If `costheta=8/(17)` , find the other five trigonometric ratios.

Explanation

We know that , `cosB=(8)/(17)=("base")/("hypotenuse")`
Now , construct a `DeltaABC` in which
`AB=8k,BC=17kandangleBAC=90^(@)`
In `DeltaABC` , from Pythagoras theorem
`AB^(2)+AC^(2)=BC^(2)`
`rArr" " AC^(2)=BC^(2)-AB^(2)=(17k)^(2)-(8k)^(2)`
`=289k^(2)-64k^(2)=225k^(2)rArrAC=15k`
Now , we have to find the trigonometric ratios for angle C.
For `ltC, base = AC =15k`, perpendicular = AB=8k.
Therefore , " " `sinC=("perpendicular")/("hypotenuse")=(AB)/(BC)=(8k)/(17k)=(8)/(17)`
`cosC=("base")/("hypotenuse")=(AC)/(BC)=(15k)/(17k)=(15)/(17)`
`tanC=("perpendicular")/("base")=(AB)/(AC)=(8k)/(15k)=(8)/(15)`
`"cosec" C=("hypotenuse")/("perpendicular")=(BC)/(AB)=(17k)/(8k)=(17)/(8)`
`sec C=("hypotenuse")/("base")=(BC)/(AC)=(17k)/(15k)=(17)/(15)`
`cot C=("base")/("perpendicular")=(AC)/(AB)=(15k)/(8k)=(15)/(8)`