4 Marks Questions - Maths STD 10 Questions - Vidyadip

Questions

4 Marks Questions

Take a timed test

18 questions · self-marked practice — reveal the answer and mark yourself.

Question 14 Marks

If $\tan\theta=\frac{20}{21},$ show that $\frac{\big(1-\sin\theta+\cos\theta\big)}{\big(1+\sin\theta+\cos\theta\big)}=\frac37.$

Answer

Given: $\tan\theta=\frac{20}{21}=\frac{20\text{k}}{21\text{k}}$
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$

By pythagoras theorem, we have
$AC^2 = (AB^2 + BC^2)$
$= (21k)^2 + (20k)^2$
$= 441k^2 + 400k^2$
$= 841k^2$
$\therefore\text{AC}=29\text{k}$
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{20\text{k}}{29\text{k}}=\frac{20}{29},\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{21\text{k}}{29\text{k}}=\frac{21}{29}$
$\text{L.H.S.}=\frac{1-\sin\theta+\cos\theta}{1+\sin\theta+\cos\theta}=\frac{1-\frac{20}{29}+\frac{21}{29}}{1+{\frac{20}{29}+\frac{21}{29}}}=\frac{\frac{29-20+21}{29}}{\frac{29+20+21}{29}}$
$=\frac{30}{70}=\frac{3}{7}=\text{R.H.S}$

View full question & answer→

Question 24 Marks

In a $\triangle\text{ABC},\angle\text{B}=90^\circ,\text{AB}=24\text{cm}$ and BC = 7cm.
Find:

$\sin\text{A}$
$\cos\text{A}$
$\sin\text{C}$
$\cos\text{C}.$

Answer

By pythagoras theoram, we have
$\text{AC}^2 = \text{AB}^2 - \text{BC}^2$
$\Rightarrow\text{AC}^2=(24)^2+(7)^2$
$\Rightarrow\text{AC}^2=576+49=625$
$\Rightarrow\text{AC}=25\text{cm}$
Now, For T-Ratios of $\angle\text{A},$ base = AB and perpendicular = BC

$\sin\text{A}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}$
$\cos\text{A}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}$

Similarly, For T-Ratios of $\angle\text{C},$ base = BC and perpendicular = AB

$\sin\text{C}=\frac{\text{AB}}{\text{AC}}=\frac{24}{25}$
$\cos\text{C}=\frac{\text{BC}}{\text{AC}}=\frac{7}{25}.$

View full question & answer→

Question 34 Marks

If $\sin\alpha=\frac12,$ Prove that $\big(3\cos\alpha-4\cos^3\alpha\big)=0.$

Answer

$\sin\alpha=\frac12\Rightarrow\sin^2\alpha=\frac14$
$\therefore\cos^2\alpha=1-\sin^2\alpha=1-\frac14=\frac34$
$\Rightarrow\cos\alpha=\frac{\sqrt{3}}{2}$
$\therefore\ \text{L.H.S.}=3\cos\alpha-4\cos^2\alpha$
$=3\times\frac{\sqrt{3}}{2}-4\Big(\frac{\sqrt{3}}{2}\Big)^2$
$=\frac{3\sqrt{3}}{2}-4\times\frac{3\sqrt{3}}{8}$
$=\frac{3\sqrt{3}}{2}-\frac{3\sqrt{3}}{2}$
$=0$
$=\text{R.H.S}$

View full question & answer→

Question 44 Marks

If $\sin\theta=\frac{3}{4},$ show that $\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}=\frac{\sqrt{7}}{3}.$

Answer

$\sin\theta=\frac34\Rightarrow\text{cosec}\theta=\frac43$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac34$
Let $BC = 3$ and $AC = 4$
Then, by pythagoras theorem,
$AB^2 = AC^2 - BC^2$
$\Rightarrow 4^2 - 3^2 = 16 - 9 = 7$
$\Rightarrow\text{AB}=\sqrt{7}$
Now,
$\cot\theta=\frac{\text{Base}}{\text{Perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac{\sqrt{7}}{3}$
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{4}{\sqrt{7}}$
$\therefore\text{L.H.S.}=\sqrt{\frac{\text{cosec}^2\theta-\cot^2\theta}{\sec^2\theta-1}}$
$=\sqrt{\frac{\big(\frac{4}{3}\big)^2-\Big(\frac{\sqrt{7}}{3}\Big)}{\Big(\frac{4}{\sqrt{7}}\Big)^2-1}}$
$=\sqrt{\frac{\frac{\frac{16}{9}-\frac{7}{9}}{12}}{\frac{16}{7}-1}}$
$=\sqrt{\frac{1}{\frac97}}$
$=\sqrt{\frac{7}{9}}$
$=\frac{\sqrt{7}}{3}$
$=\text{R.H.S.}$

View full question & answer→

Question 54 Marks

If $\tan\theta=\frac{1}{\sqrt{7}},$ show that $\frac{\big(\text{coses}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\frac34.$

Answer

Given: $\tan\theta=\frac{\text{BC}}{\text{AB}}=\frac{1}{\sqrt{7}}$
Let BC = 1k and $\text{AB}=\sqrt{7}\text{k}$
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$

By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB}^2 + \text{BC}^2)$
$\Rightarrow\text{AC}^2=\Big[\big(\sqrt{7}\text{k}\big)^2+\big(1\text{k}\big)^2\Big]$
$= 7\text{k}^2 + 1\text{k}^2 = 8\text{k}^2$
$\Rightarrow\text{AC}=2\sqrt{2}\text{k}$
$\text{cosec}\theta=\frac{\text{AC}}{\text{BC}}=\frac{2\sqrt{2}\text{k}}{1\text{k}}={2\sqrt{2}}$
$\sec\theta=\frac{\text{AC}}{\text{AB}}=\frac{2\sqrt{2}\text{k}}{\sqrt{7}\text{k}}=\frac{2\sqrt{2}}{\sqrt{7}}$
$\Rightarrow\frac{\big(\text{cosec}^2\theta-\sec^2\theta\big)}{\big(\text{cosec}^2\theta+\sec^2\theta\big)}=\begin{bmatrix}\frac{\big(2\sqrt{2}\big)^2-\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}{\big(2\sqrt{2}\big)^2+\Big(\frac{2\sqrt{2}}{\sqrt{7}}\Big)^2}\end{bmatrix}$
$=\frac{\big(8-\frac{8}{7}\big)}{\big(8+\frac87\big)}=\frac{\big(\frac{48}{7}\big)}{\big(\frac{64}{7}\big)}=\frac{48}{64}=\frac34$
Hence, $\Big(\frac{\text{cosec}^2\theta-\sec^2\theta}{\text{cosec}^2\theta+\sec^2\theta}\Big)=\frac34$

View full question & answer→

Question 64 Marks

In a $\triangle\text{ABC},\angle\text{C}=90^\circ,\angle\text{ABC}=\theta^\circ,\text{BC}=21\text{units}$ and AB = 29units.
Show that $\big(\cos^2\theta-\sin^2\theta\big)=\frac{41}{841}.$

Answer

In $\triangle\text{ABC},\angle\text{C}=90^\circ$
AB = 29 units and BC = 21 units
By Pythagoras theorem, we have
$\text{AC}^2=\text{AB}^2-\text{BC}^2$
$=29^2-21^2=841-441=400$
$\Rightarrow\text{AC}=20\text{ units}$
$\therefore\ \sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{AC}}{\text{AB}}=\frac{20}{29}$
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AB}}=\frac{21}{29}$
$\therefore\text{L.H.S.}=\cos^2\theta-\sin^2\theta$
$=\Big(\frac{21}{29}\Big)^2-\Big(\frac{20}{29}\Big)^2$
$=\frac{441}{841}-\frac{400}{841}$
$=\frac{41}{841}$
$=\text{R.H.S.}$

View full question & answer→

Question 74 Marks

If $\cot\theta=2,$ find the value of the all T-ratios of $\theta.$

Answer

Given: $\cot\theta=\frac{\text{AB}}{\text{BC}}=\frac{2\text{k}}{1\text{k}}$
Let AB = 2k
and BC = 1k, where k is positive.
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$

By pythagoras theorem, we have
$\text{AC}^2 = (\text{AB})^2 + (\text{BC})^2 $
$=\Big[(2\text{k})^2+(1\text{k})^2\Big]$
$=\Big(4\text{k}^2+1\text{k}^2\Big)5\text{k}^2$
$\therefore\text{AC}=\sqrt{5}\text{k}^2=\sqrt{5}\text{k}$
$\therefore\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{1\text{k}}{\sqrt{5}\text{k}}=\frac{1}{\sqrt{5}}$
$\cos\theta=\frac{\text{BC}}{\text{AC}}=\frac{2\text{k}}{\sqrt{5}\text{k}}=\frac{1}{\sqrt{5}}$
$\tan\theta=\frac{1}{\cot\theta}=\frac12;\cot\theta=2$ (Given)
$\text{cosec}\theta=\frac{1}{\sin\theta}=\sqrt{5}$
$\sec\theta=\frac{1}{\sin\theta}=\frac{\sqrt{5}}{2}$

View full question & answer→

Question 84 Marks

If $3\cot\theta=4,$ show that $\frac{(1-\tan^2\theta)}{(1+\tan^2\theta)}=\big(\text{cos}^2\theta-\sin^2\theta\big).$

Answer

$3\cot\theta=4\Rightarrow\cot\theta=\frac43\Rightarrow\tan\theta=\frac34$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\cot\theta=\frac{\text{Base}}{\text{Perpendicular}}=\frac{\text{AB}}{\text{BC}}=\frac43$
Let AB = 4 and BC = 3
Then, by pythagoras theoram,
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$=4^3+3^2=16+9=25$
$\Rightarrow\text{AC} = 5$
Now,
$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac35$
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac45$
$\therefore\text{L.H.S.}=\frac{\big(1-\tan^2\theta\big)}{\big(1+\tan^2\theta\big)}$
$=\frac{1-\big(\frac{3}{4}\big)^2}{1+\big(\frac{3}{4}\big)^2}$
$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}=\frac{\frac{8}{16}}{\frac{25}{16}}=\frac{8}{25}$
$\text{R.H.S.}=\cos^2\theta-\sin^2\theta=\Big(\frac45\Big)^2=\Big(\frac{3}{5}\Big)^2$
$=\frac{16}{25}-\frac{9}{25}=\frac{8}{25}$
Hence, $\frac{(1-\tan^2\theta)}{(1+\tan^2\theta)}=\big(\text{cos}^2\theta-\sin^2\theta\big).$

View full question & answer→

Question 94 Marks

In the adjoining figure, $\angle\text{B}=90^\circ,\angle\text{BAC}=\theta,\text{BC}=\text{CD}=4\text{cm}$ and AD = 10cm. Find

$\sin\theta$
$\cos\theta.$

Hint: $\text{AB}^2=\big(\text{AD}^2-\text{BD}^2\big)=36\text{cm}^2$ $\therefore\text{AB}=6\text{cm}.$ $\text{AC}^2=\big(\text{AB}^2+\text{BC}^2\big)=52\text{cm}^2$ $\therefore\text{AC}=2\sqrt{13}\text{cm}.$ Thus, AB = 6cm and $\text{AC}=2\sqrt{13}\text{cm}.$

Answer

In $\triangle\text{ABC},$ $\angle\text{B}=90^\circ$
$\text{AD}=10\text{cm}$
$\text{BD}=\text{BC}+\text{CD}$
$=4+4=8\text{cm}$
By pythagoras theoram, we have
$\text{AB}^2 = \text{AD}^2 - \text{BD}^2$
$=10^2-8^2=100-64=36$
$\Rightarrow\text{AB}=6\text{cm}$
Now, in $\triangle\text{ABC},\angle\text{B}=90^\circ.$
By Pythagoras theorem, we have
$\text{AC}^2 = \text{AB}^2 - \text{BC}^2$
$=6^2+4^2=36+16=52$
$\Rightarrow\text{AC}=2\sqrt{13}\text{cm}$

$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}$

$=\frac{\text{BC}}{\text{AC}}=\frac{4}{2\sqrt{13}}=\frac{2}{\sqrt{13}}=\frac{2\sqrt{13}}{13}$

$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}$

$=\frac{\text{AB}}{\text{AC}}=\frac{6}{2\sqrt{13}}=\frac{3}{\sqrt{13}}=\frac{3\sqrt{13}}{13}.$

View full question & answer→

Question 104 Marks

If $\tan\theta=\frac{4}{3},$ show that $(\sin\theta+\cos\theta)=\frac75.$

Answer

Given: $\tan\theta=\frac{\text{BC}}{\text{AB}}=\frac{4}{3}$
Let BC = 4k and AB = 3k
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$

By Pythagoras theorem, we get
$\big(\text{AC}\big)^2=\big(\text{AB}\big)^2+\big(\text{BC}\big)^2$
$\Rightarrow\big(\text{AC}\big)^2=\Big[\big(3\text{k}\big)^2+\big(4\text{k}\big)^2\Big]$
$\Rightarrow\big(\text{AC}\big)^2=\big(9\text{k}^2+16\text{k}^2\big)=25\text{k}^2$
$\therefore\text{AC}=\sqrt{9\text{k}^2}=5\text{k}$
$\sin\theta=\frac{4\text{k}}{5\text{k}}=\frac45$
$\cos\theta=\frac{3\text{k}}{5\text{k}}=\frac{3}{5}$
$\Rightarrow(\sin\theta+\cos\theta)=\Big(\frac{4}{5}+\frac35\Big)=\frac75$
Hence, $(\sin\theta+\cos\theta)=\frac75.$

View full question & answer→

Question 114 Marks

If $\text{x}=\cot\text{A}+\cos\text{A}$ and $\text{y}=\cot\text{A}-\cos\text{A},$ Prove that $\Big(\frac{\text{x}-\text{y}}{\text{x}+\text{y}}\Big)^2+\Big(\frac{\text{x}-\text{y}}{2}\Big)^2=1.$

Answer

$\text{x}=\cot\text{A}+\cos\text{A}$ and $\text{y}=\cot\text{A}-\cos\text{A}$
Thus, we have
$\text{x}+\text{y}=(\cot\text{A}+\cos\text{A})\\+(\cot\text{A}-\cos\text{A})=2\cos\text{A}$
$\text{x}-\text{y}=(\cot\text{A}+\cos\text{A})\\-(\cot\text{A}-\cos\text{A})=2\cos\text{A}$
$\text{L.H.S.}=\Big(\frac{\text{x}-\text{y}}{\text{x}+\text{y}}\Big)^2+\Big(\frac{\text{x}-\text{y}}{2}\Big)^2-1$
$=\Big(\frac{2\cos\text{A}}{2\cot\text{A}}\Big)^2+\Big(\frac{2\cos\text{A}}{2}\Big)^2$
$=\Big(\frac{\cos\text{A}}{\cot\text{A}}\Big)^2+(\cos\text{A})^2$
$=\Bigg(\frac{\cos\text{A}}{\frac{\cos\text{A}}{\sin\text{A}}}\Bigg)^2+(\cos\text{A})^2$
$=(\sin\text{A})^2+(\cos\text{A})^2$
$=\sin^2\text{A}+\cos^2\text{A}$
$=1$
$=\text{R.H.S}$

View full question & answer→

Question 124 Marks

If $3\cot\theta=2,$ show that $\frac{(4\sin\theta-3\cos\theta)}{(2\sin\theta+6\cos\theta)}=\frac13.$

Answer

Given:
$\cot\theta=2,\ \therefore\cot\theta=\frac23=\frac{2\text{k}}{3\text{k}}$
Let us draw a $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$

By pythagoras theoram, we have
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$=\big(2\text{k}\big)^2+\big(3\text{k}\big)^2$
$=4\text{k}^2+9\text{k}^2=13\text{k}^2$
$\Rightarrow\text{AC}=\sqrt{13}\text{k}$
$\therefore\sin\theta=\frac{3\text{k}}{\sqrt{13\text{k}}}=\frac{3}{\sqrt{13}}$
$\cos\theta=\frac{2\text{k}}{\sqrt{13}\text{k}}=\frac{2}{\sqrt{13}}$
$\therefore\text{L.H.S.}=\frac{(4\sin\theta-3\cos\theta)}{(2\sin\theta+6\cos\theta)}$
$=\frac{4\times\frac{3}{\sqrt{13}}-3\times\frac{2}{\sqrt{13}}}{2\times\frac{3}{\sqrt{13}}+6\times\frac{2}{\sqrt{13}}}$
$=\frac{\frac{12-6}{\sqrt{13}}}{\frac{6+12}{\sqrt{13}}}$
$=\frac{6}{18}=\frac13$
$=\text{R.H.S.}$

View full question & answer→

Question 134 Marks

If $\sin\theta=\frac{\text{a}}{\text{b}},$ show that $(\sec\theta+\tan\theta)=\sqrt{\frac{\text{b}+\text{a}}{\text{b}-\text{a}}}.$

Answer

$\sin\theta=\frac{\text{a}}{\text{b}}$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{\text{a}}{\text{b}}$
Let BC = a and AC = b
Then, by pythagoras theorem,
$\text{AB}^2 = \text{AC}^2 - \text{BC}^2 = \text{b}^2 - \text{a}^2$
$\Rightarrow\text{AB}=\sqrt{\text{b}^2-\text{a}^2}$
Now,
$\sec\theta=\frac{\text{Hypotenuse}}{\text{Base}}=\frac{\text{AC}}{\text{AB}}=\frac{\text{b}}{\sqrt{\text{b}^2-\text{a}^2}}$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
$\therefore\text{L.H.S.}=(\sec\theta+\tan\theta)$
$=\frac{\text{b}}{\sqrt{\text{b}^2-\text{a}^2}}+\frac{\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
$=\frac{\text{b}+\text{a}}{\sqrt{\text{b}^2-\text{a}^2}}$
$\frac{\text{b}+\text{a}}{\sqrt{(\text{b}+\text{a})(\text{b}-\text{a})}}$
$=\frac{{\sqrt{\text{b}+\text{a}}}\times\sqrt{\text{b}+\text{a}}}{\sqrt{(\text{b}+\text{a})}\times\sqrt{\text{b}-\text{a}}}$
$=\sqrt{\frac{\text{b}+\text{a}}{\text{b}-\text{a}}}$
$=\text{R.H.S.}$

View full question & answer→

Question 144 Marks

If $3\tan\theta=4,$ show that $\frac{(4\cos\theta-\sin\theta)}{(2\cos\theta+\sin\theta)}=\frac45.$

Answer

$3\tan\theta=4\Rightarrow\tan\theta=\frac43$
Consider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$ and $\angle\text{A}=\theta$
Then, $\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac43$
Let $BC = 4$ and $AB = 3$
Then, by pythagoras theoram,
$AC^2 = AB^2 + BC^2$
$= 3^2 + 4^2 = 9 + 16 = 25$
$\Rightarrow AC = 5$
Now,
$\sin\theta=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac45$
$\cos\theta=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac35$
$\therefore\text{L.H.S.}=\frac{(4\cos\theta-\sin\theta)}{(2\cos\theta+\sin\theta)}$
$=\frac{4\times\frac35-\frac45}{2\times\frac{3}{5}+\frac{4}{5}}$
$=\frac{\frac85}{\frac{10}{5}}$
$=\frac{8}{5}\times\frac12$
$=\frac45$
$=\text{R.H.S.}$

View full question & answer→

Question 154 Marks

If $\sin\text{A}=\frac{9}{41},$ find the value of $\cos\text{A}$ and $\tan\text{A}.$

Answer

Cosider $\triangle\text{ABC},$ where $\angle\text{B}=90^\circ$
Then, $\sin\text{A}=\frac{\text{Perpendicular}}{\text{Hypotenuse}}=\frac{\text{BC}}{\text{AC}}=\frac{9}{41}$
Let $BC = 9$ and $AC = 41$
Then, by Pythagoras theoram,
$AC^2 = (AB)^2 + (BC)^2$
$\Rightarrow (AB)^2 = (AC)^2 - (BC)^2$
$= 41^2 - 9^2 = 1681 - 81 = 1600$
$\Rightarrow AB = 40$
Now,
$\cos\text{A}=\frac{\text{Base}}{\text{Hypotenuse}}=\frac{\text{AB}}{\text{AC}}=\frac{40}{41}$
$\tan\text{A}=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{BC}}{\text{AB}}=\frac{9}{40}$

View full question & answer→

Question 164 Marks

If $\cos\theta=\frac{3}{5},$ show that $\frac{(\sin\theta-\cos\theta)}{2\tan\theta}=\frac{3}{160}.$

Answer

$\cos\theta=\frac{3}{5}\Rightarrow\cos^2\theta=\frac{9}{25}$
$\therefore\sin^2\theta=1-\cos^2\theta=1-\frac{9}{25}=\frac{25-9}{25}=\frac{16}{25}$
$\Rightarrow\sin\theta=\frac45$
$\Rightarrow\tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{\frac{4}{5}}{\frac{3}{5}}=\frac43$
$\Rightarrow\cot\theta=\frac{1}{\tan\theta}=\frac34$
Thus,
$\text{L.H.S.}=\frac{(\sin\theta-\cot\theta)}{2\tan\theta}$
$=\frac{\frac{4}{5}-\frac{3}{4}}{2\times\frac{4}{3}}$
$=\frac{\frac{16-15}{20}}{\frac{8}{3}}$
$=\frac{1}{20}\times\frac38$
$=\frac{3}{160}$
$=\text{R.H.S.}$

View full question & answer→

Question 174 Marks

If $\cos\theta=\frac{7}{25},$ find the value of the all T-ratios of $\theta.$

Answer

Given: $\cos\theta=\frac{7}{25}$
Let AB = 7k and AC = 25k
Where k is positive
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$

By pythagoras theorem, we have
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2$
$\Rightarrow\text{BC}^2 = \text{AC}^2 - \text{BC}^2$
$\text{AB}^2=\Big[(25\text{k})^2=\big(\sqrt{7}\text{k}\big)^2\Big]$
$=\big(625\text{k}^2-49\text{k}^2\big)$
$=576\text{k}^2$
$\Rightarrow\text{BC}=\sqrt{576\text{k}^2}=24\text{k}$
$\therefore\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{24\text{k}}{25\text{k}}=\frac{24}{25},\cos\theta=\frac{7}{25}$ (Given)
$\tan\theta=\frac{\sin\theta}{\cos\theta}=\Big(\frac{24}{25}\times\frac{25}{7}\Big)=\frac{24}{7}$
$\text{cosec}\theta=\frac{1}{\sin\theta}=\frac{25}{24},$
$\sec\theta=\frac{1}{\cos\theta}=\frac{25}{7}$ and $\cot \theta=\frac{1}{\tan\theta}=\frac{7}{24}$

View full question & answer→

Question 184 Marks

If $\theta=\frac{15}{8},$ find the value of the all T-ratios of $\theta.$

Answer

Given: $\tan\text{A}=\frac{\text{BC}}{\text{AB}}=\frac{15}{8}$
Let BC = 15k and AB = 8k, where k is positive.
Let us draw a $\triangle\text{ABC}$ in which $\angle\text{B}=90^\circ$ and $\angle\text{BAC}=\theta$
By pythagoras theorem, we have
$\text{AC}^2 = \text{AB}^2 + \text{BC}^2 $
$= (8\text{k})^2 + (15\text{k})^2 $
$= 64\text{k}^2 + 225\text{k}^2 = 289\text{k}^2$
$\Rightarrow\text{AC} = 17\text{k}$
Thus, we have
$\sin\theta=\frac{\text{BC}}{\text{AC}}=\frac{15\text{k}}{17\text{k}}=\frac{15}{17}$
$\cos\theta=\frac{\text{AB}}{\text{AC}}=\frac{8\text{k}}{17\text{k}}=\frac{8}{17}$
$\tan\theta=\frac{15}{8}$ (Given)
$\text{cosec}\theta=\frac{1}{\sin\theta}=\frac{17}{15}$
$\sec\theta=\frac{1}{\cos\theta}=\frac{17}{15}$
$\cot\theta=\frac{1}{\tan\theta}=\frac{8}{15}$

View full question & answer→

Generate Paper Free All Trigonometric Ratios topics