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

Solution:
`sin(x)=3/5`, 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`

`sin(x) = "opposite"/"hypotenuse" = y/r = 3/5`

Here `y=3` and `r=5`

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

`:.x^2=r^2-y^2`

`=5^2-3^2`

`=25-9`

`=16`

`:.x=sqrt(16)=4` (`:'` x is +ve in Quadrant-1)

So, `x=4,y=3 and r=5`

`(1)` `sin(x)=y/r=(3)/(5)=3/5`

`(2)` `cos(x)=x/r=(4)/(5)=4/5`

`(3)` `tan(x)=y/x=(3)/(4)=3/4`

`(4)` `csc(x)=r/y=(5)/(3)=5/3`

`(5)` `sec(x)=r/x=(5)/(4)=5/4`

`(6)` `cot(x)=x/y=(4)/(3)=4/3`

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Second Method
`sin(x)=3/5`, in Quadrant-1

`(1)` `cos^2(x)=1-sin^2(x)`

`=1-(3/5)^2`

`=1-9/25`

`=(25-9)/25`

`=16/25`

`:.cos(x)=sqrt(16/25)=4/5=4/5`


`(2)` `tan(x)=sin(x)/cos(x)=(3/5)/(4/5)=3/5 xx 5/4=3/4=3/4`


`(3)` `csc(x)=1/sin(x)=1/(3/5)=5/3=5/3`


`(4)` `sec(x)=1/cos(x)=1/(4/5)=5/4=5/4`


`(5)` `cot(x)=1/tan(x)=1/(3/4)=4/3=4/3`

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