Question

Prove that $\frac{sin ^{2}A+cos ^{2}A }{sec ^{2}A-tan ^{2}A } =1$

Answer 1

See the diagram.  ABC is a right angle triangle.

sin A = a / c                 cos A = b / c
So  Sin² A + Cos² A  = a²/c²  + b2 / c²  = ( a2+b² ) / c²

According to pythogoras theorem, a² + b² = c²

So Sin² A + Cos² A =c² / c²  = 1

sec A = c/b      tan A = a/b

So Sec²  A - tan² A  = c²/ b² – a² / b²

           = (c² – a² ) / b²   = b² / b² = 1       as per pythogoras theorem

Hence answer is  1/1  = 1

Answer 2

$\frac{sin^2A+cos^2A}{sec^2A-tan^2A}=1\\\\L=\frac{1}{\frac{1}{cos^2A}-\frac{sin^2A}{cos^2A}}=1:\left(\frac{1-sin^2A}{cos^2A}\right)=1:\frac{cos^2A}{cos^2A}=1:1=1=R\\\\----------------------------\\\\secx=\frac{1}{cosx}\\\\tanx=\frac{sinx}{cosx}\\\\sin^2x+cos^2x=1\to cos^2x=1-sin^2x$