math...trig identites..?

Question

Prove:
csc^2x-1=(csc^2x)(cos^2x)
can someone show me the steps of solving this question

Answer 1

csc^2 x * cos^2 x = (1/sin^2 x) * cos^2 x
csc^2 x * cos^2 x = cot^2 x

cot^2 x = csc^2 x - 1

adj^2 + opp^2 = hyp^2
adj^2 / opp^2 + 1 = hyp^2 / opp^2
cot^2 x + 1 = csc^2 x
cot^2 x = csc^2 x - 1

Answer 2

ok so beta is an perspective between 0 and a couple of pi. From the trig identities all of us understand sin^2 B + cos^2 B = a million (a million) Squaring the two sides provides sin^2 B = cos 2B*cos 2B (2) subbing in a million- cos^2 B (a rearrangement of equation a million provides: a million- cos^2 B = cos^2 2B fixing for beta provides an perspective of 15` or a million/12 pi :) desire that facilitates

Answer 3

(csc(x))^2(cos(x))^2 = (cos(x)/sin(x))^2 = (cot(x))^2
= (csc(x))^2 - 1 by a familiar Pythagorean identity.

Answer 4

csc^2x = 1/sin^2x

csc^2x * cos^2x = cos^2x / sin^2x

1/sin^2x - 1 = cos^2x / sin^2x <--multiply both sides by sin^2x

1 - sin^2x = cos^2x

1 = cos^2x + sin^2x, which is known.

Answer 5

csc^2x-1=(csc^2x)(cos^2x)
a trig identity states that csc^2x-1=cot^2x so . . .
cot^2x=(csc^2x)(cos^2x)
cot^2x is really just cos^2x/sin^2x so. . .
cos^2x/sin^2x=(csc^2x)(cos^2x)
if you spit it up it equals. . .
cos^2x/1 times 1/sin^2x=(csc^2x)(cos^2x)
another trig identity states that 1/sin^2x=csc^2x so. . .
cos^2x(csc^2x)=(csc^2x)(cos^2x)
I,KEA