how can you prove the identity of sec^4 0/tan^2 0= csc^2 0 +sec^2 0?

Answer 1

taking the angle as x,

L.H.S = (sec²x)² / (tan²x)

= (1+ tan²x)² / (tan²x)

=(1 + 2tan ² x +tan^4x ) / (tan²x)

= cot ² x + 2 + tan ² x {dividing term by term}

= ( csc ² x -1 ) +2 + (sec ² x -1)

= csc ² x + sec ² x

Answer 2

[(1 / COS) (1 / COS) (1 / COS) (1 / COS)] / [(SIN / COS) (SIN / COS)]

(1 / COS) (1 / COS) (1 / COS) (1 / COS) (COS / SIN) (COS / SIN)

cancel the (1 / COS) with the (COS /SIN) so:

(1 / COS) (1 / COS) (1 / SIN) (1 / SIN)

(1 / COS^2) (1 / SIN^2)

(SEC^2 X) (CSC^2 X)

Humm... iono how to do this

Answer 3

((sec0)^4)/((tan0)^2) = (csc0)^2 + (sec0)^2
((1/cos0)^4)/((sin0/cos0)^2) = (csc0)^2 + (sec0)^2
(1/(cos0^4))*((cos0^2)/(sin0^2)) = (csc0)^2 + (sec0)^2
(cos0^2)/(sin0^2 * cos0^4) = (csc0)^2 + (sec0)^2
1/(sin0^2 * cos0^2) = (csc0)^2 + (sec0)^2

(csc0)^2 + (sec0)^2
(1/sin0)^2 + (1/cos0)^2
(cos0^2 + sin0^2)/(sin0^2 * cos0^2)
(1 - sin0^2 + sin0^2)/(sin0^2 * cos0)^2
1/(sin0^2 * cos0^2)

so

1/(sin0^2 * cos0^2) = 1/(sin0^2 * cos0^2)