Question

Prove that 1 - tan^2theeta / 1+tan^2 theete = (cos^2 theeta - sin^2 theeta)

Answer 1

Hey!

I'm using α in place of theta.

$\frac{1-tan^{2}\alpha }{1+tan^{2}\alpha } = cos^{2}\alpha - sin^{2} \alpha$

Solving L.H.S,

We know that,

$tan\alpha =\frac{sin\alpha }{cos\alpha }$

Now,

$\frac{1-(\frac{sin\alpha }{cos\alpha })^{2} }{1+(\frac{sin\alpha }{cos\alpha })^{2} } \\\frac{1-(\frac{sin^{2}\alpha }{cos^{2}\alpha}) }{1+(\frac{sin^{2}\alpha }{cos^{2}\alpha})}\\ \frac{\frac{cos^{2}\alpha-sin^{2}\alpha }{cos^{2}\alpha } }{\frac{cos^{2}\alpha+sin^{2}\alpha }{cos^{2}\alpha } } \\\frac{cos^{2}\alpha-sin^{2}\alpha }{cos^{2}\alpha+sin^{2}\alpha }$

We know that,

$sin^{2} \alpha +cos^{2}\alpha =1$

Therefore,

$cos^{2}\alpha -sin^{2}\alpha$ = R.H.S