Question

1-tg^2a=(-cos2a)/sin^2a

Answer 1

$1-tg^2a=frac{-cos2a}{sin^2a}$

$1-frac{sin^2a}{cos^2a}+frac{cos2a}{sin^2a}=0$

$sin^2acos^2a-sin^4a+cos2acos^2a=0$

$(1-cos^2a)cos^2a-(1-cos^2a)^2+(2cos^2a-1)cos^2a=0$

$cos^2a-cos^4a-1+2cos^2a-cos^4a+2cos^4a-cos^2a=0$

$2cos^2a-1=0$

$cos^2a=frac12$

$cosa=frac{1}{sqrt2}$

 $a=frac{pi}{4}+2pi n$,n∈Z

$cosa=-frac{1}{sqrt2}$

$a=frac{3pi}{4}+2pi n$,n∈Z