
Hence the correct option is $ (c)$
MCQ 171 Mark
If $\sec\theta=\frac{25}{7}$ then $\sin\theta=?$
- A$\frac{7}{24}$
- B$\frac{24}{7}$
- ✓$\frac{24}{25}$
- DNone of these.
Answer
View full question & answer→Correct option: C.
$\frac{24}{25}$
Consider $\triangle\text{ABC}$ where $\angle\text{B}=90^\circ,\angle\text{A}=\theta.$
Then, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$=\frac{\text{AC}}{\text{AB}}=\frac{25}{7}$
Let $\text{AC}=25\text{k}$ and $\text{AB}=7\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(25\text{k})^2=(7\text{k})^2+\text{BC}^2$
$\Rightarrow\text{BC}^2=\text{625k}^2-\text{49k}^2=\text{576k}^2$
$\Rightarrow\text{BC}=\text{24k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\frac{\text{BC}}{\text{AC}}=\frac{24\text{k}}{\text{25k}}=\frac{24}{25}$
Then, $\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}$
$=\frac{\text{AC}}{\text{AB}}=\frac{25}{7}$
Let $\text{AC}=25\text{k}$ and $\text{AB}=7\text{k},$ where $k$ is positive.
By Pythagoras Theorem,
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(25\text{k})^2=(7\text{k})^2+\text{BC}^2$
$\Rightarrow\text{BC}^2=\text{625k}^2-\text{49k}^2=\text{576k}^2$
$\Rightarrow\text{BC}=\text{24k}$
Now, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$
$\frac{\text{BC}}{\text{AC}}=\frac{24\text{k}}{\text{25k}}=\frac{24}{25}$
MCQ 181 Mark
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$ is equal to :
- A$0$
- B$1$
- ✓$-1$
- DNone of these.
Answer
View full question & answer→Correct option: C.
$-1$
$2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$
$=2\big[(\sin^2\theta)^3+(\cos^2\theta)^3\big]-3\big[(\sin^2\theta)^2+(\cos^2\theta)^2\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)\big]$
$=-3\big[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta\big]$
$\{\because\ \text{a}^3+\text{b}^2=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})\}$
$=2\big[1(\sin^2\theta)^2+(\cos^2\theta)^2+2\sin^2\theta\cos^2\theta-3\sin^2\theta+\cos^2\theta\big]$
$=-3\big[(1)^2-2\sin^2\theta\cos^2\theta\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)^2-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2\big[1-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2-6\sin^2\theta\cos^3\theta-3+6\sin^2\theta\cos^2\theta$
$=-1$
$=2\big[(\sin^2\theta)^3+(\cos^2\theta)^3\big]-3\big[(\sin^2\theta)^2+(\cos^2\theta)^2\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)(\sin^4\theta+\cos^4\theta-\sin^2\theta\cos^2\theta)\big]$
$=-3\big[(\sin^2\theta+\cos^2\theta)^2-2\sin^2\theta\cos^2\theta\big]$
$\{\because\ \text{a}^3+\text{b}^2=(\text{a}+\text{b})^3-3\text{ab}(\text{a}+\text{b})\}$
$=2\big[1(\sin^2\theta)^2+(\cos^2\theta)^2+2\sin^2\theta\cos^2\theta-3\sin^2\theta+\cos^2\theta\big]$
$=-3\big[(1)^2-2\sin^2\theta\cos^2\theta\big]$
$=2\big[(\sin^2\theta+\cos^2\theta)^2-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2\big[1-3\sin^2\theta\cos^2\theta\big]-3\big[1-2\sin^2\theta\cos^2\theta\big]$
$=2-6\sin^2\theta\cos^3\theta-3+6\sin^2\theta\cos^2\theta$
$=-1$
MCQ 191 Mark
If $\cos\theta=\frac{2}{3},$ then $2\sec^2\theta+2\tan^2\theta-7$ is equal to :
- A$1$
- ✓$0$
- C$3$
- D$4$
Answer
View full question & answer→Correct option: B.
$0$
Given that $\cos\theta=\frac{2}{3}$
We have to find $2\sec^2\theta+2\tan^2\theta-7$
As we are given
$\cos\theta=\frac{2}{3}$
$\Rightarrow\text{Base}=2$
$\Rightarrow\text{Hypotenuse}=3$
$\Rightarrow\text{Perpendicular}=\sqrt{(3)^2-(2)^2}$
$\Rightarrow\text{Perpendicular}=\sqrt{5}$
We know that :
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
$\tan\theta=\frac{\text{Perpindicular}}{\text{Base}}$
Now we have to find $2\sec^2\theta+2\tan^2\theta-7$
So,
$2\sec^2\theta+2\tan^2\theta-7$
$=2\Big(\frac{3}{2}\Big)^2+2\Big(\frac{\sqrt{5}}{2}\Big)^2-7$
$=\frac{18}{4}+\frac{10}{4}-7$
$=\frac{18+10-28}{4}$
$=0$
Hence the correct option is $(b)$
We have to find $2\sec^2\theta+2\tan^2\theta-7$
As we are given
$\cos\theta=\frac{2}{3}$
$\Rightarrow\text{Base}=2$
$\Rightarrow\text{Hypotenuse}=3$
$\Rightarrow\text{Perpendicular}=\sqrt{(3)^2-(2)^2}$
$\Rightarrow\text{Perpendicular}=\sqrt{5}$
We know that :
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$
$\tan\theta=\frac{\text{Perpindicular}}{\text{Base}}$
Now we have to find $2\sec^2\theta+2\tan^2\theta-7$
So,
$2\sec^2\theta+2\tan^2\theta-7$
$=2\Big(\frac{3}{2}\Big)^2+2\Big(\frac{\sqrt{5}}{2}\Big)^2-7$
$=\frac{18}{4}+\frac{10}{4}-7$
$=\frac{18+10-28}{4}$
$=0$
Hence the correct option is $(b)$
MCQ 201 Mark
If $\text{a}\cos\theta+\text{b}\sin\theta=4$ and ${a}\sin\theta-\text{b}\cos\theta=3,$ then $a^2+b^2=0$
- A$7$
- B$12$
- ✓$25$
- DNone of these.
Answer
View full question & answer→Correct option: C.
$25$
Given,
$\text{a}\cos\theta+\text{b}\sin\theta=4,$
$\text{a}\sin\theta-\text{b}\cos\theta=3$
Squaring and then adding the above two equations, we have
$(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2=(4)^2+(3)^2$
$\Rightarrow\ (\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta)+2\times\text{a}\cos\theta\times\sin\theta)$
$=(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{a}\sin\theta\times\text{b}\cos\theta)=16+9$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta$
$=\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta=25$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta=25$
$\Rightarrow\ ( \text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta)+(\text{b}^2\sin^2\theta+\text{b}^2\cos^2\theta)=25$
$\Rightarrow\ \text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)=25$
$\Rightarrow\ \text{a}^2(1)+\text{b}^2(1)=25$
$\Rightarrow\ \text{a}^2+\text{b}^2=25$
Hence, the correct option is $(C).$
$\text{a}\cos\theta+\text{b}\sin\theta=4,$
$\text{a}\sin\theta-\text{b}\cos\theta=3$
Squaring and then adding the above two equations, we have
$(\text{a}\cos\theta+\text{b}\sin\theta)^2+(\text{a}\sin\theta-\text{b}\cos\theta)^2=(4)^2+(3)^2$
$\Rightarrow\ (\text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta)+2\times\text{a}\cos\theta\times\sin\theta)$
$=(\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{a}\sin\theta\times\text{b}\cos\theta)=16+9$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+2\text{ab}\sin\theta\cos\theta$
$=\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta-2\text{ab}\sin\theta\cos\theta=25$
$\Rightarrow\ \text{a}^2\cos^2\theta+\text{b}^2\sin^2\theta+\text{a}^2\sin^2\theta+\text{b}^2\cos^2\theta=25$
$\Rightarrow\ ( \text{a}^2\cos^2\theta+\text{a}^2\sin^2\theta)+(\text{b}^2\sin^2\theta+\text{b}^2\cos^2\theta)=25$
$\Rightarrow\ \text{a}^2(\cos^2\theta+\sin^2\theta)+\text{b}^2(\sin^2\theta+\cos^2\theta)=25$
$\Rightarrow\ \text{a}^2(1)+\text{b}^2(1)=25$
$\Rightarrow\ \text{a}^2+\text{b}^2=25$
Hence, the correct option is $(C).$
MCQ 211 Mark
If $3\cot\theta=4$ then $\frac{(5\sin\theta+3\cos\theta)}{(5\sin\theta-3\cos\theta)} =?$
- A$\frac{1}{3}$
- B$3$
- C$\frac{1}{9}$
- ✓$9$
Answer
View full question & answer→Correct option: D.
$9$
Given, $3\cot\theta=4$
Now, $\frac{(5\sin\theta+3\cos\theta)}{(5\sin\theta-3\cos\theta)} $
$=\frac{\frac{5\sin\theta}{\sin\theta}+\frac{3\cos\theta}{\sin\theta}}{\frac{5\sin\theta}{\sin\theta}-\frac{3\cos\theta}{\sin\theta}}$
$=\frac{5+3\cot\theta}{5-3\cot\theta}$
$=\frac{5+4}{5-4}=\frac{9}{1}=9$
Now, $\frac{(5\sin\theta+3\cos\theta)}{(5\sin\theta-3\cos\theta)} $
$=\frac{\frac{5\sin\theta}{\sin\theta}+\frac{3\cos\theta}{\sin\theta}}{\frac{5\sin\theta}{\sin\theta}-\frac{3\cos\theta}{\sin\theta}}$
$=\frac{5+3\cot\theta}{5-3\cot\theta}$
$=\frac{5+4}{5-4}=\frac{9}{1}=9$
MCQ 221 Mark
Choose the correct answer from the given four options. If $\sin\text{A}=\frac{1}{2},$ then the value of $\cot\text{A}$ is :
- ✓$\sqrt{3}$
- B$\frac{1}{\sqrt{3}}$
- C$\frac{\sqrt{3}}{2}$
- D$1$
Answer
View full question & answer→Correct option: A.
$\sqrt{3}$
Given, $\sin\text{A}=\frac{1}{2}$
$\therefore\ \cos\text{A}=\sqrt{1-\sin^2\text{A}}=\sqrt{1-\Big(\frac{1}{2}\Big)^2}$
$=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$
$[\because\sin^2\text{A}+\cos^2=1$
$\Rightarrow\cos\text{A}=\sqrt{1-\sin^2\text{A}}]$
Now $, \cot\text{A}=\frac{\cos\text{A}}{\sin\text{A}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}.$
$\therefore\ \cos\text{A}=\sqrt{1-\sin^2\text{A}}=\sqrt{1-\Big(\frac{1}{2}\Big)^2}$
$=\sqrt{1-\frac{1}{4}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}$
$[\because\sin^2\text{A}+\cos^2=1$
$\Rightarrow\cos\text{A}=\sqrt{1-\sin^2\text{A}}]$
Now $, \cot\text{A}=\frac{\cos\text{A}}{\sin\text{A}}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}.$
MCQ 231 Mark
If $\text{x}=\text{r}\sin\theta\cos\phi,\text{y}=\text{r}\sin\phi$ and ${z}=\text{r}\cos\theta,$ then :
- ✓$x^2+y^2+z^2=r^2 $
- B$ x^2+y^2-z^2=r^2 $
- C$ x^2-y^2+z^2=r^2 $
- D$ z^2+y^2-x^2=r^2 $
Answer
View full question & answer→Correct option: A.
$x^2+y^2+z^2=r^2 $
$\text{x}=\text{r}\sin\theta\cos\phi$
$\text{y}=\text{r}\sin\theta\sin\phi$
$\text{z}=\text{r}\cos\theta$
Squaring and adding these equations, we get
$\text{x}^2+\text{y}^2+\text{z}^2=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$
$\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=(\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta(1)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2(\sin^2\theta+\cos^2\theta)$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2(1)$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2$
Hence, the correct option is $(A).$
$\text{y}=\text{r}\sin\theta\sin\phi$
$\text{z}=\text{r}\cos\theta$
Squaring and adding these equations, we get
$\text{x}^2+\text{y}^2+\text{z}^2=(\text{r}\sin\theta\cos\phi)^2+(\text{r}\sin\theta\sin\phi)^2+(\text{r}\cos\theta)^2$
$\Rightarrow\text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=(\text{r}^2\sin^2\theta\cos^2\phi+\text{r}^2\sin^2\theta\sin^2\phi)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta(\cos^2\phi+\sin^2\phi)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta(1)+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2\sin^2\theta+\text{r}^2\cos^2\theta$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2(\sin^2\theta+\cos^2\theta)$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2(1)$
$\Rightarrow\ \text{x}^2+\text{y}^2+\text{z}^2=\text{r}^2$
Hence, the correct option is $(A).$
MCQ 241 Mark
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})=$
- A$\sec\text{A}$
- B$\sin\text{A}$
- C$\text{cosec A}$
- ✓$\cos\text{A}$
Answer
View full question & answer→Correct option: D.
$\cos\text{A}$
The given expression is $(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
Simplifying the given expression, we have
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)\times(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})(1-\sin\text{A})}{\cos\text{A}}$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
Therefore, the correct option is $(d).$
Simplifying the given expression, we have
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A})$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A})$
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)\times(1-\sin\text{A})$
$=\frac{(1+\sin\text{A})(1-\sin\text{A})}{\cos\text{A}}$
$=\frac{1-\sin^2\text{A}}{\cos\text{A}}$
$=\frac{\cos^2\text{A}}{\cos\text{A}}$
$=\cos\text{A}$
Therefore, the correct option is $(d).$
MCQ 251 Mark
If $7\tan\theta=4$ then $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} =?$
- ✓$\frac{1}{7}$
- B$\frac{5}{7}$
- C$\frac{3}{7}$
- D$\frac{5}{14}$
Answer
View full question & answer→Correct option: A.
$\frac{1}{7}$
Given, $7\tan\theta=4$
Now, $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} $
$=\frac{\frac{7\sin\theta}{\cos\theta}-\frac{3\cos\theta}{\cos\theta}}{\frac{7\sin\theta}{\cos\theta}+\frac{3\cos\theta}{\cos\theta}}$
$=\frac{7\tan\theta-3}{7\tan\theta+3}$
$=\frac{4-3}{4+3}=\frac{1}{7}$
Now, $\frac{(7\sin\theta-3\cos\theta)}{(7\sin\theta+3\cos\theta)} $
$=\frac{\frac{7\sin\theta}{\cos\theta}-\frac{3\cos\theta}{\cos\theta}}{\frac{7\sin\theta}{\cos\theta}+\frac{3\cos\theta}{\cos\theta}}$
$=\frac{7\tan\theta-3}{7\tan\theta+3}$
$=\frac{4-3}{4+3}=\frac{1}{7}$
MCQ 261 Mark
If $A, B$ and $C$ are interior angles of a triangle $\text{ABC},$ then $\sin\Big(\frac{\text{B}+\text{C}}{2}\Big)=$
- A$\sin\frac{\text{A}}{2}$
- ✓$\cos\frac{\text{A}}{2}$
- C$-\sin\frac{\text{A}}{2}$
- D$-\cos\frac{\text{A}}{2}$
Answer
View full question & answer→Correct option: B.
$\cos\frac{\text{A}}{2}$
We know tht in triangle $\text{ABC}$
$\text{A+B+C}=180^\circ$
$\Rightarrow\text{B+C}=180^\circ-\text{A}$
$\Rightarrow\frac{\text{B+C}}{2}=\frac{90^\circ}{2}-\frac{\text{A}}{2}$
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
Since $\sin(90^\circ-\text{A})=\cos{\text{A}}$
So,
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Hence the correct option is $(b)$
$\text{A+B+C}=180^\circ$
$\Rightarrow\text{B+C}=180^\circ-\text{A}$
$\Rightarrow\frac{\text{B+C}}{2}=\frac{90^\circ}{2}-\frac{\text{A}}{2}$
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\sin\Big(90^\circ-\frac{\text{A}}{2}\Big)$
Since $\sin(90^\circ-\text{A})=\cos{\text{A}}$
So,
$\Rightarrow\sin\Big(\frac{\text{B+C}}{2}\Big)=\cos\frac{\text{A}}{2}$
Hence the correct option is $(b)$
MCQ 271 Mark
If $\cos\theta=\frac{4}{5}$ then $\tan\theta=?$
- ✓$\frac{3}{4}$
- B$\frac{4}{3}$
- C$\frac{3}{5}$
- D$\frac{5}{3}$
Answer
View full question & answer→Correct option: A.
$\frac{3}{4}$
$\sin^2\theta+\cos^2\theta=1$
$\Rightarrow\sin^2\theta+\Big(\frac{4}5{}\Big)^2=1$
$\Rightarrow\sin^2\theta=1-\frac{16}{25}=\frac{9}{25}$
$\Rightarrow\sin\theta=\frac{3}{5}$
$\therefore\tan\theta=\frac{\sin\theta}{\cos\theta}$
$=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
$\Rightarrow\sin^2\theta+\Big(\frac{4}5{}\Big)^2=1$
$\Rightarrow\sin^2\theta=1-\frac{16}{25}=\frac{9}{25}$
$\Rightarrow\sin\theta=\frac{3}{5}$
$\therefore\tan\theta=\frac{\sin\theta}{\cos\theta}$
$=\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
MCQ 281 Mark
Find $\tan\theta,$ if $\theta=45^\circ:$
- ✓$1$
- B$2$
- C$3$
- DNone of these
Answer
View full question & answer→Correct option: A.
$1$
Given : $\theta=45^\circ$
Therefore, $\tan\theta=\tan45^\circ=1$
Therefore, $\tan\theta=\tan45^\circ=1$
MCQ 291 Mark
$\sec^210^\circ-\cot^280^\circ=?$
- ✓$1$
- B$0$
- C$\frac{3}{2}$
- D$\frac{1}2{}$
Answer
View full question & answer→Correct option: A.
$1$
$\sec^210^\circ-\cot^280^\circ$
$=\sec^210^\circ-\cot^2(90^\circ-10^\circ)$
$=\sec^210^\circ-\tan^210^\circ$
$=1$
$=\sec^210^\circ-\cot^2(90^\circ-10^\circ)$
$=\sec^210^\circ-\tan^210^\circ$
$=1$
MCQ 301 Mark
Choose the correct answer from the given four options. Given that $\sin\alpha=\frac{1}{2}$ and $\cos\beta=\frac{1}{2}$ then the value of $(\alpha+\beta)$ is :
- A$0^\circ$
- B$30^\circ$
- C$60^\circ$
- ✓$90^\circ$
Answer
View full question & answer→Correct option: D.
$90^\circ$
Given $,\sin\alpha=\frac{1}{2}=\sin30^\circ$
$\Big[\because\sin30^\circ=\frac{1}{2}\Big]$
$\Rightarrow\ \alpha=30^\circ$
and $\cos\beta=\frac{1}{2}=\cos60^\circ$
$\Big[\because\cos60^\circ=\frac{1}{2}\Big]$
$\Rightarrow\ \beta=60^\circ$
$\therefore\ \alpha+\beta=30^\circ+60^\circ=90^\circ$
$\Big[\because\sin30^\circ=\frac{1}{2}\Big]$
$\Rightarrow\ \alpha=30^\circ$
and $\cos\beta=\frac{1}{2}=\cos60^\circ$
$\Big[\because\cos60^\circ=\frac{1}{2}\Big]$
$\Rightarrow\ \beta=60^\circ$
$\therefore\ \alpha+\beta=30^\circ+60^\circ=90^\circ$
MCQ 311 Mark
If $\theta=45$ then $\frac{2\tan\theta}{1+\tan^2\theta}\text{is}:$
- ✓$1$
- B$0$
- C$2$
- D$3$
Answer
View full question & answer→Correct option: A.
$1$
$\frac{2\tan\theta}{1+\tan^2\theta}$
$=\frac{2\tan45^\circ}{1+\tan^245^\circ}$
$=\frac{2\times1}{1+1}=\frac{2}{2}=1$
$=\frac{2\tan45^\circ}{1+\tan^245^\circ}$
$=\frac{2\times1}{1+1}=\frac{2}{2}=1$
MCQ 321 Mark
If $\sec\text{4A}=\text{cosec}(\text{A}-10^\circ)$ and $\text{4A}$ is acute then $\angle\text{A}=?$
- ✓$20^\circ$
- B$30^\circ$
- C$40^\circ$
- D$50^\circ$
Answer
View full question & answer→Correct option: A.
$20^\circ$
$\sec\text{4A}=\text{cosec}(\text{A}-10^\circ)$
$\Rightarrow\text{cosec}(90^\circ-\text{4A})=\text{cosec}(\text{A}-10^\circ)$
$\Rightarrow90^\circ-\text{4A}=\text{A}-10^\circ$
$\Rightarrow\text{5A}=100^\circ$
$\Rightarrow\text{A}=20^\circ$
$\Rightarrow\text{cosec}(90^\circ-\text{4A})=\text{cosec}(\text{A}-10^\circ)$
$\Rightarrow90^\circ-\text{4A}=\text{A}-10^\circ$
$\Rightarrow\text{5A}=100^\circ$
$\Rightarrow\text{A}=20^\circ$
MCQ 331 Mark
The value of $\Big(\sin^230^\circ\cos^245^\circ+4\tan^230^\circ+\frac{1}{2}\sin^290^\circ+\frac{1}{8}\cot^260^\circ\Big)=?$
- A$\frac{3}8{}$
- B$\frac{5}{8}$
- C$6$
- ✓$2$
Answer
View full question & answer→Correct option: D.
$2$
$\Big(\sin^230^\circ\cos^245^\circ+4\tan^230^\circ+\frac{1}{2}\sin^290^\circ+\frac{1}{8}\cot^260^\circ\Big)$
$=\frac{1}{2^2}\times\frac{1}{\big(\sqrt2\big)^2}+4\times\frac{1}{\big(\sqrt3\big)^2}+\frac{1}{2}+1^2+\frac{1}{8}\times\frac{1}{\big(\sqrt3\big)^2}$
$\begin{bmatrix}\because\sin30^\circ=\frac{1}{2}\ \text{and }\cos45^\circ=\frac{1}{\sqrt2}\\\text{and }\tan30^\circ=\frac{1}{2 }\ \text{and }\cot60^\circ=\frac{1}{\sqrt3}\end{bmatrix}$
$=\frac{1}{4}\times\frac{1}{2}+4\times\frac{1}{3}+\frac{1}{2}+\frac{1}{24}$
$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}$
$=\frac{3+32+12+1}{24}$
$=\frac{48}{24}=2$
$=\frac{1}{2^2}\times\frac{1}{\big(\sqrt2\big)^2}+4\times\frac{1}{\big(\sqrt3\big)^2}+\frac{1}{2}+1^2+\frac{1}{8}\times\frac{1}{\big(\sqrt3\big)^2}$
$\begin{bmatrix}\because\sin30^\circ=\frac{1}{2}\ \text{and }\cos45^\circ=\frac{1}{\sqrt2}\\\text{and }\tan30^\circ=\frac{1}{2 }\ \text{and }\cot60^\circ=\frac{1}{\sqrt3}\end{bmatrix}$
$=\frac{1}{4}\times\frac{1}{2}+4\times\frac{1}{3}+\frac{1}{2}+\frac{1}{24}$
$=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}$
$=\frac{3+32+12+1}{24}$
$=\frac{48}{24}=2$
MCQ 341 Mark
If $5\tan\theta-4=0,$ then the value of $\frac{5\sin\theta-4\cos\theta}{5\sin\theta+4\cos\theta}$ is :
- A$\frac{5}{3}$
- B$\frac{5}{6}$
- ✓$\ 0$
- D$\frac{1}{6}$
Answer
View full question & answer→Correct option: C.
$\ 0$
Given that : $5\tan\theta-4=0.$
We have to find the value of the following expression
$\frac{5\sin\theta-4\cos\theta}{5\sin\theta+4\cos\theta}$
Since $5\tan\theta-4=0$
$\Rightarrow \tan\theta =\frac{4}{5}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
We know that :
$\Rightarrow \text{Base}=5$
$\Rightarrow{\text{perpendicular}}=4$
$\Rightarrow{\text{Hypotenuse=}\sqrt{\text{(Perpendicular)}^2+(\text{Base)}^2}}$
$\Rightarrow{\text{Hypotenuse=}}\sqrt{16+25}$
$\Rightarrow{\text{Hypotenuse=}}\sqrt{41}$
Since $\sin\theta =\frac{{\text{Perpendicular}}}{\text{Hypotenuse}}$ and $\cos\theta=\frac{\text{Base}}{{\text{Hypotenuse}}}$
Now we find
$\frac{5\sin\theta-4\cos\theta}{5\sin\theta+4\cos\theta}$
$=\frac{5\times\frac{4}{\sqrt{41}}-4\times\frac{5}{\sqrt{41}}}{5\times\frac{4}{\sqrt{41}}+4\times\frac{5}{\sqrt{41}}}$
$=\frac{\frac{20}{\sqrt{41}}-\frac{20}{\sqrt{41}}}{\frac{20}{\sqrt{41}}+\frac{20}{\sqrt{41}}}$
$=0$
Hence the correct option is $(c)$
We have to find the value of the following expression
$\frac{5\sin\theta-4\cos\theta}{5\sin\theta+4\cos\theta}$
Since $5\tan\theta-4=0$
$\Rightarrow \tan\theta =\frac{4}{5}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}$
We know that :
$\Rightarrow \text{Base}=5$
$\Rightarrow{\text{perpendicular}}=4$
$\Rightarrow{\text{Hypotenuse=}\sqrt{\text{(Perpendicular)}^2+(\text{Base)}^2}}$
$\Rightarrow{\text{Hypotenuse=}}\sqrt{16+25}$
$\Rightarrow{\text{Hypotenuse=}}\sqrt{41}$
Since $\sin\theta =\frac{{\text{Perpendicular}}}{\text{Hypotenuse}}$ and $\cos\theta=\frac{\text{Base}}{{\text{Hypotenuse}}}$
Now we find
$\frac{5\sin\theta-4\cos\theta}{5\sin\theta+4\cos\theta}$
$=\frac{5\times\frac{4}{\sqrt{41}}-4\times\frac{5}{\sqrt{41}}}{5\times\frac{4}{\sqrt{41}}+4\times\frac{5}{\sqrt{41}}}$
$=\frac{\frac{20}{\sqrt{41}}-\frac{20}{\sqrt{41}}}{\frac{20}{\sqrt{41}}+\frac{20}{\sqrt{41}}}$
$=0$
Hence the correct option is $(c)$
MCQ 351 Mark
If $\sin\text{3A}=\cos(\text{A}-10^\circ)$ and $\text{3A}$ is acute then $\angle\text{A}=?$
- A$35^\circ$
- ✓$25^\circ$
- C$20^\circ$
- D$45^\circ$
Answer
View full question & answer→Correct option: B.
$25^\circ$
$\sin\text{3A}=\cos(\text{A}-10^\circ)$
$\Rightarrow\cos(90^\circ-\text{3A})=\cos(\text{A}-10^\circ)$
$\Rightarrow90^\circ-\text{3A}=\text{A}-10^\circ$
$\Rightarrow\text{4A}=100^\circ$
$\Rightarrow\text{A}=25^\circ$
$\Rightarrow\cos(90^\circ-\text{3A})=\cos(\text{A}-10^\circ)$
$\Rightarrow90^\circ-\text{3A}=\text{A}-10^\circ$
$\Rightarrow\text{4A}=100^\circ$
$\Rightarrow\text{A}=25^\circ$
MCQ 361 Mark
If angles $A, B, C $ to a $\triangle\text{ABC}$ from an increasing $AP,$ then $\sin B =$
- A$\frac{1}{2}$
- ✓$\frac{\sqrt{3}}{2}$
- C$1$
- D$\frac{1}{\sqrt{2}}$
Answer
View full question & answer→Correct option: B.
$\frac{\sqrt{3}}{2}$
Let the angles $A, B , C$ of $\triangle\text{ABC}$
$\angle\text{A}=(\text{a}-\text{d})$
$\angle\text{B}=\text{a}$
$\angle\text{C}=\text{a}+\text{d}$
from an increasing $A.P$
then sum of the all there angles of $\triangle\text{ABC}$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow(\text{a} -\text{d})+\text{a}+(\text{a+d})=180^\circ$
$\Rightarrow\text{3a}=180^\circ $
$\Rightarrow\text{a}=60^\circ =\angle\text{B}$
then $\sin\text{b=}\sin\text{a}=\sin60^\circ\ ($from the table$)$
$=\frac{\sqrt{3}}{2}$
Hence the correct option is $(b)$
$\angle\text{A}=(\text{a}-\text{d})$
$\angle\text{B}=\text{a}$
$\angle\text{C}=\text{a}+\text{d}$
from an increasing $A.P$
then sum of the all there angles of $\triangle\text{ABC}$
$\Rightarrow\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow(\text{a} -\text{d})+\text{a}+(\text{a+d})=180^\circ$
$\Rightarrow\text{3a}=180^\circ $
$\Rightarrow\text{a}=60^\circ =\angle\text{B}$
then $\sin\text{b=}\sin\text{a}=\sin60^\circ\ ($from the table$)$
$=\frac{\sqrt{3}}{2}$
Hence the correct option is $(b)$
MCQ 371 Mark
Choose the correct answer from the given four options. If $4\tan\theta=3,\text{then}\Big(\frac{4\sin\theta-\cos\theta}{4\sin\theta+\cos\theta}\Big)$ is equal to :
- A$\frac{2}{3}$
- B$\frac{1}{3}$
- ✓$\frac{1}{2}$
- D$\frac{3}{4}$
Answer
View full question & answer→Correct option: C.
$\frac{1}{2}$
Given $,4\tan\theta=3$
$\Rightarrow\tan\theta=\frac{3}{4}\ \ ...(\text{i})$
$\therefore\ \frac{4\sin\theta-\cos\theta}{4\sin\theta+\cos\theta}=\frac{4\frac{\sin\theta}{\cos\theta}-1}{4\frac{\sin\theta}{\cos\theta}+1}$
$[$divide by $\cos\theta$ in both numerator and denominator$]$
$=\frac{4\tan\theta-1}{4\tan\theta+1}$
$\bigg[\because\ \tan\theta=\frac{\sin\theta}{\cos\theta}\bigg]$
$=\frac{4\big(\frac{3}{4}\big)-1}{4\big(\frac{3}{4}\big)+1}$
$=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}\ [$put the value from Eq.$(i)]$
$\Rightarrow\tan\theta=\frac{3}{4}\ \ ...(\text{i})$
$\therefore\ \frac{4\sin\theta-\cos\theta}{4\sin\theta+\cos\theta}=\frac{4\frac{\sin\theta}{\cos\theta}-1}{4\frac{\sin\theta}{\cos\theta}+1}$
$[$divide by $\cos\theta$ in both numerator and denominator$]$
$=\frac{4\tan\theta-1}{4\tan\theta+1}$
$\bigg[\because\ \tan\theta=\frac{\sin\theta}{\cos\theta}\bigg]$
$=\frac{4\big(\frac{3}{4}\big)-1}{4\big(\frac{3}{4}\big)+1}$
$=\frac{3-1}{3+1}=\frac{2}{4}=\frac{1}{2}\ [$put the value from Eq.$(i)]$
MCQ 381 Mark
$\sec70^\circ\sin20^\circ+\cos20^\circ\text{cosec}70^\circ=?$
- A$0$
- B$1$
- C$-1$
- ✓$2$
Answer
View full question & answer→Correct option: D.
$2$
$\sec70^\circ\sin20^\circ+\cos20^\circ\text{cosec}70^\circ$
$=\frac{1}{\cos70^\circ}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\sin70^\circ}$
$=\frac{1}{\cos(90^\circ-20^\circ)}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\sin(90^\circ-20^\circ)}$
$=\frac{1}{\sin20^\circ}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\cos20^\circ}$
$=1+1$
$=2$
$=\frac{1}{\cos70^\circ}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\sin70^\circ}$
$=\frac{1}{\cos(90^\circ-20^\circ)}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\sin(90^\circ-20^\circ)}$
$=\frac{1}{\sin20^\circ}\times\sin20^\circ+\cos20^\circ\times\frac{1}{\cos20^\circ}$
$=1+1$
$=2$
MCQ 391 Mark
Choose the correct option. Justify your choice : $(\sec\text{A}+\tan\text{A})(1-\sin\text{A})= $
- A$\sec \text{A}$
- B$\sin \text{A}$
- C$\text{cosec A}$
- ✓$\cos \text{A}$
Answer
View full question & answer→Correct option: D.
$\cos \text{A}$
$(\sec\text{A}+\tan\text{A})(1-\sin\text{A}) $
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\frac{1-\sin^2\text{A}}{\cos\text{A}} \ \ [\text{since} (\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
$=\frac{\cos^2\text{A}}{\cos\text{A}} $
$=\cos\text{A} \ \ [\because1-\sin^2 \text{A}=\cos^2 \text{A}]$
$=\Big(\frac{1}{\cos\text{A}}+\frac{\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\Big(\frac{1+\sin\text{A}}{\cos\text{A}}\Big)(1-\sin\text{A}) $
$=\frac{1-\sin^2\text{A}}{\cos\text{A}} \ \ [\text{since} (\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2]$
$=\frac{\cos^2\text{A}}{\cos\text{A}} $
$=\cos\text{A} \ \ [\because1-\sin^2 \text{A}=\cos^2 \text{A}]$
MCQ 401 Mark
In Fin. the value of $\cos\phi$ is :
- A$\frac{5}{4}$
- B$\frac{5}{3}$
- C$\frac{3}{5}$
- ✓$\frac{4}{5}$
Answer
View full question & answer→Correct option: D.
$\frac{4}{5}$
We should proceed with the fact that sum of angles on one side of a straight line is $180^\circ $
So from the given figure,
$\theta+\phi+90^\circ=180^\circ$
So, $\theta=90^\circ-\phi\ \dots(1)$
Now from the triangle $\triangle\text{ABC},$
$\sin\theta=\frac{4}{5}$
Now we will use equation $(1)$ in the above,
$\sin(90^\circ-\phi)=\frac{4}{5}$
Therefore, $\cos\phi=\frac{4}{5}$
So the answer is $(d)$
So from the given figure,
$\theta+\phi+90^\circ=180^\circ$
So, $\theta=90^\circ-\phi\ \dots(1)$
Now from the triangle $\triangle\text{ABC},$
$\sin\theta=\frac{4}{5}$
Now we will use equation $(1)$ in the above,
$\sin(90^\circ-\phi)=\frac{4}{5}$
Therefore, $\cos\phi=\frac{4}{5}$
So the answer is $(d)$
MCQ 411 Mark
What is the value of $\sin30^\circ\times\text{cosec}\ 30^\circ?$
- ✓$1$
- B$0$
- C$\frac{1}{2}$
- D$2$
Answer
View full question & answer→Correct option: A.
$1$
$\sin30^\circ\times\text{cosec}\ 30^\circ$
$=\frac{1}{2}\times\frac{2}{1}=\frac{1}{1}=1$
$=\frac{1}{2}\times\frac{2}{1}=\frac{1}{1}=1$
MCQ 421 Mark
If $\cos(\alpha+\beta)=0$ then $\sin(\alpha-\beta)=?$
- A$\sin\alpha$
- B$\cos\beta$
- C$\sin2\alpha$
- ✓$\cos2\beta$
Answer
View full question & answer→Correct option: D.
$\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\cos(\alpha+\beta)=\cos90^\circ$
$\Rightarrow\alpha+\beta=90^\circ$
$\Rightarrow\alpha=90^\circ-\beta$
Now, $\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
$\Rightarrow\cos(\alpha+\beta)=\cos90^\circ$
$\Rightarrow\alpha+\beta=90^\circ$
$\Rightarrow\alpha=90^\circ-\beta$
Now, $\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)$
$=\sin(90^\circ-2\beta)$
$=\cos2\beta$
MCQ 431 Mark
Choose the correct answer from the given four options. $\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$ is equal to :
- A$2\cos\theta$
- ✓$0$
- C$2\sin\theta$
- D$1$
Answer
View full question & answer→Correct option: B.
$0$
$\sin(45^\circ+\theta)-\cos(45^\circ-\theta)$
$=\cos[90^\circ-(45^\circ+\theta)]-\cos(45^\circ-6)$
$[\therefore\cos(90^\circ-\theta)=\sin0]$
$=\cos(45^\circ-0)-\cos(45^\circ-0)$
$=0$
$=\cos[90^\circ-(45^\circ+\theta)]-\cos(45^\circ-6)$
$[\therefore\cos(90^\circ-\theta)=\sin0]$
$=\cos(45^\circ-0)-\cos(45^\circ-0)$
$=0$
MCQ 441 Mark
If $\sin\theta=\frac{\text{a}}{\text{b}}$ then $\cos\theta=?$
- A$\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
- ✓$\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
- C$\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
- D$\frac{\text{b}}{\text{a}}$
Answer
View full question & answer→Correct option: B.
$\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
$\cos^2\theta=1-\sin^2\theta$
$=1-\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=1-\frac{\text{a}^2}{\text{b}^2}$
$=\frac{\text{b}^2-\text{a}^2}{\text{b}^2}$
$\Rightarrow\cos\theta=\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
$=1-\Big(\frac{\text{a}}{\text{b}}\Big)^2$
$=1-\frac{\text{a}^2}{\text{b}^2}$
$=\frac{\text{b}^2-\text{a}^2}{\text{b}^2}$
$\Rightarrow\cos\theta=\frac{\sqrt{\text{b}^2-\text{a}^2}}{\text{b}}$
MCQ 451 Mark
$\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta + 1}$ is equal to :
- A$2\tan\theta$
- B$2\sec\theta$
- ✓$2\text{cosec }\theta$
- D$2\tan\theta\sec\theta$
Answer
View full question & answer→Correct option: C.
$2\text{cosec }\theta$
The givne expression is $\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
Simplifying the given expression, we have
$\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
$=\frac{\tan\theta(\sec\theta+1)+\tan\theta(\sec\theta-1)}{(\sec\theta-1)(\sec\theta+1)}$
$=\frac{\tan\theta\sec\theta+\tan\theta+\tan\theta\sec\theta-\tan\theta}{\sec^2\theta-1}$
$=\frac{2\tan\theta\sec\theta}{\tan^2\theta}$
$=\frac{2\sec\theta}{\tan\theta}$
$=\frac{2\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=2\frac{1}{\sin\theta}$
$=2\text{cosec }\theta$
Therefore, the correct option is $(c).$
Simplifying the given expression, we have
$\frac{\tan\theta}{\sec\theta-1}+\frac{\tan\theta}{\sec\theta+1}$
$=\frac{\tan\theta(\sec\theta+1)+\tan\theta(\sec\theta-1)}{(\sec\theta-1)(\sec\theta+1)}$
$=\frac{\tan\theta\sec\theta+\tan\theta+\tan\theta\sec\theta-\tan\theta}{\sec^2\theta-1}$
$=\frac{2\tan\theta\sec\theta}{\tan^2\theta}$
$=\frac{2\sec\theta}{\tan\theta}$
$=\frac{2\frac{1}{\cos\theta}}{\frac{\sin\theta}{\cos\theta}}$
$=2\frac{1}{\sin\theta}$
$=2\text{cosec }\theta$
Therefore, the correct option is $(c).$
MCQ 461 Mark
If $16\cot\times=12,$ then $\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}} $ equals :
- ✓$\frac{1}{7}$
- B$\frac{3}{7}$
- C$\frac{2}{7}$
- D$0$
Answer
View full question & answer→Correct option: A.
$\frac{1}{7}$
We are given $16\cot\text{x}=12.$We are asked to find the following
$\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}} $
We know that: $\cot\text{x}=\frac{\text{Base}}{\text{Perpendicular}}$
$\Rightarrow{\text{Base}}=3$
$\Rightarrow\text{Perpendicular}=4$
$\Rightarrow \text{Hypotenuse}=\sqrt{(\text{Perpendicular)}^2+\text{(Base)}^2}$
$\Rightarrow \text{Hypotenuse}=\sqrt{16+9}$
$\Rightarrow \text{Hypotenuse}=5$
Now we have
$16\cot\text{x}=12$
$\cot\text{x}=\frac{12}{16}$
$\cot\text{x}=\frac{3}{4}$
We know $\sin\text{x}=\frac{ \text{Perpendicular}}{ \text{Hypotenuse}}$ and $\cos\text{x}=\frac{ \text{Base}}{ \text{Hypotenuse}}$
Now we find
$\frac{\sin\text{x-\cos}\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}$
$=\frac{\frac{1}{5}}{\frac{7}{5}}$
$=\frac{1}{7}$
Hence the correct option is $(a)$
$\frac{\sin\text{x}-\cos\text{x}}{\sin\text{x}+\cos\text{x}} $
We know that: $\cot\text{x}=\frac{\text{Base}}{\text{Perpendicular}}$
$\Rightarrow{\text{Base}}=3$
$\Rightarrow\text{Perpendicular}=4$
$\Rightarrow \text{Hypotenuse}=\sqrt{(\text{Perpendicular)}^2+\text{(Base)}^2}$
$\Rightarrow \text{Hypotenuse}=\sqrt{16+9}$
$\Rightarrow \text{Hypotenuse}=5$
Now we have
$16\cot\text{x}=12$
$\cot\text{x}=\frac{12}{16}$
$\cot\text{x}=\frac{3}{4}$
We know $\sin\text{x}=\frac{ \text{Perpendicular}}{ \text{Hypotenuse}}$ and $\cos\text{x}=\frac{ \text{Base}}{ \text{Hypotenuse}}$
Now we find
$\frac{\sin\text{x-\cos}\text{x}}{\sin\text{x}+\cos\text{x}}$
$=\frac{\frac{4}{5}-\frac{3}{5}}{\frac{4}{5}+\frac{3}{5}}$
$=\frac{\frac{1}{5}}{\frac{7}{5}}$
$=\frac{1}{7}$
Hence the correct option is $(a)$
MCQ 471 Mark
In right triangle $\text{ABC},$ right angled at $C,$ if $\tan A = 1,$ then the value of $2 \sin A \cos A$ is :
- A$-1$
- ✓$1$
- C$2$
- D$0$
Answer
View full question & answer→Correct option: B.
$1$
Let $BC = k$ and $AC = k$
$\therefore\text{AB} = \sqrt{{\text{(k})}^{2}+(\text{k})^{2}}$
$\sqrt{{\text{k}}^{2}+\text{k}^{2}}$
$\Rightarrow\text{AB} = \sqrt{2\text{k}^{2}}$
$=\sqrt{2\text{k}}$
$\therefore{2}\sin\text{A}\cos\text{A} = 2\times\frac{\text{k}}{\sqrt2\text{k}}\times\frac{\text{k}}{\sqrt2\text{k}}$
$=2\times\frac{1}{2} = 1$
$\therefore\text{AB} = \sqrt{{\text{(k})}^{2}+(\text{k})^{2}}$
$\sqrt{{\text{k}}^{2}+\text{k}^{2}}$
$\Rightarrow\text{AB} = \sqrt{2\text{k}^{2}}$
$=\sqrt{2\text{k}}$
$\therefore{2}\sin\text{A}\cos\text{A} = 2\times\frac{\text{k}}{\sqrt2\text{k}}\times\frac{\text{k}}{\sqrt2\text{k}}$
$=2\times\frac{1}{2} = 1$
MCQ 481 Mark
If $3\cos\theta=5\sin\theta,$ then the value of $\frac{5\sin\theta-2\sec^3\theta+2\cos\theta}{5\sin\theta+2\sec^3\theta-2\cos\theta}$ is :
- ✓$\frac{271}{979}$
- B$\frac{316}{2937}$
- C$\frac{542}{2937}$
- D$\text{None of these}$
Answer
View full question & answer→Correct option: A.
$\frac{271}{979}$
We have,
$3 \cos\theta=5\sin\theta$
$\frac{\cos\theta}{\sin\theta}=\frac{5}{3 }$
${\cot\theta}=\frac{5}{3} $
In $\triangle\text{ABC}, $
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow \text{AC}^2=(3)^2+(5)^2$
$\Rightarrow \text{AC}^2=9+25$
$\Rightarrow \text{AC}^2=34$
$\Rightarrow \text{AC}=\sqrt{34}$
$\therefore \sin\theta=\frac{3}{\sqrt{34}}\ \cos\theta=\frac{5}{\sqrt{34}}\ \sec\theta=\frac{\sqrt{34}}{{5}}$
Now, $\frac{5\sin\theta-2\sec^3+2\cos}{5\sin\theta+2\sec^3-2\cos}$
$=\frac{5\times\frac{3}{\sqrt{34}}-2\Big(\frac{\sqrt{34}}{5}\Big)^3+2\times\frac{5}{\sqrt{34}}}{5\times\frac{3}{\sqrt{34}}+2\Big(\frac{\sqrt{34}}{5}\Big)^3-2\times\frac{5}{\sqrt{34}}}$
$=\frac{\frac{125\times15-2\times34\times34+10\times125}{125\sqrt{34}}}{\frac{125\times15+2\times34\times34-10\times125}{125\sqrt{34}}}$
$=\frac{1875-2312+1250}{1875+2312-1250}$
$=\frac{813}{2937} $
$=\frac{271}{979} $
Thus, $\frac{5\sin\theta-2\sec\theta+2\cos\theta}{5\sin\theta+2\sec\theta-2\cos\theta}=\frac{271}{979}$
Hence the correct option is $(a)$
$3 \cos\theta=5\sin\theta$
$\frac{\cos\theta}{\sin\theta}=\frac{5}{3 }$
${\cot\theta}=\frac{5}{3} $
In $\triangle\text{ABC}, $
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow \text{AC}^2=(3)^2+(5)^2$
$\Rightarrow \text{AC}^2=9+25$
$\Rightarrow \text{AC}^2=34$
$\Rightarrow \text{AC}=\sqrt{34}$
$\therefore \sin\theta=\frac{3}{\sqrt{34}}\ \cos\theta=\frac{5}{\sqrt{34}}\ \sec\theta=\frac{\sqrt{34}}{{5}}$
Now, $\frac{5\sin\theta-2\sec^3+2\cos}{5\sin\theta+2\sec^3-2\cos}$
$=\frac{5\times\frac{3}{\sqrt{34}}-2\Big(\frac{\sqrt{34}}{5}\Big)^3+2\times\frac{5}{\sqrt{34}}}{5\times\frac{3}{\sqrt{34}}+2\Big(\frac{\sqrt{34}}{5}\Big)^3-2\times\frac{5}{\sqrt{34}}}$
$=\frac{\frac{125\times15-2\times34\times34+10\times125}{125\sqrt{34}}}{\frac{125\times15+2\times34\times34-10\times125}{125\sqrt{34}}}$
$=\frac{1875-2312+1250}{1875+2312-1250}$
$=\frac{813}{2937} $
$=\frac{271}{979} $
Thus, $\frac{5\sin\theta-2\sec\theta+2\cos\theta}{5\sin\theta+2\sec\theta-2\cos\theta}=\frac{271}{979}$
Hence the correct option is $(a)$
MCQ 491 Mark
If $\sin\theta = \frac{5}{13}$ then $\cos\theta =$
- A$\frac{\sqrt{5}}{13}$
- ✓$\frac{12}{13}$
- C$\frac{13}{12}$
- D$\frac{12}{5}$
Answer
View full question & answer→Correct option: B.
$\frac{12}{13}$
Let $AB = 5k$ and $AC = 13k$
$\therefore\text{BC} = \sqrt{(13\text{k})^{2}-(5\text{k})^{2}} $
$= \sqrt{{169}\text{k}^{2}-25\text{k}^{2}}$
$\Rightarrow \text{BC}= \sqrt{144}\text{k}^{2} = 12\text{k}$
$\therefore\cos\theta=\frac{\text{BC}}{\text{AC}} = \frac{\text{12k}}{\text{13k}} = \frac{12}{13}$
$\therefore\text{BC} = \sqrt{(13\text{k})^{2}-(5\text{k})^{2}} $
$= \sqrt{{169}\text{k}^{2}-25\text{k}^{2}}$
$\Rightarrow \text{BC}= \sqrt{144}\text{k}^{2} = 12\text{k}$
$\therefore\cos\theta=\frac{\text{BC}}{\text{AC}} = \frac{\text{12k}}{\text{13k}} = \frac{12}{13}$
MCQ 501 Mark
If $\cos(\alpha+\beta)=0,$ then $\sin(\alpha-\beta)$ can be reduced to :
- A$\cos\beta$
- ✓$\cos2\beta$
- C$\sin\alpha$
- D$\sin2\alpha$
Answer
View full question & answer→Correct option: B.
$\cos2\beta$
$\cos(\alpha+\beta)=0$
$\Rightarrow\ \alpha+\beta=90^\circ \big[\because\ \cos90^\circ=0\big]$
$\Rightarrow\ \alpha=90^\circ-\beta\ .....(\text{i})$
$\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ \big[\text{using (i)}\big]$
$=\sin\big(90^\circ-2\beta\big)$
$=\cos2\beta\ \big[\because \sin(90^\circ -\theta)=\cos\theta\big]$
$\Rightarrow\ \alpha+\beta=90^\circ \big[\because\ \cos90^\circ=0\big]$
$\Rightarrow\ \alpha=90^\circ-\beta\ .....(\text{i})$
$\sin(\alpha-\beta)=\sin(90^\circ-\beta-\beta)\ \big[\text{using (i)}\big]$
$=\sin\big(90^\circ-2\beta\big)$
$=\cos2\beta\ \big[\because \sin(90^\circ -\theta)=\cos\theta\big]$