Question

1- tan^2x÷cot^ 2x -1 =tan ^2x

Answer 1

from given equation

Step-by-step explanation:

we know that n square theta is equal to sin square theta upon cos square theta and Cos square theta is equal to cos square theta upon sin square theta

Answer 2

Answer:

LHS =$\frac{1-tan^2x}{cot^2x-1}$

       =$\frac{1-\frac{sin^2x}{cos^2x} }{\frac{cos^2x-}{sin^2x}-1 }$

       =$\frac{\frac{cos^2x-sin^2x}{cos^2x}}{\frac{cos^2x-sin^2x}{sin^2x}}$

       =$\frac{\frac{1}{cos^2x} }{\frac{1}{sin^2x} }$

       =$\frac{sin^2x}{cos^2x}$

       =$tan^2x$

       =RHS proved