Prove that:

Sec^4A (1-Sin^4A) -2tan^2A =1

Dear Student,

We can rewrite the problem as

=$$\begin{gathered} =se{c}^{4}A(1-{\sin }^{4}A)- 2{\tan }^{2}A \\ =\frac{(1-{\sin }^{4}A)}{{\cos }^{4}A}- 2{\tan }^{2}A \\ =\frac{(1-{\sin }^{2}A)(1+{\sin }^{2}A)}{{\cos }^{4}A}- 2 {\tan }^{2}A (a^2-b^2=(a-b\left)\right(a+b\left)\right) \left(secA= \frac{1}{\cos A}\right) \\ =\frac{{\cos }^{2}A(1+{\sin }^{2}A)}{{\cos }^{4}A}- 2 {\tan }^{2}A \\ =\frac{(1+{\sin }^{2}A)}{{\cos }^{2}A}-2 {\tan }^{2}A \\ =\frac{1}{{\cos }^{2}A}+\frac{{\sin }^{2}A}{{\cos }^{2}A}-2 {\tan }^{2}A \\ = se{c}^{2}A+ {\tan }^{2}A- 2 {\tan }^{2}A \left(\frac{\sin A}{\cos A}= \tan A\right) \\ = se{c}^{2}A- {\tan }^2=1 \\ =RHS \end{gathered}$$

Regards