10th Class Mathematics Introduction to Trigonometry Chart for the Sign of Different Trigonometrical

       Chart for the Sign of Different Trigonometrical  

Quadrant\[\to \] Ratios  \[\downarrow \]	I	II	III	IV
\[\sin \theta \]	+	+	-	-
\[\cos \theta \]	+	-	-	+
\[\tan \theta \]	+	-	+	-
\[\cot \theta \]	+	-	+	-
\[\sec \theta \]	+	-	-	+
\[co\sec \theta \]	+	+	-	-

            Value of Trigonometrical Ratios for Some Special Angles  

Angle \[\to \] Rations \[\downarrow \]	\[{{0}^{o}}\]	\[{{30}^{o}}\]	\[{{45}^{o}}\]	\[{{60}^{o}}\]	\[{{90}^{o}}\]
\[\sin \theta \]	0	\[\frac{1}{2}\]	\[\frac{1}{\sqrt{2}}\]	\[\frac{\sqrt{3}}{2}\]	1
\[\cos \theta \]	1	\[\frac{\sqrt{3}}{2}\]	\[\frac{1}{\sqrt{2}}\]	\[\frac{1}{2}\]	0
\[\tan \theta \]	0	\[\frac{1}{\sqrt{3}}\]	1	\[\sqrt{3}\]	Not Defined
\[\cot \theta \]	Not Defined	\[\sqrt{3}\]	1	\[\frac{1}{\sqrt{3}}\]	0
\[\sec \theta \]	1	\[\frac{2}{\sqrt{3}}\]	\[\sqrt{2}\]	2	Not Defined
\[co\sec \theta \]	Not Defined	2	\[\sqrt{2}\]	\[\frac{2}{\sqrt{3}}\]	1

 

From the above table

\[{{\sin }^{o}}=0,\sin {{30}^{o}}=\frac{1}{2},\sin {{45}^{o}}=\frac{1}{\sqrt{2}}\] and so on.  

 

 

• In ancient times probably trigonometry was invented for astronomy.
• Do you know that this field of mathematics originated from the civilization of Egypt, Mesopotamia and the Indus valley, more then 4000 years ago.
• Do you know that the sulba sutras written in India between 800 BC and 500 BC by which mathematician are able to, computer correctly the value of sin45.  

 

 

• \[\sin \theta =\frac{p}{h},\cos \theta =\frac{b}{h},\tan \theta =\frac{p}{b},\cot \theta =\frac{b}{p},\] \[\sec \theta =\frac{h}{b},\cos ec\theta =\frac{h}{p}\]
• \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\]
• \[{{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1\]
• \[co{{\sec }^{2}}\theta -{{\cot }^{2}}\theta =1\]
• \[-1\le \sin \theta \le 1\]and\[-1\le \cos \theta \le 1\]
• \[-\infty \le \tan \theta \le \infty \]and \[-\infty \le \cot \theta \le \infty \]
• \[\sec \theta \le -1\]or \[\sec \theta \ge 1\]and\[co\sec \theta \le -1\]or\[co\sec \theta \ge 1\]     

 

 

 

 

  The value of  \[\frac{{{\sin }^{2}}{{45}^{o}}+{{\cos }^{2}}{{45}^{o}}}{{{\sin }^{2}}{{30}^{o}}}\] is________.

(a) \[\frac{1}{2}\]                                            

(b) \[\frac{1}{4}\]           

(c) 4                                                      

(d) 1

(e) None of these  

 

Answer: (c)  

Explanation:

We know that

\[\sin {{45}^{o}}=\frac{1}{\sqrt{2}}=\cos {{45}^{o}}\] and \[\sin {{30}^{o}}=\frac{1}{2}\]

\[=\frac{{{\sin }^{2}}{{45}^{o}}+{{\cos }^{2}}{{45}^{o}}}{{{\sin }^{2}}{{30}^{o}}}=\]\[\frac{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}}{{{\left( \frac{1}{\sqrt{2}} \right)}^{2}}}\]

\[=\frac{\frac{1}{2}+\frac{1}{2}}{\frac{1}{4}}\]    \[=\frac{1}{\frac{1}{4}}\]              \[=4\]     

 

 

  The value of \[\text{2(si}{{\text{n}}^{\text{2}}}\text{45}{}^\circ +\text{co}{{\text{t}}^{\text{2}}}\text{3}0{}^\circ )-\text{6}(\text{co}{{\text{s}}^{\text{2}}}\text{45}{}^\circ \] \[-\text{co}{{\text{t}}^{\text{2}}}\text{6}0{}^\circ )\]is _____.

(a) 3                                      

(b) 5             

(c) 4                                      

(d) \[\text{ta}{{\text{n}}^{\text{2}}}\text{6}0{}^\circ +\text{co}{{\text{t}}^{\text{2}}}\text{3}0{}^\circ \]

(e) None of these  

 

Answer: (d)

Explanation:

Given trigonometrical expression is:

\[\text{2(si}{{\text{n}}^{\text{2}}}\text{45}{}^\circ +\text{co}{{\text{t}}^{\text{2}}}\text{3}0{}^\circ )-\text{6}(\text{co}{{\text{s}}^{\text{2}}}\text{45}{}^\circ -\text{co}{{\text{t}}^{\text{2}}}\text{6}0{}^\circ )\]

\[=2\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \sqrt{3} \right)}^{2}} \right]-6\left[ {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \frac{1}{\sqrt{3}} \right)}^{2}} \right]\]

\[=2\left[ \frac{1}{2}+3 \right]-6\left[ \frac{1}{2}+\frac{1}{3} \right]\]\[=\bcancel{2}\left[ \frac{7}{\bcancel{2}} \right]-\bcancel{6}\left[ \frac{1}{\bcancel{6}} \right]\]

\[=\text{7}-\text{1}=\text{6}=\text{3}+\text{3}=\text{ ta}{{\text{n}}^{\text{2}}}\text{6}0{}^\circ +\text{co}{{\text{t}}^{\text{2}}}\text{3}0{}^\circ \]  

 

 

  If \[2\sin 2\theta =\sqrt{3}\] then the value of \[\theta \] is _____.

(a) \[\frac{\pi }{3}\]                                                        

(b) \[\frac{\pi }{6}\]

(c) \[\frac{\pi }{4}\]                                                        

(d) \[\frac{\pi }{2}\]

(e) None of these

 

Answer: (b)                      

 

 

  If \[\text{cos2}x=\text{ sin6}0{}^\circ .\text{ cos3}0{}^\circ -\text{cos6}0{}^\circ .\text{ sin3}0{}^\circ \] then the value of \[x\] is _____.

(a) \[\frac{\pi }{6}\]                                                        

(b) \[\frac{\pi }{3}\]       

(c) \[\frac{2\pi }{3}\]                                                      

(d) \[\frac{3\pi }{6}\]

(e) None of these  

 

Answer: (a)    

 

                                  

  If \[\tan 5\theta =1\] then the value of \[\theta \] is _____.

(a) \[-\pi <\theta <0\]                                   

(b) \[\frac{\pi }{3}<\theta <\frac{\pi }{2}\]

(c) \[0<\theta <\frac{\pi }{6}\]                                  

(d) \[\frac{\pi }{6}<\theta <\frac{\pi }{3}\]

(e) None of these  

 

Answer: (c)                  

 

 

If \[\sec \alpha +\tan \alpha =p\] then the value of \[\tan \alpha \] is_____.

(a) \[\frac{p-1}{2p}\]                     

(b) \[\frac{{{p}^{2}}-1}{2p}\]

(c) \[{{p}^{2}}-1\]                            

(d) \[\frac{{{p}^{2}}-1}{{{p}^{2}}+1}\]                    

(e) None of these

 

Answer: (b)  

 

 

  If \[\text{cosec}\theta -\text{sin}\theta =\text{m}\] and \[\text{sec}\theta -\text{cos}\theta =\text{n}\] then

(a) \[{{({{m}^{2}}n)}^{\frac{2}{3}}}+{{(m{{n}^{2}})}^{\frac{2}{3}}}=1\]                    

(b) \[{{({{m}^{2}}{{n}^{2}})}^{\frac{1}{3}}}+{{(m{{n}^{2}})}^{\frac{1}{3}}}=1\]

(c) \[{{(m{{n}^{2}})}^{\frac{1}{3}}}+{{({{m}^{2}}{{n}^{2}})}^{\frac{2}{3}}}=1\]                     

(d) \[{{({{m}^{2}}{{n}^{2}})}^{\frac{1}{3}}}+{{({{m}^{2}}{{n}^{2}})}^{\frac{1}{3}}}=1\]

(e) None of these  

 

Answer: (a)

Explanation:

Here given that

\[\cos ec\theta -\sin \theta =m\] and \[sec\theta -\cos \theta =n\]

\[cosec\theta -\sin \theta =m\]

\[\Rightarrow \] \[\frac{1}{\sin \theta }-\sin \theta =m\]

\[\Rightarrow \]\[\frac{{{\cos }^{2}}\theta }{\sin \theta }=m\]                                    ....(i)

Similarly

\[\Rightarrow \]               \[\frac{{{\sin }^{2}}\theta }{\cos \theta }=n\]                      ....(ii)

\[={{({{m}^{2}}n)}^{\frac{2}{3}}}={{\left[ \left( \frac{{{\cos }^{2}}\theta }{\sin \theta } \right).\frac{{{\sin }^{2}}\theta }{\cos \theta } \right]}^{\frac{2}{3}}}\] \[={{[{{\cos }^{3}}\theta ]}^{\frac{2}{3}}}\]          \[={{\cos }^{2}}\theta \]

Similarly

\[{{(m{{n}^{2}})}^{\frac{2}{3}}}={{\sin }^{2}}\theta \]

\[\Rightarrow \]\[{{({{m}^{2}}n)}^{\frac{2}{3}}}+{{(m{{n}^{2}})}^{\frac{2}{3}}}={{\cos }^{2}}\theta +{{\sin }^{2}}\theta \] \[\Rightarrow \]\[{{({{m}^{2}}n)}^{\frac{2}{3}}}+{{(m{{n}^{2}})}^{\frac{2}{3}}}=1\]  

 

 

  Which one of the following identities is incorrect?

(a) \[{{\sin }^{4}}\theta +{{\cos }^{4}}\theta =1-2{{\sin }^{2}}\theta .{{\cos }^{2}}\theta \]

(b) \[{{\sin }^{4}}\theta +{{\cos }^{4}}\theta =1-2{{\sin }^{2}}\theta .{{\cos }^{2}}\theta \]

(c) \[{{\sin }^{6}}\theta +{{\cos }^{6}}\theta =1-4{{\sin }^{2}}\theta .{{\cos }^{2}}\theta \]

(d) \[{{\sec }^{4}}\theta -{{\sec }^{2}}\theta ={{\tan }^{4}}\theta +{{\tan }^{2}}\theta \]

(e) None of these

 

Answer: (c)  

 

 

  If \[(\sec \theta +\tan \theta )(\sec \alpha +\tan \alpha )(\sec \beta +\tan \beta )\] is equal to \[(\sec \theta -\tan \theta )\]\[(\sec \alpha -\tan \alpha )\]\[(\sec \beta -\tan \beta )\] then the each of the sides is equal to... .

(a) \[\sec \theta +\sec \alpha +\sec \beta \]                      

(b) \[\pm ({{\sec }^{2}}\theta -{{\tan }^{2}}\theta )\]

(c) \[\tan \theta +\tan \alpha +\tan \beta \]                       

(d) \[\sec \theta .\tan \theta +\sec \alpha .\tan \alpha .\tan \alpha +\sec \beta .\tan \beta \]

(e) None of these

 

Answer: (b)  

 

 

  If \[p=\tan \alpha +\sin \alpha \] and \[q=\tan \alpha -\sin \alpha \] then \[({{p}^{2}}-{{q}^{2}})\] is equal to.....

(a) \[pq\]                                                            

(b) \[\sqrt{pq}\]

(c) \[{{(pq)}^{\frac{2}{3}}}\]                                       

(d) \[4{{(pq)}^{\frac{1}{2}}}\]

(e) None of these  

 

Answer: (d)