Proving Trig identity?

Question

Hi,
I was just wondeirng if i could get some help solving this trig identity Prove:
sin^2x sec^2x = sec^2x - 1
i dont know how to prove LS = RS plz help

~Thnx

Answer 1

sin^2(x)sec^2(x)

substitute sin^2(x) = 1- cos^2(x)

sec^2x(1 - cos^2(x)

=>sec^2(x) - sec^2x cos^2x

=>sec^2(x) - cos^2(x)/cos^2x) (since sec^2(x) = 1/cos^2(x))

=>sec^2(x) - 1

so LHS = RHS

Answer 2

In this problem, I believe you need to use the Pythagorean, Reciprocal, and Quotient Identities...as far as a textbook is concerned. Practically, here -
take each side of the identity at a time:
Sin^2(x) Sec^2(x) can be simplified to Tan^2 (x) because secant is 1/cosine, thus you get Sin^2(x) / Cos^2(x).

so it is now Tan^2(x) = sec^2(x) -1. According to the Pythag ID, 1+tan^2(x) = Sec^2(x). This is easily modified by subtracting '1' from both sides. This property illustrates the end of the proof, ending with:
tan^2(x) = tan^2(x)

Answer 3

sin^2x.sec^2x=sin^2x/cos^2x = tan^2x = sec^2x-1 (because 1+tan^2x=sec^2x - Identity)

Answer 4

Prove the identity.

(sin²x)(sec²x) = sec²x - 1
_____________

Left Hand Side = (sin²x)(sec²x) = (1 - cos²x)(1/cos²x)

= (1 - cos²x)/cos²x = 1/cos²x - cos²x/cos²x

= sec²x - 1 = Right Hand Side

Answer 5

sec^2x = 1/(1-sin^2x) - Basic identity

sin^2x * sec^2x = sec^2x - 1
sin^2x/(1-sin^2x) = 1/(1-sin^2x) - 1 :Multiple everything by (1-sin^2x)
sin^2x = 1 - 1 + sin^2x
sin^2x = sin^2x

Answer 6

1) cos(x)(1 + tan^2(x)) = cos(x)sec^2(x) = cos(x)/cos^2(x) = 1/cos(x) = sec(x) 2) 1 - cot(x) What else is there to say?

Answer 7

since
sec^2x=1/cos^2x

so
sin^2xsec^2x=sin^2x/cos^2x

as
sin^2x+cos^2x=1

therefore
sin^2x=1-cos^2x

sin^2xsec^2x=(1-cos^2x)/cos^2x

sin^2xsec^2x=(1/cos^2x)-(cos^2x/cos^2x)

sin^2xsec^2x=sec^2x-1

hence proved LS=RS