Question

Use basic identities to simplify the expression.

Answer 1

\[\frac{ \cos ^{2} \theta }{\sin ^{2}\theta } +\csc \theta \sin \theta \]

Answer 2

answers are
A. csc2θ

B. sec2θ

C. 1

D. tan2θ

Answer 3

keep in mind that \(\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad tan^2(\theta)=\cfrac{sin^2(\theta)}{cos^2(\theta)}\\ \quad \\ \quad \\
csc(\theta)=\cfrac{1}{sin(\theta)}\)

so replace respectively

Answer 4

notice that \(\bf csc(\theta)sin(\theta)\implies \cfrac{1}{\cancel{sin(\theta)}}\cancel{sin(\theta)}\)

Answer 5

Also can someone \[\cos x-\sin ^{2}x-1\]Factor the algebraic expression below in terms of a single trigonometric function.

Answer 6

the answer to the first one is D correct

Answer 7

use the pythagorean identities

Answer 8

the what?

Answer 9

http://web.mit.edu/wwmath/trig/eq1.gif

Answer 10

but how can u explain the second one

Answer 11

notice the 1st identity
if you solve for \(\bf sin^2(\theta)\) what would you get?

Answer 12

1?

Answer 13

1? whatever happened to the cosine?

Answer 14

oh yeah im so confused

Answer 15

ok... lemme put it this way

a + b = 1

solve for "a" what do you get?

Answer 16

b-1

Answer 17

1-b

Answer 18

positive?

Answer 19

right, so \(\bf \begin{array}{llll}
sin^2(\theta)+&cos^2(\theta)=&1\\
a+&b=&1\\
\end{array}\qquad
\begin{array}{llll}
\color{red}{sin^2(\theta)}=&1-&cos^2(\theta)
\\a=&1-&b\\
\end{array}\\ \quad \\ \quad \\
cos(x)-sin^2(x)-1\implies cos(x)-[\color{red}{1-cos^2}]-1\)

Answer 20

well. sorta missed, the "x", but anyhow, just distribute and simplify

Answer 21

and the answer is csc^2(x)

Answer 22

or
(cosx+2) (cosx-1)