Question

Докажите тождество$frac{tg ^{2} alpha -sin ^{2} alpha }{ctg ^{2} alpha -cos ^{2} alpha } =tg ^{2} alpha$

Answer 1

$frac{tg^2 alpha -sin^2 alpha }{ctg^2 alpha -cos^2a} =tg^6 alpha \ \ frac{ frac{sin^2 alpha }{cos^2 alpha } -sin^2 alpha }{ frac{cos^2 alpha }{sin^2 alpha } -cos^2a} =tg^6 alpha \ \ frac{ frac{sin^2 alpha -sin^2 alpha *cos^2 alpha }{cos^2 alpha } }{ frac{cos^2 alpha -cos^2 alpha *sin^2 alpha }{sin^2 alpha } } =tg^6 alpha \ \ frac{(sin^2 alpha -sin^2 alpha *cos^2 alpha)*sin^2 alpha }{(cos^2 alpha -cos^2 alpha *sin^2 alpha)*cos^2 alpha } =tg^6 alpha$

$frac{sin^2 alpha (1-cos^2 alpha )sin^2 alpha }{cos^2 alpha(1- sin^2 alpha )cos^2 alpha } =tg^6 alpha \ \ frac{sin^2 alpha*sin^2 alpha*sin^2 alpha}{cos^2 alpha*cos^2 alpha*cos^2 alpha} =tg^6 alpha \ \ frac{sin^6 alpha }{cos^6 alpha } =tg^6 alpha \ \ tg^6 alpha =tg^6 alpha$