Find the identical expression for: cos^4t-sin^4t/cos^2t?

(cos⁴t - sin⁴t) / cos²t

= [(cos²t) - (sin²t)²] / cos²t

= (cos²t + sin²t)(cos²t - sin²t) / cos²t ...... for a² - b² = (a + b)(a - b)

= (cos²t - sin²t) / cos²t ...... for cos²t + sin²t = 1

= (cos²t / cos²t) - (sin²t / cos²t)

= 1 - (sint / cost)²

= 1 - tan²t

Hello,

(cos⁴t - sin⁴t) /cos²t =

let's factor the numerator as a difference of squares:

[(cos²t)² - (sin²t)²] /cos²t =

[(cos²t + sin²t)(cos²t - sin²t)] /cos²t =

let's apply the fundamental identity cos²t + sin²t = 1:

[(1)(cos²t - sin²t)] /cos²t =

(cos²t - sin²t) /cos²t =

(distributing and simplifying)

(cos²t /cos²t) - (sin²t /cos²t) =

1 - (sin t /cos t)² =

(being sin t /cos t = tan t)


1 - (tan t)²


then, in conclusion:

(cos⁴t - sin⁴t) /cos²t = 1 - tan²t


I hope it's helpful