# sec B=(-17/8), pi<B<(3pi/2) find sin B, cos B, and tan B....help please!

*print*Print*list*Cite

### 2 Answers

sec B = -17/8 and pi < B < 3pi/2

In the given range of B, both sin and cos have negative values and therefore, tan has a positive value.

now, sec is inverse of cos.

Thus, sec B = 1/cosB = -17/8

means, **cos B = -8/17**

sin B = `sqrt (1-cos^2B) = sqrt(1-(-8/17)^2) = - 15/17`

and tan B = sinB/cosB = `(-15/17)/(-8/17) = 15/8`

Hope this helps.

From the below tables givne in the attachments we can easily find the values of **sin B, cos B, tan B**

**Given ,**

`sec B=(-17/8), pi<B<(3pi/2) ` ie the **angle B** is in the **Third quadrant .**

From the table I , we get the values and from the table II , we assign the signs that is the final answer

Now , Finding

1)

`sin B = (+- (sqrt(sec^2 B -1 )/(sec B)))`

= `(+- (sqrt((-17/8)^2 -1)/(-17/8)))`

= `(+- (sqrt((17^2 - 8^2)/8^2)/(-17/8)))`

=`(+- (sqrt((17^2 - 8^2))/8)/(-17/8))`

= `(+- (sqrt((17^2 - 8^2))/(-17)))`

= `(+- (sqrt((289 - 64))/(-17)))`

= `(+- (sqrt((225))/(-17)))`

= `(+-(-15)/17)`

As the angle is in the **third quadrent** **sin is negative** (see table II )

so ,

`sin B =((-15)/17)`

2)

Cos B = `(1/ sec B)`

= `(1/((-17)/8))`

= `((-8)/17)`

3)

Tan B = `(+- sqrt(sec^2 B -1 ))`

= `(+- sqrt(((-17)/8)^2 -1 ))`

= `(+- sqrt(((289)/64) -1 ))`

= `(+- sqrt(((289-64)/64) ))`

= `(+- sqrt(((225)/64) ))`

= `(+- (15/8) )`

As the angle is in the **Third quadrent** **tan is positive** (see table II )

so,

`Tan B =(15/8)`

simple :)

**Images:**