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

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