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# TRIGO question: show that sec2x - tan2x = (cosx-sinx)/(cosx+sinx)?

help me with this question with necessary workings?

### 1 Answer

- Seamus OLv 79 years agoFavorite Answer
LHS = sec (2x) - tan (2x)

= [1 / cos (2x)] - [sin (2x) / cos (2x)]

= [1 - sin (2x)] / cos (2x) ... [taking a common denominator]

= [1 - 2 sinx cosx] / [cos² x - sin² x]

= [1 - 2 sinx cosx] / [(cosx - sinx)(cosx + sinx)] ... [using the difference of two squares rule]

= [cos² x + sin² x - 2 sinx cosx] / [(cosx - sinx)(cosx + sinx)] ... [b/c cos² x + sin² x = 1]

= [cos² x - 2 sinx cosx + sin² x] / [(cosx - sinx)(cosx + sinx)] ... [rearranging numerator]

= [(cosx - sinx)²] / [(cosx - sinx)(cosx + sinx)] ... [Factoring the numerator using the rule (a - b)² = a² - 2ab + b² in reverse]

= (cosx - sinx) / (cosx + sinx) ... [the (cosx - sinx) in the denominator cancels with one of the (cosx - sinx) factors in the numerator

= RHS and it's done