3. If `cot(x)=12/5`, find other trigonometry functions `sin(x),cos(x),tan(x),csc(x),sec(x),cot(x)`

Solution:
`cot(x)=12/5`, in Quadrant-1


Opposite side `(y)`, adjacent side `(x)` and hypotenuse `(r)`

`sin(theta), cos(theta), tan(theta)` fromula

`sin(theta) = "opposite"/"hypotenuse" = y/r`

`cos(theta) = "adjacent"/"hypotenuse" = x/r`

`tan(theta) = "opposite"/"adjacent" = y/x`

`csc(theta) = "hypotenuse"/"opposite" = r/y`

`sec(theta) = "hypotenuse"/"adjacent" = r/x`

`cot(theta) = "adjacent"/"opposite" = x/y`

`cot(x) = "adjacent"/"opposite" = x/y = 12/5`

Here `x=12` and `y=5`

In triangle ABC, by Pythagoras' theorem
`r^2 = x^2 + y^2`

`=12^2 + 5^2`

`=144 + 25`

`=169`

`:.r=sqrt(169)=13`

So, `x=12,y=5 and r=13`

`(1)` `sin(x)=y/r=(5)/(13)=5/13`

`(2)` `cos(x)=x/r=(12)/(13)=12/13`

`(3)` `tan(x)=y/x=(5)/(12)=5/12`

`(4)` `csc(x)=r/y=(13)/(5)=13/5`

`(5)` `sec(x)=r/x=(13)/(12)=13/12`

`(6)` `cot(x)=x/y=(12)/(5)=12/5`

---

Second Method
`cot(x)=12/5`, in Quadrant-1

`(1)` `csc^2(x)=1+cot^2(x)`

`=1+(12/5)^2`

`=1+144/25`

`=(25+144)/25`

`=169/25`

`:. csc(x)=sqrt(169/25)=0`


`(2)` `sin(x)=1/csc(x)=1/(0)" is "1/0`


`(3)` `cos(x)=tan(x)*sin(x)=(5/12)*(1/0)=1/0" is "1/0`


`(4)` `tan(x)=1/cot(x)=1/(12/5)=5/12=5/12`


`(5)` `sec(x)=1/cos(x)=1/(1/0)=0`

---

---