# 10th Class Mathematics Introduction to Trigonometry Trigonometric Ratio

Trigonometric Ratio

Category : 10th Class

Trigonometric Ratio

Trigonometric Ratio

1. (i) $\sin \theta \,\,=\,\,\frac{Perpendicular}{Hypotenuse}\,\,=\,\,\frac{y}{r}$

(ii)  $\cos \theta \,\,=\,\,\frac{Base}{Hypotenuse}\,\,=\,\,\frac{x}{r}$

(iii) $\tan \,\,\theta \,\,=\,\,\frac{Perpendicular}{Base}\,\,=\,\,\frac{y}{x}$

(iv) $\cos ec\,\,\theta \,\,=\,\,\frac{Hypotenuse}{Perpendicular}\,\,=\,\,\frac{r}{y}$

(v) $sec\,\,\theta \,\,=\,\,\frac{Hypotenuse}{Base}\,\,\,=\,\,\frac{r}{x}$

(vi) $\cot \,\,\theta \,\,=\,\,\frac{Base}{Perpendicular}\,\,\,=\,\,\frac{x}{y}$

1. (i) $\cos ec\,\,\theta \,\,=\,\,\frac{1}{\sin \,\theta }\,\,$

(ii)  $sec\,\,\theta \,\,=\,\,\frac{1}{\cos \,\theta }\,\,$

(iii) $\cot \,\,\theta \,\,=\,\,\frac{1}{\tan \,\theta }\,\,$

(iv) $\tan \,\,\theta \,\,=\,\,\frac{\sin \,\theta }{\cos \,\theta }\,\,$

(v) $\cot \,\,\theta \,\,=\,\,\frac{\cos \,\theta }{\sin \,\theta }\,\,$

1. (i) ${{\sin }^{2}}+\theta +{{\cos }^{2}}\theta \,\,=\,\,1\,$

(ii) $1\,\,+\,\,{{\tan }^{2}}\,\,\theta \,\,=\,\,{{\sec }^{2}}\,\,\theta \,\,for\,\,0{}^\circ \,\,\le \,\,\theta \,\,<\,\,90{}^\circ$

(iii) $1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta \,\,for\,\,0{}^\circ <\theta \le 90{}^\circ$

1. (i) $sin\,\,\left( 90{}^\circ -\theta \right)=\,\,\cos \,\,\theta$

(ii) $\cos (90{}^\circ -\theta )\,\,=\,\,sin\theta$

(iii) $\tan (90{}^\circ \,\,-\,\,\theta )\,\,=\,\,cot\theta$

(iv) $\cot (90{}^\circ \,\,-\,\,\theta )\,\,=\,\,\tan \,\theta$

(v) $\sec (90{}^\circ \,\,-\,\,\theta )\,\,=\,\,\cos ec\,\theta$

(vi) $co\sec \,(90{}^\circ \,\,-\,\,\theta )\,\,=\,\,sec\,\theta$

1. The value of sin or cos never exceeds 1, whereas the value of sec or cosec is always greater or equal to 1.

6.         Table for T- ratios of $0{}^\circ ,\text{ }30{}^\circ ,\text{ }45{}^\circ ,\text{ }60{}^\circ ,\text{ }90{}^\circ$.

 $\theta$ sin $\theta$ cos $\theta$ tan $\theta$ cosec $\theta$ sec $\theta$ cot $\theta$ $0{}^\circ$ 0 1 0 not defined 1 not defined $30{}^\circ$ $\frac{1}{2}$ $\frac{\sqrt{3}}{2}$ $\frac{1}{\sqrt{3}}$ 2 $\frac{2}{\sqrt{3}}$ $\sqrt{3}$ $45{}^\circ$ $\frac{1}{\sqrt{2}}$ $\frac{1}{\sqrt{2}}$ 1 $\sqrt{2}$ $\sqrt{2}$ 1 $60{}^\circ$ $\frac{\sqrt{3}}{2}$ $\frac{1}{2}$ $\sqrt{3}$ $\frac{2}{\sqrt{3}}$ 2 $\frac{1}{\sqrt{3}}$ $90{}^\circ$ 1 0 not defined 1 not defined 0

Snap Test

1. ${{(\mathbf{sec}~\theta -\text{ }\mathbf{tan}\theta )}^{\mathbf{2}}}=$ ?

(a) $\frac{1+\sin \theta }{1-\sin \theta }$

(b) $\frac{2+\sin \theta }{2-\sin \theta }$

(c) $\frac{2-\sin \theta }{2+\sin \theta }$

(d) $\frac{1-\sin \theta }{1+\sin \theta }$

(e) None of these

Ans.     (d)

Explanation: ${{(sec\,\theta -tan\,\theta )}^{2}}$= ${{\left( \frac{1-\sin \theta }{\cos \theta } \right)}^{2}}=\frac{{{(1-\sin \theta )}^{2}}}{{{\cos }^{2}}\theta }\,\,=\,\,\frac{{{(1-\sin \theta )}^{2}}}{1-{{\sin }^{2}}\theta }\,\,=\,\,\frac{1-\sin \theta }{1+\sin \theta }$

1. If $\mathbf{4}\,\mathbf{tan}\,\theta \,\,~=\text{ }\mathbf{3}$, find the value of $\frac{\mathbf{4}\,\,\mathbf{sin}\theta \,\,\mathbf{-}\,\,\mathbf{2}\,\,\mathbf{cos}\theta }{\mathbf{4}\,\,\mathbf{sin}\theta \,\,\mathbf{+}\,\,\mathbf{3}\,\,\mathbf{cos}\theta }$.

(a) $\frac{1}{4}$           (b) $\frac{1}{6}$

(c) $\frac{1}{8}$           (d) $\frac{3}{8}$

(e) None of these

Ans.     (b)

Explanation: We have $4\text{ }tan\,\theta \,\,~=\text{ }3$

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

=         $\frac{4\tan \theta -2}{4\tan \theta +3}$

=         $\frac{3-2}{3+3}=\frac{1}{6}$.

1. If $\mathbf{tan}\theta \mathbf{=}\frac{\mathbf{a}}{\mathbf{b}}$, then $\frac{\mathbf{a}\,\mathbf{sin}\theta \mathbf{-b}\,\mathbf{cos}\theta }{\mathbf{a}\,\mathbf{sin}\theta \mathbf{+b}\,\mathbf{cos}\theta }\mathbf{=?}$

(a) $\frac{{{a}^{2}}\,\,+\,\,{{b}^{2}}}{a\,\,+\,\,b}$

(b) $\frac{{{a}^{2}}\,\,-\,\,{{b}^{2}}}{a{{\,}^{2}}\,-\,\,b}$

(c) $\frac{{{a}^{2}}\,\,-\,\,{{b}^{2}}}{{{a}^{2}}\,\,+\,\,{{b}^{2}}}$

(d) $\frac{{{a}^{2}}\,\,-\,\,{{b}^{2}}}{a\,\,-\,\,b}$

(e) None of these

Ans.     (c)

Explanation: We have $\tan \theta =\frac{a}{b}$

Now, $\frac{a\,\sin \theta -b\,\cos \,\theta }{a\,\sin \theta +b\,\cos \theta }$ =         $\frac{a.\frac{\sin \theta }{\cos \theta }-b}{a.\frac{\sin \theta }{\cos \theta }+b}$

= $\frac{a\,\tan \theta -b}{a\,\tan \theta +b}$                 =         $\frac{a.\frac{a}{b}-b}{a.\frac{a}{b}+b}$   $\left[ \therefore \tan \theta =\frac{a}{b}(given) \right]$

$=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}$

1. Evaluate:

$\mathbf{Si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{30{}^\circ co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{45{}^\circ +4ta}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{30{}^\circ +}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{90{}^\circ -2co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{90{}^\circ +}\frac{\mathbf{1}}{\mathbf{24}}$

(a) 3                              (b) 6

(c) 4                              (d) 2

(e) None of these

Ans.     (d)

Explanation: $si{{n}^{2}}30{}^\circ \text{ }cos+\text{ }45{}^\circ \text{ }+\text{ }4\text{ }ta{{n}^{2}}30{}^\circ \text{ }+\frac{1}{2}\,\,si{{n}^{2}}90{}^\circ \text{ }-\text{ }2\text{ }co{{s}^{2}}90{}^\circ +\frac{1}{24}$

$={{\left( \frac{1}{2} \right)}^{2}}\times {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+4\times {{\left( \frac{1}{\sqrt{3}} \right)}^{2}}+\left( \frac{1}{2} \right)\times {{1}^{2}}-2\times {{(0)}^{2}}+\frac{1}{24}$

$=\,\,\left( \frac{1}{4}\times \frac{1}{2} \right)+\left( 4\times \frac{1}{3} \right)+\left( \frac{1}{2}+\frac{1}{24} \right)=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}$

$=\,\,\frac{3+32+12+1}{24}\,\,=\,\,\frac{48}{24}\,\,=\,\,2$

1. In a $\Delta \,\mathbf{ABC},\text{ }\mathbf{if}\,\,\angle \mathbf{A}\text{ }=\text{ }\mathbf{30}{}^\circ ,\,\,\,\angle \mathbf{B}\text{ }=\text{ }\mathbf{90}{}^\circ \text{ }\mathbf{and}\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{10}\text{ }\mathbf{cm}$, then find AB and BC.

(a) $5~\sqrt{3}\,cm,\text{ }5\text{ }cm$

(b) $5\sqrt{2}~cm,\text{ }5\text{ }cm$

(c) $5~\sqrt{2}\,cm,\text{ }3\text{ }cm$

(d) $5\sqrt{2}~cm,\text{ }8\text{ }cm$

(e) None of these

Ans.     (a)

Explanation: It is given that in $\Delta \,ABC$ $\angle \,A\text{ }=\text{ }30{}^\circ$, and $AC\text{ }=\text{ }10\text{ }cm$.

We have:

(i)    $\frac{AB}{AC}\text{ }=\text{ }cos\text{ }A\text{ }=\text{ }cos\text{ }30{}^\circ$

$\Rightarrow \,\,\frac{AB}{10}=\frac{\sqrt{3}}{2}\Rightarrow AB=\left( 10\times \frac{\sqrt{3}}{2} \right)cm\,\,5\sqrt{3}\,cm.$

(ii)  $\frac{BC}{AC}=\sin \,A=\sin 30{}^\circ \,\,\,\,\,\,\Rightarrow \,\,\,\,\,\frac{BC}{10}=\frac{1}{2}\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,BC=\frac{10}{2}\,\,cm\,\,=\,\,5\,cm$

6.         In the given figure, $\angle \,\mathbf{ABC}=\mathbf{90}{}^\circ ,\,\,\angle \,\mathbf{BAC}=\theta ,\,\,\angle \,\mathbf{ADC}\text{ }=\phi .\text{ }\mathbf{BC}\text{ }=\text{ }\mathbf{5}\text{ }\mathbf{cm},\text{ }\mathbf{AC}\text{ }=\text{ }\mathbf{13}\text{ }\mathbf{cm}$ and $\mathbf{AD}\text{ }=\text{ }\mathbf{14}$          cm. Also, $\angle \,\mathbf{BAD}\text{ }=\text{ }\mathbf{90}{}^\circ$. Find the values of cosec $\phi$ and tan$\phi$..

(a) $\frac{12}{13},\,\frac{4}{3}$

(b) $\frac{5}{4},\,\frac{4}{3}$

(c) $\frac{4}{3},\,\frac{4}{5}$

(d) $\frac{4}{3},\,\frac{7}{5}$

(e) None of these

Ans.     (b)

Explanation: In right $\Delta$ABC we have: $A{{C}^{2}}=\text{ }A{{B}^{2}}+\text{ }B{{C}^{2}}$

$\Rightarrow \,\,\,\,\,\,AB\,\,\,\,\,\,\,\,\,\,=\,\,\,\,\,\sqrt{A{{C}^{2}}-B{{C}^{2}}}\,\,\,\,\,=\,\,\,\,\sqrt{{{(13)}^{2}}-{{(5)}^{2}}}$

$=\,\,\,\,\,\sqrt{169-25}\text{ }=\text{ }\sqrt{144}\text{ }=\text{ }12\text{ }cm.$

Now,  $EC~~~~~~=~~AB\text{ }=\text{ }12\text{ }cm.$

and $DE~~~~~~~~=~~AD\text{ }\text{ }EA\text{ }=\text{ }AD\text{ }-\text{ }BC$

$=~~\left( 14\text{ }-\text{ }5 \right)\text{ }cm\text{ }=\text{ }9\text{ }cm.$

In right $\Delta \,CDE$ we have:

$C{{D}^{2}}=\text{ }D{{E}^{2}}+\text{ }E{{C}^{2}}=\text{ }{{9}^{2}}+\text{ }{{12}^{2}}=\text{ }81\text{ }+\text{ }144\text{ }=\text{ }225$

$\Rightarrow \,\,CD=\sqrt{225}=15\,cm.$

(i) $cosec\,\phi \,\,\,\,=\,\,\,\frac{CD}{EC}=\frac{15}{12}=\frac{5}{4}.$

(ii) $\tan \,\phi \,\,\,\,=\,\,\,\frac{EC}{DE}=\frac{12}{9}=\frac{4}{3}.$

1. ${{\left( \mathbf{cosec}\text{ }\theta \text{ }-\text{ }\mathbf{cot}\text{ }\theta \right)}^{\mathbf{2}}}\mathbf{is}\text{ }\mathbf{equal}\text{ }\mathbf{to}:$

(a) $\frac{2-\,\cos \theta }{1+\cos \,\theta }$

(b) $\frac{1}{1\,\,-\,\,\cos \theta }$

(c) $\frac{1-\cos \,\theta }{1\,\,+\,\,\cos \,\theta }$

(d) $\frac{2+\cos \,\theta }{1\,\,+\,\,\cos \,\theta }$

(e) None of these

Ans.     (c)

Explanation: ${{\left( cosec\text{ }\theta \text{ }-\text{ }cot\text{ }\theta \right)}^{2}}={{\left( \frac{1}{\sin \,\theta }-\frac{\cos \,\theta }{\sin \,\theta } \right)}^{2}}={{\left( \frac{1-\cos \theta }{\sin \,\theta } \right)}^{2}}$

$=\,\,\,\,\frac{{{(1-\cos \,\theta )}^{2}}}{(1-co{{s}^{2}}\theta )}$

$=\,\,\,\,\frac{{{(1-\cos \,\theta )}^{2}}(1-cos\,\theta )}{(1+cos\theta )(1-cos\theta )}=\frac{1-\cos \,\theta }{1+\cos \,\theta }$

1. $\frac{\mathbf{sin}\,\,\theta }{\mathbf{1+cos}\,\,\theta }\mathbf{+}\frac{\mathbf{1+cos}\,\,\theta }{\mathbf{sin}\,\,\theta }\,\,\,\mathbf{=}$

(a) $2\text{ }cosec\,\theta$

(b) $2\text{ }cos\,\theta$

(c) $2\text{ }sec\,\theta ~$

(d) $2\text{ }sin\,\theta$

(e) None of these

Ans.     (a)

Explanation: $\frac{{{\sin }^{2}}\,\theta }{1+\cos \theta }+\frac{1+\cos \,\theta }{sin\,\theta }=\frac{{{\sin }^{2}}\theta +{{(1+cos\theta )}^{2}}}{\sin \theta (1+cos\,\theta )}$

= $\frac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta +1+2\,\cos \theta }{\sin \theta (1+cos\,\theta )}=\frac{2+2\cos \,\theta }{\sin \theta (1+cos\,\theta )}$

= $\frac{2(1+cos\,\theta )}{\sin \theta (1+cos\,\theta )}=\frac{2}{\sin \,\theta }=2\,\,\cos ec\,\theta$

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