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

Chart for the Sign of Different Trigonometrical

Category : 10th Class

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

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

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

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

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

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

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

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

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

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