How do you find the other five trigonometric values given csc b = -6/5 and cot b > 0?

Question

How do you find the other five trigonometric values given csc b = -6/5 and cot b > 0?

1 Answer

Impact of this question

5052 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-03T08:02:57

$sin b = - \frac{5}{6}$ $sinb=-5/6$ , $cos b = - \frac{\sqrt{11}}{6}$ $cosb=-sqrt11/6$ , $tan b = \frac{5}{\sqrt{11}}$ $tanb=5/sqrt11$ ,

$cot b = \frac{\sqrt{11}}{5}$ $cotb=sqrt11/5$ and $sec b = - \frac{6}{\sqrt{11}}$ $secb=-6/sqrt11$ .

Explanation:

As $csc b = - \frac{6}{5}$ $cscb=-6/5$ $\Rightarrow$ $=>$ ${csc}^{2} b = \frac{36}{25}$ $csc^2b=36/25$ and using ${csc}^{2} b = 1 + {cot}^{2} b$ $csc^2b=1+cot^2b$ , we have

$1 + {cot}^{2} b = \frac{36}{25}$ $1+cot^2b=36/25$ and $cot b = \sqrt{\frac{36}{25} - 1} = \frac{\sqrt{11}}{5}$ $cotb=sqrt(36/25-1)=sqrt11/5$ (as $cot b > 0$ $cotb>0$ )

As $cot b = \frac{\sqrt{11}}{5}$ $cotb=sqrt11/5$ , $tan b = \frac{1}{cot b} = \frac{5}{\sqrt{11}}$ $tanb=1/cotb=5/sqrt11$

As $csc b = - \frac{6}{5}$ $cscb=-6/5$ , $sin b = \frac{1}{csc b} = - \frac{5}{6}$ $sinb=1/cscb=-5/6$

and $cos b = \frac{cos b}{sin b} \times sin b = cot b \times sin b$ $cosb=cosb/sinbxxsinb=cotbxxsinb$

= $\frac{\sqrt{11}}{5} \times (- \frac{5}{6}) = - \frac{\sqrt{11}}{6}$ $sqrt11/5xx(-5/6)=-sqrt11/6$

and $sec b = \frac{1}{cos b} = - \frac{6}{\sqrt{11}}$ $secb=1/cosb=-6/sqrt11$ .

Hence, we have $sin b = - \frac{5}{6}$ $sinb=-5/6$ , $cos b = - \frac{\sqrt{11}}{6}$ $cosb=-sqrt11/6$ , $tan b = \frac{5}{\sqrt{11}}$ $tanb=5/sqrt11$ ,

$cot b = \frac{\sqrt{11}}{5}$ $cotb=sqrt11/5$ and $sec b = - \frac{6}{\sqrt{11}}$ $secb=-6/sqrt11$ .

Science

Math

Humanities

... and beyond

How do you find the other five trigonometric values given csc b = -6/5 and cot b > 0?

1 Answer

Explanation:

Related questions

Impact of this question