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
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\text{x}^2=(24)^2+(7)^2$ $\text{x}^2=576+49=62.5$ $\text{x}=25$ $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}$ $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}$ $\sin\theta+\cos\theta=\frac{24}{25}+\frac{7}{25}$ $=\frac{31}{25}$ View full question & answer→Question 523 Marks
If A = 30° and B = 60°, verify that.
$\cos(\text{A}+\text{B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$
AnswerWe 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 $\text{B}=15^\circ$
View full question & answer→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$
View full question & answer→Question 543 Marks
Evaluate the following:
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
AnswerWe 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}$
View full question & answer→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}$
View full question & answer→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}$ View full question & answer→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$
View full question & answer→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}$
AnswerWe 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}$
View full question & answer→Question 593 Marks
Given 15 cot A = 8, find sin A and sec A.
AnswerWe 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}$ View full question & answer→Question 603 Marks
If $3\cot\theta=4,$ find the value of $\frac{4\cos\theta-\sin\theta}{2\cos+\sin\theta}.$
AnswerWe 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}$
View full question & answer→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}$
View full question & answer→Question 623 Marks
If $\sec\theta=\frac{13}{5},$ show that $\frac{\sin\theta-3\cos\theta}{4\sin\theta-9\cos\theta}=3.$
AnswerWe 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$ View full question & answer→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}$ View full question & answer→Question 643 Marks
If $\tan\theta=\frac{4}{5},$ find the value of $\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}.$
AnswerWe have, $\tan\theta=\frac{4}{5}$ In $\triangle\text{ABC},$
$\text{AC}^2=\text{AB}^2+\text{BC}^2$ $\Rightarrow\text{AC}^2=(4)^2+(5)^2$ $\Rightarrow\text{AC}^2=16+25$ $\Rightarrow\text{AC}^2=41$ $\Rightarrow\text{AC}=\sqrt{41}$ $\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{5}{\sqrt{14}}$ and $\sin\theta=\frac{\text{AB}}{\text{AC}}=\frac{4}{\sqrt{41}}$ Now, $\frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta}=\frac{\frac{5}{\sqrt{41}}-\frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}}+\frac{4}{\sqrt{41}}}$ $=\frac{\frac{5-4}{\sqrt{41}}}{\frac{5+4}{\sqrt{41}}}$ $=\frac{1}{9}$ View full question & answer→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}$
View full question & answer→Question 663 Marks
In $\triangle\text{PQR},$ right-angled at Q, PQ = 3cm and PR = 6cm. 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$ View full question & answer→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}$ View full question & answer→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.
AnswerWe 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}$ View full question & answer→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}}$
View full question & answer→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}$
View full question & answer→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.
AnswerIn 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 $\Rightarrow\frac{1}{2}\times\frac{1}{2}+\frac{\sqrt{3}}{2}\times\frac{\sqrt{3}}{2}$ $\Rightarrow\frac{1}{4}+\frac{\sqrt{3}}{4}\times(\sqrt{3})$ $=\frac{\sqrt{3}\times\sqrt{3}+1}{4}=\frac{3+1}{4}=\frac{4}{4}=1$ View full question & answer→