Respuesta :
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/ opposite
so, hyp=5
opposite =3
now, we can find adjacent
we 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^2x
we 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/hyp
so, opp=5
hyp=13
now, 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/adj
so, we get
[tex]sec(\theta)=\frac{13}{12}[/tex]
