Trig Identities

[ssb]

PROVE: (secx - cosx) (cscx - sinx) = tanx / 1+tan^2x

Steps:

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

= (1 - cos^2(x))(1 - sin^2(x)) / sin x cos x

= sin^2(x)cos^2(x) / sin x cos x

= sin x cos x
**stuck right here, what should be my next step?

 

[Ishuda]

PROVE: (secx - cosx) (cscx - sinx) = tanx / 1+tan^2x

Steps:

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

= (1 - cos^2(x))(1 - sin^2(x)) / sin x cos x

= sin^2(x)cos^2(x) / sin x cos x

= sin x cos x
**stuck right here, what should be my next step?

In a lot of these kind of problems, it is easier to work backward or both backwards and forwards. For example, look at the answer
$\displaystyle \dfrac{tan(x)}{1+tan^2(x)} = \dfrac{\dfrac{sin(x)}{cos(x)}}{1 + [\frac{sin(x)}{cos(x)}]^2}$