Trigonometric identity question?

Question

Trigonometric identity question?
Could someone pls help me with this? Thx.
The identity [(1/sin x) - sin x] is equivalent to:

A) cos x / sin x
B) tan x / cos x
C) sin x /cos x
D) 1

Answer 1

1/sin(x) - sin(x)
Make a common denominator of sin(x):
= 1/sin(x) - sin^2(x)/sin(x)
= [1 - sin^2(x)] / sin(x)
Recall that cos^2(x) + sin^2(x) = 1:
= cos^2(x) / sin(x)
Separate squared term:
= cos(x) * cos(x)/sin(x)
Recall that cos(x)/sin(x) = cot(x):
= cos(x) * cot(x)

none of the above

Answer 2

For the reasons the first couple of responders show, I'll wager you wrote one of the answer choices incorrectly. Either A) (cos x)^2 / sin x or B) cos x / tan x. Either one would be something you can transform the expression you gave into.

Answer 3

None of them are true.

Let x = π/4.

Then we have:

[(1/sin x) - sin x] = 1/(1/√2) - 1/√2 = √2 - 1/√2

A) cosx / sinx = (1/√2) / (1/√2) = 1
B) tanx / cosx = 1/(1/√2) = √2
C) sinx / cosx = (1/√2) / (1/√2) = 1
D) 1

As you can see the answer is none of the above.

Answer 4

(1 / sin) - sin =
(common demon) = sin

(1 / sin) - (sin^2 / sin) =

1 - sin^2 / sin =

so ... cos^2 / sin (almost "A")