Q:

# 1. If cscθ=5/3 then secθ=a. 25/16b. 5/4c. 16/25d. 4/52. Simplify the expression (cos x) (sec x) - (sin2 x)a. cos xb. cos2 xc. sec2 x3. If θ is in Quadrant IV and sinθ=-5/4 then secθ=a. -13/12b. -12/3c. 12/13d. 13/12

Accepted Solution

A:
Answer:(1)b. $$sec(\theta)=\frac{5}{4}$$(2)b. $$cos^2(x)$$(3)$$sec(\theta)=\frac{13}{12}$$
Step-by-step explanation:(1)we are given $$csc(\theta)=\frac{5}{3}$$we can use triangle method we know that csc=hyp/ oppositeso, hyp=5opposite =3now, we can find adjacentwe can use Pythagoras theorem$$hyp^2=opp^2+adj^2$$$$5^2=3^2+adj^2$$$$adj=4$$now, we can find sec$$sec(\theta)=\frac{5}{4}$$
(2)we are given $$(cosx)(secx)-sin^2(x)$$we can simplify it $$(cosx)\times (\frac{1}{cosx})-sin^2(x)$$$$1-sin^2(x)$$now, we can replace 1 as sin^2x +cos^2xwe get $$sin^2(x)+cos^2(x)-sin^2(x)$$$$=cos^2(x)$$(3)we are given θ is in Quadrant IV$$sin(\theta)=\frac{-5}{13}$$we know that sin =opp/hypso, opp=5hyp=13now, we can use Pythagoras theorem$$hyp^2=opp^2+adj^2$$now, we can plug values$$13^2=5^2+adj^2$$$$adj=12$$sec=hyp/adjso, we get $$sec(\theta)=\frac{13}{12}$$