# need help very urgent

• May 11th 2006, 06:12 AM
cheesepie
need help very urgent
can show mi the step...thanx

if sin x=-0.6 and tan x = -0.75,then sec x is

1)-1.25
2)1.25
3)0.80
4)-1.35
• May 11th 2006, 06:23 AM
earboth
Quote:

Originally Posted by cheesepie
can show mi the step...thanx

if sin x=-0.6 and tan x = -0.75,then sec x is

1)-1.25
2)1.25
3)0.80
4)-1.35

Hello,

1. You know that $\tan(x)=\frac{\sin(x)}{\cos(x)}$ and

$\sec(x)=\frac{1}{\cos(x)}$ so

$\sec(x)=\frac{\tan(x)}{\sin(x)}$

Therefore: $\frac{-\frac{3}{4}}{-\frac{3}{5}}=\frac{5}{4}$

Greetings

EB
• May 11th 2006, 06:24 AM
CaptainBlack
Quote:

Originally Posted by cheesepie
can show mi the step...thanx

if sin x=-0.6 and tan x = -0.75,then sec x is

1)-1.25
2)1.25
3)0.80
4)-1.35

$\tan(x)=\sin(x)/\cos(x)=\sin(x)\sec(x)$,

so:

$
\sec(x)=\tan(x)/\sin(x)=(-0.75)/(-0.6)=1.25
$

RonL
• May 11th 2006, 06:35 AM
earboth
Quote:

Originally Posted by CaptainBlack
$\tan(x)=\sin(x)/\cos(x)=\sin(x)\sec(x)$,

so:

$
\sec(x)=\tan(x)/\sin(x)=-0.75/0.6=-1.25
$

RonL

Hello,

I'm a little bit confused, because I read a sin x = -0.6 in the problem of cheesepie.

Greetings

EB
• May 11th 2006, 07:09 AM
cheesepie
can u show mi the step of this question ? thanx

if cos x=0.6 and tan x is negative ,then csc x is

1)1.25
2)-1.25
3)0.80
4)0.75
• May 11th 2006, 08:07 AM
CaptainBlack
Quote:

Originally Posted by earboth
Hello,

I'm a little bit confused, because I read a sin x = -0.6 in the problem of cheesepie.

Greetings

EB

Sorry I misread the value :o

RonL
• May 11th 2006, 12:37 PM
earboth
Quote:

Originally Posted by cheesepie
can u show mi the step of this question ? thanx
if cos x=0.6 and tan x is negative ,then csc x is
1)1.25
2)-1.25
3)0.80
4)0.75

Hello,

This problem is a tricky one, because you didn't give a value for tan(x). To be able to calculate a value I take the value from the previous problem: tan(x)=-.75

1. You know that $\cot(x)={\cos(x) \over \sin(x)}$ and

$\csc(x)={1 \over \sin(x)}$ and $\tan(x)={\sin(x) \over \cos(x)}$ So you get:

$\csc(x)={{\cos(x) \over \sin(x)} \over \cos(x)}}$ = ${{1 \over \tan(x)} \over \cos(x)}}$ = ${{1 \over -.75} \over .6}}$ = $-{20 \over 9} \approx -2.22$

This answer doesn't exist. So I presume, that my asumption about the value of tan(x) was incorrect. But maybe you can use my explanations when you plug in the right value for tan(x).

Greetings

EB
• May 11th 2006, 04:33 PM
topsquark
Quote:

Originally Posted by cheesepie
can u show mi the step of this question ? thanx

if cos x=0.6 and tan x is negative ,then csc x is

1)1.25
2)-1.25
3)0.80
4)0.75

We know that cos x = 0.6
Thus: $sin \,x = \pm \sqrt{1-0.6^2}=\pm 0.8$

Since tan x is negative, sin x = -0.8

Thus csc x = 1/sin x = 1/-0.8 = -1.25

-Dan