Please help!
\(tan(x+y)tan(x-y)=\frac{sin^2x-sin^2y}{cos^2x-sin^2y}\)
and this one:
\(\frac{1+cosx}{sinx}=cot\left(\frac{x}{2}\right)\)
Thank you so much!!
for the second 1 multiply top and bottom by 1-cos x
For: tan(x + y)tan(x - y) = ( sin2x - sin2y ) / ( cos2x - sin2y )
= [ (sinx + siny)(sinx - siny) ] / [ (cosx + cosy)(cosx - cosy) ]
= [ (sinx + siny) / (cosx + cosy) ] · [ (sinx - siny) / (cosx - cosy) ]
= tan(x + y) · tan(x - y)
