# Thread: need help very urgent

1. ## 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

2. 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 $\displaystyle \tan(x)=\frac{\sin(x)}{\cos(x)}$ and

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

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

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

Thus the answer is 2)

Greetings

EB

3. 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
$\displaystyle \tan(x)=\sin(x)/\cos(x)=\sin(x)\sec(x)$,

so:

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

RonL

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

so:

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

RonL
Hello,

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

Greetings

EB

5. 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

6. 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

RonL

7. 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 $\displaystyle \cot(x)={\cos(x) \over \sin(x)}$ and

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

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

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

8. 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: $\displaystyle sin \,x = \pm \sqrt{1-0.6^2}=\pm 0.8$

Since tan x is negative, sin x = -0.8

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

-Dan