Let θ be an angle in quadrant II such that cotθ=32. Find the exact values of sinθ and secθ.

Question

Let θ be an angle such that secθ=85 and cotθ>0 Find the exact values of tanθ and cscθ

I have 4 more if that's ok

Answer 1

cotangents are all negative in Quadrant II, so cotT=32 would be in Quadrant I or III

maybe you left out a negative sign? If so then cotT =-32

= adjacent side over opposite side, hypotenuse = square root of (32^2 +1^2) = sqr1025 = 5sqr41= about 32.02

sinT =opposite side over hypotenuse = 1/5sqr41

secT = hypotenuse over adjacent side = -5sqr41/32

secT= 84, cotT>0 means quadrant I, with cosT>0

adjacent side = 1, hypotenuse = 84, opposite side=sqr(84^2-1^2) = sqr7057

cosT = 1/84

sinT = sqr7057/84

cscT =1/sinT = 84/sqr7057

tanT = sinT/cosT = 1/sqr7057