Question

Prove the following identity: (sinx - tanx)(cosx - cotx) = (sinx - 1)(cosx - 1).

Answer 1

Try rewriting it with tan and cot replaced by their equivalents in sin and cos. tan = sin/cos and cot = cos/sin.

Answer 2

Then expand each side of the equation.

Answer 3

(sin x - sin x/cos x)(cos x - cos x/sin x) =
= (sin x*cos x- sin x)(sin x*cos x - cos x)/sin x*cos x =
= (sin^2 x*cos^2 x - sin x*cos^2 x - sin^2 x*cos x + sin x*cos x)sin x*cos x
Simplify by sin x*cos x
= sin x*cos x - cos x - sin x +1 = sin x(cos x - 1) - (cos x - 1) =
= (cos x - 1)(sin x - 1)

Answer 4

For a lot of these trig identity problems, it's easier to substitute \(s = \sin x\) and \(c = \cos x\), do the algebra, then undo the substitution. Here we would have \[(s-\frac{s}{c})(c-\frac{c}{s}) = (s-1)(c-1)\]\[s*c - \frac{s*c}{s} - \frac{s*c}{c}+\frac{s*c}{c*s} = s*c - s -c + 1\]\[s*c-c-s+1 = s*c -s -c + 1\]\[0=0\]Done.