How do you prove that: #(sin^2 x+ cos^2 x + cot^2 x) / (1+tan^2 x) = cot^2 x# ?
1 Answer
We wish to prove that:
# (sin^2 x+ cos^2 x + cot^2 x) / (1+tan^2 x) -= cot^2 x #
We can utilise the identities:
# sin^2A + cos^2A -= 1 #
# 1 + tan^2A -= sec^2 A #
# 1 + cot^2A -= csc^2 A #
Consider the LHS:
# LHS = (sin^2 x+ cos^2 x + cot^2 x) / (1+tan^2 x) #
# \ \ \ \ \ \ \ \ = (1 + cot^2 x) / (1+tan^2 x) #
# \ \ \ \ \ \ \ \ = (csc^2 x) / (sec^2 x) #
# \ \ \ \ \ \ \ \ = ((csc x) / (sec x))^2 #
# \ \ \ \ \ \ \ \ = ((1/sinx) / (1/cosx))^2 #
# \ \ \ \ \ \ \ \ = (cosx/sinx )^2 #
# \ \ \ \ \ \ \ \ = (cotx)^2 #
# \ \ \ \ \ \ \ \ = cot^2x #
# \ \ \ \ \ \ \ \ = RHS \ \ \ \ # QED
