proving identities in trigo, help?

Question

[(sec^4 x + tan^4 x) / (sec^2 x tan^2 x)] - [(cos^ 4 x)/(sin^2 x)] = 2

thanks!

Answer 1

Prove the identity.

[(sec^4 x + tan^4 x) / (sec²x tan²x)] - [(cos^ 4 x)/(sin²x)] = 2

Left Hand Side

= [(sec^4 x + tan^4 x) / (sec²x tan²x)] - [(cos^ 4 x)/(sin²x)]

= [(cos^4 x)(sec^4 x + tan^4 x) / (cos^4 x sec²x tan²x)]
- [(cos^ 4 x)/(sin²x)]

= [(1 + sin^4 x) / (sin²x)] - [(cos^ 4 x)/(sin²x)]

= [1 + sin^4 x - cos^4 x] / sin²x

= [1 + (sin²x + cos²x)(sin²x - cos²x)] / sin²x

= [1 + sin²x - cos²x] / sin²x

= [(1 - cos²x) + sin²x] / sin²x

= [2sin²x] / sin²x = 2 = Right Hand Side

Answer 2

[(sec^4 x + tan^4 x) / (sec^2 * tan^2 x)] - [(cos^ 4 x)/(sin^2 x)] = 2
LHS = [(sec^2 x)^2 + tan^4 x) / (sec^2 x tan^2 x)] - [1/((1/cos^2 x)(sin^2 x/ cos ^2 x))]
LHS = [(1 + tan^2 x)^2 + tan^4 x) / (sec^2 * tan^2 x)] - [1/(sec^2 * tan^2 x)]
LHS=(1 + 2tan^2 x + tan^4 x + tan^4 x - 1) /(sec^2 * tan^2 x)
LHS = (2tan^ 2 + 2 tan^4 x) / (sec^2 * tan^2 x)
LHS = 2tan^2 x * (1 + tan^2 x) / (sec^2 * tan^2 x)
LHS = 2(1 + tan^2 x) / sec^2 x
LHS = 2 sec^2 x / sec^2 x
LHS = 2 = RHS

LHS = left hand side
RHS = right hand side