Chapter 3: Trigonometric Functions

Ans:

Here given that,  $\text{cos x=-}\frac{\text{1}}{\text{2}}$

Therefore we have,

$\text{sec x=}\frac{\text{1}}{\text{cos x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}$

$\text{=-2}$

Now we know that, $\text{si}{{\text{n}}^{\text{2}}}\text{x+co}{{\text{s}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{si}{{\text{n}}^{\text{2}}}\text{x=1-co}{{\text{s}}^{\text{2}}}\text{x}$

Substituting  $\text{cos x=-}\frac{\text{1}}{\text{2}}$ in the formula, we obtain,

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}{{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}^{\text{2}}}$

$\text{si}{{\text{n}}^{\text{2}}}\text{x=1-}\frac{\text{1}}{\text{4}}$

$\text{=}\frac{\text{3}}{\text{4}}$

$\text{sin x= }\!\!\pm\!\!\text{ }\frac{\sqrt{\text{3}}}{\text{2}}$

Since $\text{x}$ lies in the ${{\text{3}}^{\text{rd}}}$quadrant, the value of $\sin x$ will be negative.

$\text{sin x=-}\frac{\sqrt{\text{3}}}{\text{2}}$

Therefore, $\text{cosec x=}\frac{\text{1}}{\text{sin x}}$ 

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\sqrt{\text{3}}}{\text{2}} \right)}$ 

$\text{=-}\frac{\text{2}}{\sqrt{\text{3}}}$ 

Hence ,

$\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

$\text{=}\frac{\left( \text{-}\frac{\sqrt{\text{3}}}{\text{2}} \right)}{\left( \text{-}\frac{\text{1}}{\text{2}} \right)}$

$\text{=}\sqrt{\text{3}}$

And 

$\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=}\frac{\text{1}}{\sqrt{\text{3}}}$

Ans:

Here given that,  $\text{sin x=}\frac{\text{3}}{\text{5}}$

Therefore we have,

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$=\frac{1}{\left( \frac{3}{5} \right)}$

$=\frac{5}{3}$

Now we know that , ${{\sin }^{2}}x+{{\cos }^{2}}x=1$

Therefore we have, $\text{co}{{\text{s}}^{\text{2}}}\text{x=1-si}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\sin x=\frac{3}{5}$  in the formula, we obtain,

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}{{\left( \frac{\text{3}}{\text{5}} \right)}^{\text{2}}}$

$\text{co}{{\text{s}}^{\text{2}}}\text{x=1-}\frac{\text{9}}{\text{25}}$

$\text{=}\frac{\text{16}}{\text{25}}$ 

$\text{cos x= }\!\!\pm\!\!\text{ }\frac{\text{4}}{\text{5}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\cos x$ will be negative.

$\text{cos x=-}\frac{\text{4}}{\text{5}}$

Therefore, $sec x=\frac{1}{\cos x}$ 

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{4}}{\text{5}} \right)}$ 

$\text{=-}\frac{\text{5}}{\text{4}}$ 

Hence ,

$\tan x=\frac{\sin  x}{\cos x}$ 

$\text{=}\frac{\left( \frac{\text{3}}{\text{5}} \right)}{\left( \text{-}\frac{\text{4}}{\text{5}} \right)}$

$\text{=-}\frac{\text{3}}{\text{4}}$

And 

$\cot x=\frac{1}{\tan x}$

$\text{=-}\frac{\text{4}}{\text{3}}$

Ans:

Here given that,  $\cot x=\frac{3}{4}$

Therefore we have,

$\tan x=\frac{1}{\cot x}$

$=\frac{1}{\left( \frac{3}{4} \right)}$

$=\frac{4}{3}$

Now we know that , \[\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}\]

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=}\frac{\text{4}}{\text{3}}$  in the formula, we obtain,

${{\sec }^{2}}x=1+{{\left( \frac{4}{3} \right)}^{2}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\frac{\text{16}}{\text{9}}$

$=\frac{25}{9}$ 

$\text{sec x= }\!\!\pm\!\!\text{ }\frac{\text{5}}{\text{3}}$

Since $x$ lies in the ${{3}^{rd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\frac{\text{5}}{\text{3}}$

Therefore, $\cos  x=\frac{1}{\sec x}$ 

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{5}}{\text{3}} \right)}$ 

$\text{=-}\frac{\text{3}}{\text{5}}$ 

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

Hence we have, $\text{sin x=}\frac{\text{4}}{\text{3}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{3}}{\text{5}} \right)$   

\[\text{=}\left( \text{-}\frac{\text{4}}{\text{5}} \right)\] 

And 

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=-}\frac{\text{5}}{\text{4}}$

Ans:

Here given that,  $\sec x=\frac{13}{5}$

Therefore we have,

$\cos x=\frac{1}{\sec x}$

$=\frac{1}{\left( \frac{13}{5} \right)}$

$=\frac{5}{13}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{ta}{{\text{n}}^{\text{2}}}\text{x=se}{{\text{c}}^{\text{2}}}\text{x-1}$

Substituting  \[\text{sec x=}\frac{\text{13}}{\text{5}}\]  in the formula, we obtain,

\[\text{ta}{{\text{n}}^{\text{2}}}\text{x=}{{\left( \frac{\text{13}}{\text{5}} \right)}^{\text{2}}}\text{-1}\]

$\text{ta}{{\text{n}}^{\text{2}}}\text{x=}\frac{\text{169}}{\text{25}}\text{-1}$

$\text{=}\frac{\text{144}}{\text{25}}$ 

\[\text{tanx= }\!\!\pm\!\!\text{ }\frac{\text{12}}{\text{5}}\]

Since $x$ lies in the ${{4}^{th}}$ quadrant, the value of $\tan x$ will be negative.

$\text{tan x=-}\frac{\text{12}}{\text{5}}$

Therefore, \[\text{cot x=}\frac{\text{1}}{\text{tan x}}\] 

$\text{=-}\frac{\text{5}}{\text{12}}$ 

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

Hence we have, $\text{sin x=}\frac{\text{5}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{12}}{\text{5}} \right)$   

$\text{=}\left( \text{-}\frac{\text{12}}{\text{13}} \right)$ 

And 

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=-}\frac{\text{13}}{\text{12}}$

Ans:

Here given that,  $\text{tan x=-}\frac{\text{5}}{\text{12}}$

Therefore we have,

$\text{cot x=}\frac{\text{1}}{\text{tan x}}$

$\text{=}\frac{\text{1}}{\left( \text{-}\frac{\text{5}}{\text{12}} \right)}$

$\text{=-}\frac{\text{12}}{\text{5}}$

Now we know that , $\text{se}{{\text{c}}^{\text{2}}}\text{x-ta}{{\text{n}}^{\text{2}}}\text{x=1}$

Therefore we have, $\text{se}{{\text{c}}^{\text{2}}}\text{x=1+ta}{{\text{n}}^{\text{2}}}\text{x}$

Substituting  $\text{tan x=-}\frac{\text{5}}{\text{12}}$  in the formula, we obtain,

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}{{\left( \text{-}\frac{\text{5}}{\text{12}} \right)}^{\text{2}}}$

$\text{se}{{\text{c}}^{\text{2}}}\text{x=1+}\frac{\text{25}}{\text{144}}$

$=\frac{169}{144}$ 

$\text{sec x= }\!\!\pm\!\!\text{ }\frac{\text{13}}{\text{12}}$

Since $x$ lies in the ${{2}^{nd}}$ quadrant, the value of $\sec x$ will be negative.

$\text{sec x=-}\frac{\text{13}}{\text{12}}$

Therefore, $\text{cos x=}\frac{\text{1}}{\text{sec x}}$ 

$\text{=-}\frac{\text{12}}{\text{13}}$ 

Now  , $\text{tan x=}\frac{\text{sin x}}{\text{cos x}}$ 

Therefore, $\text{sin x=tan xcos x}$ 

Hence we have, $\text{sin x=}\left( \text{-}\frac{\text{5}}{\text{12}} \right)\text{ }\!\!\times\!\!\text{ }\left( \text{-}\frac{\text{12}}{\text{13}} \right)$   

$=\left( \frac{5}{13} \right)$ 

And 

$\text{cosec x=}\frac{\text{1}}{\text{sin x}}$

$\text{=}\frac{\text{13}}{\text{5}}$

Ans: We know that the values of $\sin x$ repeat after an interval of $2\pi $ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin76}{{\text{5}}^{\text{o}}}\text{=sin}\left( \text{2 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

$\text{=sin4}{{\text{5}}^{\text{o}}}$

$\text{=}\frac{\text{1}}{\sqrt{\text{2}}}\text{.}$

Ans: We  know that the values of $\text{cosec x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{cosec}\left( \text{-141}{{\text{0}}^{\text{o}}} \right)\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+4 }\!\!\times\!\!\text{ 36}{{\text{0}}^{\text{o}}} \right)$

$\text{=cosec}\left( \text{-141}{{\text{0}}^{\text{o}}}\text{+144}{{\text{0}}^{\text{o}}} \right)$

$\text{=cosec3}{{\text{0}}^{\text{o}}}$

$=2$ 

Ans: We know that the values of $\text{tan x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

Therefore we can write,

$\text{tan}\frac{\text{19 }\!\!\pi\!\!\text{ }}{\text{3}}\text{=tan6}\frac{\text{1}}{\text{3}}\text{ }\!\!\pi\!\!\text{ }$

$\text{=tan}\left( \text{6 }\!\!\pi\!\!\text{ +}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

\[\text{=tan}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}}\]

$\text{=}\sqrt{\text{3}}$      

Ans: We know that the values of $\text{sin x}$ repeat after an interval of $\text{2 }\!\!\pi\!\!\text{ }$ or ${{360}^{\circ }}$ .

Therefore we can write,

$\text{sin}\left( \text{-}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}} \right)\text{=sin}\left( \text{-}\frac{\text{11 }\!\!\pi\!\!\text{ }}{\text{3}}\text{+2 }\!\!\times\!\!\text{ 2 }\!\!\pi\!\!\text{ } \right)$ 

$\text{=sin}\left( \frac{\text{ }\!\!\pi\!\!\text{ }}{\text{3}} \right)$

$=\frac{\sqrt{3}}{2}$

Ans: We know that the values of $\text{cot x}$ repeat after an interval of $\text{ }\!\!\pi\!\!\text{ }$ or \[{{180}^{\circ }}\].

Therefore we can write,

$\text{cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}} \right)\text{=cot}\left( \text{-}\frac{\text{15 }\!\!\pi\!\!\text{ }}{\text{4}}\text{+4 }\!\!\pi\!\!\text{ } \right)$ 

$\text{=cot}\frac{\text{ }\!\!\pi\!\!\text{ }}{\text{4}}$

$=1$