We know that , `sinA=(4)/(5)=("Perpendicular")/("hypotenuse")`
`:.` Now construct a `DeltaABC` in which
`angleB=90^(@), BC=4kandAC=5k`.
In `DeltaABC` , from Pythagoras theorem
`AB^(2)+BC^(2)=AC^(2) rArr AB^(2)=AC^(2)-BC^(2)-(5k)^(2)-(4k)^(2)`
`rArr=25k^(2)-16k^(2)=9k^(2)`
`rArrAB=3k`
Now , `cosA=("base")/("hypotenuse")=(AB)/(AC)=(3k)/(5k)=(3)/(5)`
` tanA=("perpendicular")/("base")=(BC)/(AB)=(4k)/(3k)=(4)/(3)`
`"cosec"A=("hypotenuse")/("perpendicular")=(AC)/(BC)=(5k)/(4k)=(5)/(4)`
`secA=("hypotenuse")/("base")=(AC)/(AB)=(5k)/(3k)=(5)/(3)`
` "cotA=("base")/("perpendicular")=(AB)/(BC)=(3k)/(4k)=(3)/(4)`
