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Maths 3 Marks Question

Trigonometric Ratios · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 71 questions.

Questions · Page 2 of 2

3 Marks Question

Question 513 Marks
If $\tan\theta=\frac{24}{7},$ find that $\sin\theta+\cos\theta.$
Answer
$\tan\theta=\frac{24}{7}$ find $\sin\theta+\cos\theta$

Let $x - 1$ be the hypotenuse By applying Pythagoras theorem we get
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\text{x}^2=(24)^2+(7)^2$
$\text{x}^2=576+49=62.5$
$\text{x}=25$
$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}$
$\sin\theta+\cos\theta=\frac{24}{25}+\frac{7}{25}$
$=\frac{31}{25}$
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Question 523 Marks
If $A = 30^\circ $ and $B = 60^\circ ,$ verify that.
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
Answer
We have,
$\sin(\text{A}-\text{B})=\frac{1}{2}$ and $\cos(\text{A}+\text{B})=\frac{1}{2}$
Now, $\sin(\text{A}-\text{B})=\frac{1}{2}$
$\Rightarrow\sin(\text{A}-\text{B})=\sin30^\circ$
$\Rightarrow(\text{A}-\text{B})=30^\circ\dots(1)$
And, $\cos(\text{A}+\text{B})=\frac{1}{2}$
$\Rightarrow\cos(\text{A}+\text{B})=\cos60^\circ$
$\Rightarrow(\text{A}+\text{B})=60^\circ\dots(2)$
Adding $(1)$ and $(2),$ we get,
$\Rightarrow2\text{A}=90^\circ$
$\Rightarrow\text{A}=\frac{90^\circ}{2}=45^\circ$
Put $\text{A}=45^\circ$ in $(2),$ we get,
$\Rightarrow45^\circ+\text{B}=60^\circ$
$\Rightarrow\text{B}=60^\circ-45^\circ$
$\Rightarrow\text{B}=15^\circ$
Thus, $\text{A}=45^\circ$ and $\text{B}=15^\circ$
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Question 533 Marks
Evaluate the following:
$ \frac{\tan^260^\circ+4\cos^245^\circ+3\sec^230^\circ+5\cos^290^\circ}{\text{cosec30}^\circ+\sec60^\circ-\cot^230^\circ}$
Answer
$ \frac{\tan^260^\circ+4\cos^245^\circ+3\sec^230^\circ+5\cos^290^\circ}{\text{cosec30}^\circ+\sec60^\circ-\cot^230^\circ}$$=\frac{(\sqrt{3})^2+4\big(\frac{1}{\sqrt{2}}\big)^2+3\big(\frac{2}{\sqrt{3}}\big)^2+5\times0}{2+2-(\sqrt{3})^2}$
$ =\frac{3+4\times\frac{1}{2}+3\times\frac{4}{3}+0}{4-3}$
$=3+2+4$
$=9$
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Question 543 Marks
Evaluate the following:
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
Answer
We have to find the value of the following expression
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ\dots(1)$
$\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2},\sin60^\circ=\frac{\sqrt3}{2}\cos60^\circ=\frac{1}{2}$
Now
So by substituting above values in equation $(1)$
We get,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
$=\frac{1}{2}\times\frac{1}{\sqrt2}-\frac{\sqrt3}{2}\times\frac{1}{\sqrt2}$
$=\frac{1}{2\sqrt2}-\frac{\sqrt3}{2\sqrt2}$
$=\frac{1-\sqrt3}{2\sqrt2}$
Therefore,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ=\frac{1-\sqrt3}{2\sqrt2}$
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Question 553 Marks
Evaluate:
$\frac{2}{3}(\cos^430^\circ-\sin^445^\circ)-3(\sin^260^\circ-\sec^245^\circ)+\frac{1}{4}\cot^230^\circ$
Answer
$\cos30^\circ=\frac{\sqrt{3}}{2},\sin60^\circ=\frac{\sqrt{3}}{2},\cot30^\circ=\sqrt{3},$ $\sin45^\circ=\frac{1}{\sqrt{2}},\sec45^\circ=\sqrt{2}$
Substituting above value in question,
$=\frac{2}{3}\bigg[\Big(\frac{\sqrt{3}}{2}\Big)^4-\Big(\frac{1}{\sqrt{2}}\Big)^4\bigg]-3\bigg[\Big(\frac{\sqrt{3}}{2}\Big)^2-\Big(\sqrt{2}\Big)^2\bigg]+\frac{1}{4}\Big(\sqrt{3}\Big)^2$
$=\frac{2}{3}\Big[\frac{9}{16}-\frac{1}{4}\Big]-3\Big[\frac{3}{4}-2\Big]+\frac{1}{4}\times3$
$=\frac{2}{3}\Big[\frac{5}{16}\Big]-3\Big[\frac{-5}{4}\Big]+\frac{3}{4}$
$=\frac{5}{24}+\frac{15}{4}+\frac{3}{4}$
$=\frac{5+90+18}{24}$
$=\frac{133}{24}$
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Question 563 Marks
In the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.
$\sin\theta=\frac{11}{15}$
Answer
$\sin\theta=\frac{11}{5}$
We know $ \sin{\theta}=\frac{\text{perpendicullar}}{\text{Hypotenuse}}=\frac{11}{15}$
Consider right angled $\triangle^\text{le}\text{ACB}.$

Let $x =$ adjacent side
By applying Pythagoras
$\text{AB}^2=\text{AC}^2+\text{BC}^2$
$225=121-\text{x}^2$
$\text{x}^2=225-121$
$\text{x}^2=104$
$\text{x}=\sqrt{104}$
$\cos\theta=\frac{\text{adjacent side}}{\text{hypotenuse}}=\frac{\sqrt{104}}{15}$
$\tan\theta=\frac{\text{opposite side}}{\text{adjacent side}}=\frac{11}{\sqrt{104}}$
$\text{cosec }\theta=\frac{1}{\sin\theta}=\frac{15}{11}$
$\sec\theta=\frac{1}{\cos\theta}=\frac{15}{\sqrt{104}}$
$\cot\theta=\frac{1}{\tan\theta}=\frac{\sqrt{104}}{11}$
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Question 573 Marks
Evaluate the following:
$2\sin^230^\circ-3\cos^245^\circ+\tan^260^\circ$
Answer
$2\sin^230^\circ-3\cos^245^\circ+\tan^260^\circ....(\text{i})$
By trigonometric ratios we have
$\sin30^\circ=\frac{1}{2},\ \cos45^\circ=\frac{1}{\sqrt{2}},\ \tan60^\circ=\sqrt{3}$
By substituing above values in (i) we get
$2\cdot\Big[\frac{1}{2}\Big]^2-3\Big[\frac{1}{\sqrt{2}}\Big]^2+\big[\sqrt{3}\big]^2$
$2\cdot\frac{1}{4}-3\cdot\frac{1}{2}+3$
$\frac{1}{2}-\frac{3}{2}+3\Rightarrow\frac{3}{2}+2=2$
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Question 583 Marks
Evaluate the following:
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
Answer
We have
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}\ \dots(1)$
Now,
$\sin45^\circ=\frac{1}{\sqrt{2}},\ \sin30^\circ=\frac{1}{2},\ \sin90^\circ=1,$ $\tan45^\circ=\cot45^\circ=1,\ \sin60^\circ=\cos30^\circ=\frac{\sqrt{3}}{2}$
$\sec60^\circ=2$
So by substituting above values in equation $(1)$
We get,
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
$ =\frac{\frac{1}{2}}{\frac{1}{\sqrt{2}}}+\frac{1}{2}-\frac{\frac{\sqrt{3}}{2}}{1}-\frac{\frac{\sqrt{3}}{2}}{1}$
Now by further simplifying
We get,
$ \frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ} $
$ =\frac{1}{\sqrt{2}\times\sqrt{2}}\times\frac{\sqrt{2}}{1}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
Now, one $\sqrt{2}$ gets cancelled and
We get,
$\frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
$ =\frac{1}{\sqrt{2}}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
Now, by taking $LCM$
We get,
$ \frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$
$=\frac{1\times\sqrt{2}}{\sqrt{2}\times\sqrt{2}}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
$=\frac{\sqrt{2}}{2}+\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}$
$ =\frac{\sqrt{2}+1-\sqrt{3}-\sqrt{3}}{2}$
$=\frac{\sqrt{2}+1-2\sqrt{3}}{2}$
Therefore,
$ \frac{\sin30^\circ}{\sin45^\circ}+\frac{\tan45^\circ}{\sec60^\circ}-\frac{\sin60^\circ}{\cot45^\circ}-\frac{\cos30^\circ}{\sin90^\circ}$ $=\frac{\sqrt{2}+1-2\sqrt{3}}{2}$
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Question 593 Marks
Given $15 \cot A = 8,$ find $\sin A$ and $\sec A.$
Answer
We have,
$15\cot\text{A}=8$
$\cot\text{A}=\frac{8}{15}=\frac{\text{Base}}{\text{Perpendicular}}$

In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow\text{AC}^2=(8)^2+(15)^2$
$\Rightarrow\text{AC}^2=64+225$
$\Rightarrow\text{AC}^2=289$
$\Rightarrow \text{AC}=17$
$\sin\text{A}=\frac{\text{perpendicular}}{\text{hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{15}{17}$
$\sec\text{A}=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{17}{8}$
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Question 603 Marks

If $3\cot\theta=4,$ find the value of $\frac{4\cos\theta-\sin\theta}{2\cos+\sin\theta}.$

Answer

We have,

$3\cot\theta=4$

$\cot\theta=\frac{4}{3}$

In $\triangle\text{ABC},$

$\text{AC}^2=\text{AB}^2+\text{BC}^2$

$\Rightarrow\text{AC}^2=(3)^2+(4)^2$

$\Rightarrow\text{AC}^2=9+16$

$\Rightarrow\text{AC}^2=25$

$\Rightarrow\text{AC}=5$

$\therefore\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{4}{5}$ and $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{3}{5}$

Now, $\frac{4\cos\theta-\sin\theta}{2\cos\theta+\sin\theta}=\frac{4\times\frac{4}{5}-\frac{3}{5}}{2\times\frac{4}{5}+\frac{3}{5}}$

$=\frac{\frac{16-3}{5}}{\frac{8+3}{5}}$

$=\frac{13}{5}\times\frac{5}{11}$

$=\frac{13}{11}$

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Question 613 Marks
Evaluate the following:
$\big(\text{cosesc}^245^\circ\sec^230^\circ\big)\big(\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ\big)$
Answer
$\big(\text{cosesc}^245^\circ\sec^230^\circ\big)\big(\sin^230^\circ+4\cot^245^\circ-\sec^260^\circ\big)\dots(\text{i})$
By trigonometric ration we have
$\text{cosec}45^\circ=\sqrt{2},\ \sec30^\circ=\frac{2}{\sqrt{3}},$ $\sin30^\circ=\frac{1}{2}\ \cot45^\circ=1\ \sec60^\circ=2$
By substituting above values in (i), we get
$\bigg[\Big(\sqrt{2}\Big)^2\cdot\Big(\frac{2}{\sqrt{3}}\Big)^2\bigg]\bigg[\Big(\frac{1}{2}\Big)^2+4(1)^2-(2)^2\bigg]$
$\Rightarrow\Big[2\cdot\frac{4}{3}\Big]\Big[\frac{1}{4}+4-4\Big]\Rightarrow2\cdot\frac{4}{3}\cdot\frac{1}{4}=\frac{2}{3}$
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Question 623 Marks
If $\sec\theta=\frac{13}{5},$ show that $\frac{\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=3.$
Answer
We have,
$\sec\theta=\frac{13}{5}$

In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$\Rightarrow(13)^2=\text{AB}^2+(5^2)$
$\Rightarrow\text{AB}^2=169-25$
$\Rightarrow\text{AB}^2=144$
$\Rightarrow\text{AB}=12$
$\therefore\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{12}{13}$ and $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{5}{13}$
Now, $\frac{2\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=\frac{2\times\frac{12}{13}-3\times\frac{5}{13}}{4\times\frac{12}{13}-9\times\frac{5}{13}}$
$=\frac{\frac{24-15}{13}}{\frac{48-45}{13}}$
$=\frac{9}{3}$
$=3$
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Question 633 Marks
Given $\tan\theta=\frac{1}{\sqrt{5}},$ what is the value of $\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}?$
Answer
$\tan\theta=\frac{1}{\sqrt{5}}=\frac{\text{Perpendicular}}{\text{Base}}$
Draw a right $\triangle\text{ABC}$ in which
$\angle\text{B}=90^\circ,\text{BC}=1,\text{AB}=\sqrt{5} \text{ units}$

By Pythagoras Theorem
$\text{AC}^2=\text{AB}^2+\text{BC}^2$
$=\big(\sqrt{5}\big)^2+(1)^2=5+1=6$
$\therefore\ \text{AC}=\sqrt{6}\text{ units }$
$\therefore\ \sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{\sqrt{6}}{\sqrt{5}}=\frac{\sqrt{6}}{\sqrt{5}}$
$\text{cosec }\theta=\frac{\text{Hypotenuse}}{\text{Perpendicular}}=\frac{\text{AC}}{\text{AB}}=\frac{\sqrt{6}}{1}=\sqrt{6}$
Now $\frac{\text{cosec }^2\theta-\sec^2\theta}{\text{cosec }^2\theta+\sec^2\theta}=\frac{\big(\sqrt{6}\big)^2-\Big(\sqrt{\frac{6}{5}}\Big)^2}{\big(\sqrt{6}\big)^2+\Big(\sqrt{\frac{6}{5}}\Big)^2}$
$=\frac{6-\frac{6}{5}}{6+\frac{6}{5}}=\frac{\frac{30-6}{5}}{\frac{30+6}{5}}=\frac{\frac{24}{5}}{\frac{36}{5}}$
$=\frac{24}{5}\times\frac{5}{36}=\frac{24}{36}=\frac{2}{3}$
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Question 653 Marks
If $\theta=30^\circ,$ verify that.
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
Answer
$\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
Substitutc $\theta=30^\circ$
$\text{L.H.S}=\cos2\theta,\ \text{R.H.S}=\frac{1-\tan^2\theta}{1+\tan^2\theta}$
$=\cos2(30^\circ)=\frac{1-\tan^230^\circ}{1+\tan^230^\circ}$
$=\cos60^\circ=\frac{1}{2}=\frac{1-\Big(\frac{1}{\sqrt{3}}\Big)^2}{1+\Big(\frac{1}{\sqrt{3}}\Big)^2}=\frac{1-\frac{1}{3}}{1+\frac{1}{3}}=\frac{\frac{2}{3}}{\frac{4}{3}}=\frac{1}{2}$
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Question 663 Marks
In $\triangle\text{PQR},$ right-angled at $Q, PQ = 3\ cm$ and $PR = 6\ cm.$ Determine $\angle\text{P}$ and $\angle\text{R}.$
Answer

From above figure
$\sin\text{R}=\frac{\text{PQ}}{\text{PR}}$
$\sin\text{R}=\frac{3}{6}=\frac{1}{2}$
$\therefore\ \sin\text{R}=\sin30^\circ$
$\text{R}=\sin30^\circ$
We Know in $\triangle\text{le }\angle\text{P}+\angle\text{Q}+\angle\text{R}=180^\circ$
$\angle\text{P}+90^\circ+30^\circ=180^\circ$
$\angle\text{P}=60^\circ$
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Question 673 Marks

If $\tan\theta=\frac{12}{13},$ find the value of $\frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}.$

Answer

$\tan\theta=\frac{\text{opposite side}}{\text{adjacent side}}$

Let x be, the hypotenuse

By pythagoras we get

$\text{AC}^2=\text{AB}^2+\text{BC}^2$

$\text{x}^2=144+169$

$\text{x}=\sqrt{313}$

$\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{12}{\sqrt{313}}$

$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{13}{\sqrt{313}}$

Substitute, $\sin\theta,\cos\theta$ in equation we get

$\frac{2\sin\theta,\cos\theta}{\cos^2\theta-\sin\theta}=\frac{2\times\frac{12}{\sqrt{313}}\times\frac{13}{\sqrt{313}}}{\frac{169}{313}-\frac{144}{313}}$

$=\frac{\frac{312}{313}}{\frac{25}{313}}=\frac{312}{25}$

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Question 683 Marks
If $\triangle\text{ABC}$ is a right triangle such that $\angle\text{C} = 90^\circ,\angle\text{A}=45^\circ$ and $BC = 7$ units. Find $\angle\text{B} AB$ and $AC.$
Answer
We have
$\angle\text{C}=90^\circ,\angle\text{A}=45^\circ$ and $\text{BC}=7\text{ units}$
In $\triangle\text{ABC},$
$\angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow45^\circ+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\angle\text{B}=180^\circ-135^\circ$
$\Rightarrow\angle\text{B}=45^\circ$

Now, $\tan\text{B}=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\tan45^\circ=\frac{\text{AC}}{7}$
$\Rightarrow\text{AC}=\frac{\text{BC}}{\tan45^\circ}$
$\Rightarrow\text{AC}=\frac{7}{1}$
$\Rightarrow\text{AC}=7\text{ units}$
And, $\cos\text{B}=\frac{\text{BC}}{\text{AB}}$
$\Rightarrow\text{AB}=\frac{\text{BC}}{\cos45^\circ}$
$\Rightarrow\text{AB}=\frac{7}{\frac{1}{\sqrt{2}}}$
$\Rightarrow\text{AB}=7\sqrt{2}\text{ units}$
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Question 693 Marks
If $\tan\theta=\frac{\text{a}}{\text{b}} ,$ find the value of $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}.$
Answer
$\tan\theta=\frac{\text{a}}{\text{b}}$ find $\frac{\cos\theta+\sin\theta}{\cos\theta-\sin\theta}$
Divide equation (i) with $\cos\theta,$ we get
$\Rightarrow\frac{\frac{\cos\theta+\sin\theta}{\cos\theta}}{\frac{\cos\theta-\sin\theta}{\cos\theta}}$
$\Rightarrow\frac{1+\frac{\sin\theta}{\cos\theta}}{ 1-\frac{\sin\theta}{\cos\theta}}$
$\Rightarrow\frac{1+\tan\theta}{1-\tan\theta}$
$=\frac{1+\frac{\text{a}}{\text{b}}}{1-\frac{\text{a}}{\text{b}}} $
$=\frac{\text{b}+\text{a}}{\text{b}-\text{a}}$
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Question 703 Marks

Evaluate the following:

$\sin 45^{\circ} \sin30^{\circ} +\cos 45^{\circ} \cos30^{\circ}$

Answer

$\sin 45^{\circ} \sin30^{\circ} +\cos 45^{\circ} \cos30^{\circ}\ \dots(1)$

We know that by trigonometric ratios we have,

$\sin45^\circ =\frac{1}{\sqrt2}\ \sin30^\circ=\frac{1}{\sqrt2}$

$\cos45^\circ=\frac{1}{\sqrt2}\ \cos30^\circ=\frac{\sqrt3}{2}$

Substituting the values in (i) we get

$\frac{1}{\sqrt2}.\frac{1}{2}+\frac{1}{\sqrt2}.\frac{\sqrt3}{2}$

$=\frac{1}{\sqrt2}.\frac{\sqrt3}{2\sqrt2}=\frac{\sqrt3+1}{2\sqrt2}$

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Question 713 Marks
In a $\triangle\text{ABC},$ right angled at $A,$ if $\tan\text{C}=\sqrt{3},$ find the value of $\sin B \cos C + \cos B \sin C.$
Answer
In a $\triangle^\text{le }\text{ABC}$ right angaled at A $\tan\text{C}=\sqrt{3}$
Find $\sin\text{B}\cos\text{C}+\cos\text{B}\sin\text{C}.$

$\tan\text{c}=\sqrt{3}$
$\tan\text{C}=\frac{\text{opposite side}}{\text{adjacent side}}$
Let x be the hypotenuse By appying Pythagoras we get
$\text{BC}^2=\text{BA}^2+\text{AC}^2$
$\text{x}^2=(\sqrt{3})^2+1^2$
$\text{x}^2=\triangle\Rightarrow\text{x}=2$
at $\angle\text{B},\sin\text{B}=\frac{\text{AC}}{\text{BC}}=\frac{1}{2}$
$\cos\text{B}=\frac{\sqrt{3}}{2}$
at $\angle\text{C},\sin=\frac{\sqrt{3}}{2}$
$\cos\text{c}=\frac{1}{2}$
On substitution we get
$\Rightarrow\frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}$
$\Rightarrow\frac{1}{4}+\frac{\sqrt{3}}{4}\times(\sqrt{3})$ $=\frac{\sqrt{3}\times\sqrt{3}+1}{4}=\frac{3+1}{4}=\frac{4}{4}=1$
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