Question

Krish 1 - tan square by 1 minus cot a square is equal to tan square,A angle A is acute

Answer 1

To prove that 
$\frac{1-tan^2A}{1-Cot^2A} = - tan^2 A\\\\LHS: \frac{1-\frac{sin^2A}{cos^2A}}{1-\frac{Cos^2A}{Sin^2A}}=\frac{cos^2A-sin^2A}{sin^2A-cos^2A}*\frac{sin^2A}{cos^2A}\\\\=- tan^2A$

Hope it is clear.