Anonymous

# Prove this trig identity?

(secA-cosA)/(cosA-sinA)

=

sinA/cotAcosA-cosA

### 2 Answers

Relevance

- TomVLv 72 years agoFavorite Answer
LHS = (secA-cosA)/(cosA-sinA)

= (1 - cos²A)/(cos²A - cosAsinA) : multiplication of denominator and numerator by cosA ≠ 0

= sin²A/(cos²a - cosAsinA) : application of Pythagorean Theorem, 1 - cos²A = sin²A

= sinA/(cos²A/sinA - cosA) : divide denominator and numerator by sinA ≠ 0

= sinA/(cotAcosA - cosA) : application of definition of cotA = cosA/sinA

= RHS

Still have questions? Get your answers by asking now.