Question

Simply (tan^2 x - sec^2 x)(sin^2 x + cos^2 x)

Answer 1

we can use pythagorean identieis

remember that $sin^2(x)+cos^2(x)=1$

also remember that $tan(x)=\frac{sin(x)}{cos{x}}$ and $sec(x)=\frac{1}{cos(x)}$

so if we take $sin^2(x)+cos^2(x)=1$ and divide both sides by $cos^2(x)$ we get

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

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

$tan^2(x)+1=sec^2(x)$

subtracting $1+sec^2(x)$ from both sides

$tan^2(x)-sec^2(x)=-1$

now subsitute into original problem

$\frac{tan^2(x)-sec^2(x)}{sin^2(x)+cos^2(x)}=$

$\frac{-1}{1}=$

$-1$

the answer is -1