Trigonometric Ratio
Category : 10th Class
Trigonometric Ratio
Trigonometric Ratio
(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) \[\cot \,\,\theta \,\,=\,\,\frac{Base}{Perpendicular}\,\,\,=\,\,\frac{x}{y}\]
(ii) \[sec\,\,\theta \,\,=\,\,\frac{1}{\cos \,\theta }\,\,\]
(iii) \[\cot \,\,\theta \,\,=\,\,\frac{1}{\tan \,\theta }\,\,\]
(iv) \[\tan \,\,\theta \,\,=\,\,\frac{\sin \,\theta }{\cos \,\theta }\,\,\]
(v) \[\cot \,\,\theta \,\,=\,\,\frac{\cos \,\theta }{\sin \,\theta }\,\,\]
(ii) \[1\,\,+\,\,{{\tan }^{2}}\,\,\theta \,\,=\,\,{{\sec }^{2}}\,\,\theta \,\,for\,\,0{}^\circ \,\,\le \,\,\theta \,\,<\,\,90{}^\circ \]
(iii) \[1+{{\cot }^{2}}\theta =\cos e{{c}^{2}}\theta \,\,for\,\,0{}^\circ <\theta \le 90{}^\circ \]
(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) \[co\sec \,(90{}^\circ \,\,-\,\,\theta )\,\,=\,\,sec\,\theta \]
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
(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: \[{{(sec\,\theta -tan\,\theta )}^{2}}\]= \[{{\left( \frac{1-\sin \theta }{\cos \theta } \right)}^{2}}=\frac{{{(1-\sin \theta )}^{2}}}{{{\cos }^{2}}\theta }\,\,=\,\,\frac{{{(1-\sin \theta )}^{2}}}{1-{{\sin }^{2}}\theta }\,\,=\,\,\frac{1-\sin \theta }{1+\sin \theta }\]
(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, \[\frac{4\sin \theta \,\,-\,\,2\cos \theta }{4\sin \theta \,\,+\,\,3\cos \theta }\] = \[\frac{4\frac{\sin \theta }{\cos \theta }-2}{4\frac{\sin \theta }{\cos \theta }+3}\]
= \[\frac{4\tan \theta -2}{4\tan \theta +3}\]
= \[\frac{3-2}{3+3}=\frac{1}{6}\].
(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, \[\frac{a\,\sin \theta -b\,\cos \,\theta }{a\,\sin \theta +b\,\cos \theta }\] = \[\frac{a.\frac{\sin \theta }{\cos \theta }-b}{a.\frac{\sin \theta }{\cos \theta }+b}\]
= \[\frac{a\,\tan \theta -b}{a\,\tan \theta +b}\] = \[\frac{a.\frac{a}{b}-b}{a.\frac{a}{b}+b}\] \[\left[ \therefore \tan \theta =\frac{a}{b}(given) \right]\]
\[=\frac{{{a}^{2}}-{{b}^{2}}}{{{a}^{2}}+{{b}^{2}}}\]
\[\mathbf{Si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{30{}^\circ co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{45{}^\circ +4ta}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{30{}^\circ +}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{si}{{\mathbf{n}}^{\mathbf{2}}}\mathbf{90{}^\circ -2co}{{\mathbf{s}}^{\mathbf{2}}}\mathbf{90{}^\circ +}\frac{\mathbf{1}}{\mathbf{24}}\]
(a) 3 (b) 6
(c) 4 (d) 2
(e) None of these
Ans. (d)
Explanation: \[si{{n}^{2}}30{}^\circ \text{ }cos+\text{ }45{}^\circ \text{ }+\text{ }4\text{ }ta{{n}^{2}}30{}^\circ \text{ }+\frac{1}{2}\,\,si{{n}^{2}}90{}^\circ \text{ }-\text{ }2\text{ }co{{s}^{2}}90{}^\circ +\frac{1}{24}\]
\[={{\left( \frac{1}{2} \right)}^{2}}\times {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+4\times {{\left( \frac{1}{\sqrt{3}} \right)}^{2}}+\left( \frac{1}{2} \right)\times {{1}^{2}}-2\times {{(0)}^{2}}+\frac{1}{24}\]
\[=\,\,\left( \frac{1}{4}\times \frac{1}{2} \right)+\left( 4\times \frac{1}{3} \right)+\left( \frac{1}{2}+\frac{1}{24} \right)=\frac{1}{8}+\frac{4}{3}+\frac{1}{2}+\frac{1}{24}\]
\[=\,\,\frac{3+32+12+1}{24}\,\,=\,\,\frac{48}{24}\,\,=\,\,2\]
(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) \[\tan \,\phi \,\,\,\,=\,\,\,\frac{EC}{DE}=\frac{12}{9}=\frac{4}{3}.\]
(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: \[{{\left( cosec\text{ }\theta \text{ }-\text{ }cot\text{ }\theta \right)}^{2}}={{\left( \frac{1}{\sin \,\theta }-\frac{\cos \,\theta }{\sin \,\theta } \right)}^{2}}={{\left( \frac{1-\cos \theta }{\sin \,\theta } \right)}^{2}}\]
\[=\,\,\,\,\frac{{{(1-\cos \,\theta )}^{2}}}{(1-co{{s}^{2}}\theta )}\]
\[=\,\,\,\,\frac{{{(1-\cos \,\theta )}^{2}}(1-cos\,\theta )}{(1+cos\theta )(1-cos\theta )}=\frac{1-\cos \,\theta }{1+\cos \,\theta }\]
(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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